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I've been using the == operator in my program to compare all my strings so far.
However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.
Is == bad? When should it and should it not be used? What's the difference?
== tests for reference equality (whether they are the same object).
.equals() tests for value equality (whether they contain the same data).
Objects.equals() checks for null before calling .equals() so you don't have to (available as of JDK7, also available in Guava).
Consequently, if you want to test whether two strings have the same value you will probably want to use Objects.equals().
// These two have the same value
new String("test").equals("test") // --> true
// ... but they are not the same object
new String("test") == "test" // --> false
// ... neither are these
new String("test") == new String("test") // --> false
// ... but these are because literals are interned by
// the compiler and thus refer to the same object
"test" == "test" // --> true
// ... string literals are concatenated by the compiler
// and the results are interned.
"test" == "te" + "st" // --> true
// ... but you should really just call Objects.equals()
Objects.equals("test", new String("test")) // --> true
Objects.equals(null, "test") // --> false
Objects.equals(null, null) // --> true
You almost always want to use Objects.equals(). In the rare situation where you know you're dealing with interned strings, you can use ==.
From JLS 3.10.5. String Literals:
Moreover, a string literal always refers to the same instance of class String. This is because string literals - or, more generally, strings that are the values of constant expressions (§15.28) - are "interned" so as to share unique instances, using the method String.intern.
Similar examples can also be found in JLS 3.10.5-1.
Other Methods To Consider
String.equalsIgnoreCase() value equality that ignores case. Beware, however, that this method can have unexpected results in various locale-related cases, see this question.
String.contentEquals() compares the content of the String with the content of any CharSequence (available since Java 1.5). Saves you from having to turn your StringBuffer, etc into a String before doing the equality comparison, but leaves the null checking to you.
== tests object references, .equals() tests the string values.
Sometimes it looks as if == compares values, because Java does some behind-the-scenes stuff to make sure identical in-line strings are actually the same object.
For example:
String fooString1 = new String("foo");
String fooString2 = new String("foo");
// Evaluates to false
fooString1 == fooString2;
// Evaluates to true
fooString1.equals(fooString2);
// Evaluates to true, because Java uses the same object
"bar" == "bar";
But beware of nulls!
== handles null strings fine, but calling .equals() from a null string will cause an exception:
String nullString1 = null;
String nullString2 = null;
// Evaluates to true
System.out.print(nullString1 == nullString2);
// Throws a NullPointerException
System.out.print(nullString1.equals(nullString2));
So if you know that fooString1 may be null, tell the reader that by writing
System.out.print(fooString1 != null && fooString1.equals("bar"));
The following are shorter, but it’s less obvious that it checks for null:
System.out.print("bar".equals(fooString1)); // "bar" is never null
System.out.print(Objects.equals(fooString1, "bar")); // Java 7 required
== compares Object references.
.equals() compares String values.
Sometimes == gives illusions of comparing String values, as in following cases:
String a="Test";
String b="Test";
if(a==b) ===> true
This is because when you create any String literal, the JVM first searches for that literal in the String pool, and if it finds a match, that same reference will be given to the new String. Because of this, we get:
(a==b) ===> true
String Pool
b -----------------> "test" <-----------------a
However, == fails in the following case:
String a="test";
String b=new String("test");
if (a==b) ===> false
In this case for new String("test") the statement new String will be created on the heap, and that reference will be given to b, so b will be given a reference on the heap, not in String pool.
Now a is pointing to a String in the String pool while b is pointing to a String on the heap. Because of that we get:
if(a==b) ===> false.
String Pool
"test" <-------------------- a
Heap
"test" <-------------------- b
While .equals() always compares a value of String so it gives true in both cases:
String a="Test";
String b="Test";
if(a.equals(b)) ===> true
String a="test";
String b=new String("test");
if(a.equals(b)) ===> true
So using .equals() is always better.
The == operator checks to see if the two strings are exactly the same object.
The .equals() method will check if the two strings have the same value.
Strings in Java are immutable. That means whenever you try to change/modify the string you get a new instance. You cannot change the original string. This has been done so that these string instances can be cached. A typical program contains a lot of string references and caching these instances can decrease the memory footprint and increase the performance of the program.
When using == operator for string comparison you are not comparing the contents of the string, but are actually comparing the memory address. If they are both equal it will return true and false otherwise. Whereas equals in string compares the string contents.
So the question is if all the strings are cached in the system, how come == returns false whereas equals return true? Well, this is possible. If you make a new string like String str = new String("Testing") you end up creating a new string in the cache even if the cache already contains a string having the same content. In short "MyString" == new String("MyString") will always return false.
Java also talks about the function intern() that can be used on a string to make it part of the cache so "MyString" == new String("MyString").intern() will return true.
Note: == operator is much faster than equals just because you are comparing two memory addresses, but you need to be sure that the code isn't creating new String instances in the code. Otherwise you will encounter bugs.
String a = new String("foo");
String b = new String("foo");
System.out.println(a == b); // prints false
System.out.println(a.equals(b)); // prints true
Make sure you understand why. It's because the == comparison only compares references; the equals() method does a character-by-character comparison of the contents.
When you call new for a and b, each one gets a new reference that points to the "foo" in the string table. The references are different, but the content is the same.
Yea, it's bad...
== means that your two string references are exactly the same object. You may have heard that this is the case because Java keeps sort of a literal table (which it does), but that is not always the case. Some strings are loaded in different ways, constructed from other strings, etc., so you must never assume that two identical strings are stored in the same location.
Equals does the real comparison for you.
Yes, == is bad for comparing Strings (any objects really, unless you know they're canonical). == just compares object references. .equals() tests for equality. For Strings, often they'll be the same but as you've discovered, that's not guaranteed always.
Java have a String pool under which Java manages the memory allocation for the String objects. See String Pools in Java
When you check (compare) two objects using the == operator it compares the address equality into the string-pool. If the two String objects have the same address references then it returns true, otherwise false. But if you want to compare the contents of two String objects then you must override the equals method.
equals is actually the method of the Object class, but it is Overridden into the String class and a new definition is given which compares the contents of object.
Example:
stringObjectOne.equals(stringObjectTwo);
But mind it respects the case of String. If you want case insensitive compare then you must go for the equalsIgnoreCase method of the String class.
Let's See:
String one = "HELLO";
String two = "HELLO";
String three = new String("HELLO");
String four = "hello";
one == two; // TRUE
one == three; // FALSE
one == four; // FALSE
one.equals(two); // TRUE
one.equals(three); // TRUE
one.equals(four); // FALSE
one.equalsIgnoreCase(four); // TRUE
I agree with the answer from zacherates.
But what you can do is to call intern() on your non-literal strings.
From zacherates example:
// ... but they are not the same object
new String("test") == "test" ==> false
If you intern the non-literal String equality is true:
new String("test").intern() == "test" ==> true
== compares object references in Java, and that is no exception for String objects.
For comparing the actual contents of objects (including String), one must use the equals method.
If a comparison of two String objects using == turns out to be true, that is because the String objects were interned, and the Java Virtual Machine is having multiple references point to the same instance of String. One should not expect that comparing one String object containing the same contents as another String object using == to evaluate as true.
.equals() compares the data in a class (assuming the function is implemented).
== compares pointer locations (location of the object in memory).
== returns true if both objects (NOT TALKING ABOUT PRIMITIVES) point to the SAME object instance.
.equals() returns true if the two objects contain the same data equals() Versus == in Java
That may help you.
== performs a reference equality check, whether the 2 objects (strings in this case) refer to the same object in the memory.
The equals() method will check whether the contents or the states of 2 objects are the same.
Obviously == is faster, but will (might) give false results in many cases if you just want to tell if 2 Strings hold the same text.
Definitely the use of the equals() method is recommended.
Don't worry about the performance. Some things to encourage using String.equals():
Implementation of String.equals() first checks for reference equality (using ==), and if the 2 strings are the same by reference, no further calculation is performed!
If the 2 string references are not the same, String.equals() will next check the lengths of the strings. This is also a fast operation because the String class stores the length of the string, no need to count the characters or code points. If the lengths differ, no further check is performed, we know they cannot be equal.
Only if we got this far will the contents of the 2 strings be actually compared, and this will be a short-hand comparison: not all the characters will be compared, if we find a mismatching character (at the same position in the 2 strings), no further characters will be checked.
When all is said and done, even if we have a guarantee that the strings are interns, using the equals() method is still not that overhead that one might think, definitely the recommended way. If you want an efficient reference check, then use enums where it is guaranteed by the language specification and implementation that the same enum value will be the same object (by reference).
If you're like me, when I first started using Java, I wanted to use the "==" operator to test whether two String instances were equal, but for better or worse, that's not the correct way to do it in Java.
In this tutorial I'll demonstrate several different ways to correctly compare Java strings, starting with the approach I use most of the time. At the end of this Java String comparison tutorial I'll also discuss why the "==" operator doesn't work when comparing Java strings.
Option 1: Java String comparison with the equals method
Most of the time (maybe 95% of the time) I compare strings with the equals method of the Java String class, like this:
if (string1.equals(string2))
This String equals method looks at the two Java strings, and if they contain the exact same string of characters, they are considered equal.
Taking a look at a quick String comparison example with the equals method, if the following test were run, the two strings would not be considered equal because the characters are not the exactly the same (the case of the characters is different):
String string1 = "foo";
String string2 = "FOO";
if (string1.equals(string2))
{
// this line will not print because the
// java string equals method returns false:
System.out.println("The two strings are the same.")
}
But, when the two strings contain the exact same string of characters, the equals method will return true, as in this example:
String string1 = "foo";
String string2 = "foo";
// test for equality with the java string equals method
if (string1.equals(string2))
{
// this line WILL print
System.out.println("The two strings are the same.")
}
Option 2: String comparison with the equalsIgnoreCase method
In some string comparison tests you'll want to ignore whether the strings are uppercase or lowercase. When you want to test your strings for equality in this case-insensitive manner, use the equalsIgnoreCase method of the String class, like this:
String string1 = "foo";
String string2 = "FOO";
// java string compare while ignoring case
if (string1.equalsIgnoreCase(string2))
{
// this line WILL print
System.out.println("Ignoring case, the two strings are the same.")
}
Option 3: Java String comparison with the compareTo method
There is also a third, less common way to compare Java strings, and that's with the String class compareTo method. If the two strings are exactly the same, the compareTo method will return a value of 0 (zero). Here's a quick example of what this String comparison approach looks like:
String string1 = "foo bar";
String string2 = "foo bar";
// java string compare example
if (string1.compareTo(string2) == 0)
{
// this line WILL print
System.out.println("The two strings are the same.")
}
While I'm writing about this concept of equality in Java, it's important to note that the Java language includes an equals method in the base Java Object class. Whenever you're creating your own objects and you want to provide a means to see if two instances of your object are "equal", you should override (and implement) this equals method in your class (in the same way the Java language provides this equality/comparison behavior in the String equals method).
You may want to have a look at this ==, .equals(), compareTo(), and compare()
Function:
public float simpleSimilarity(String u, String v) {
String[] a = u.split(" ");
String[] b = v.split(" ");
long correct = 0;
int minLen = Math.min(a.length, b.length);
for (int i = 0; i < minLen; i++) {
String aa = a[i];
String bb = b[i];
int minWordLength = Math.min(aa.length(), bb.length());
for (int j = 0; j < minWordLength; j++) {
if (aa.charAt(j) == bb.charAt(j)) {
correct++;
}
}
}
return (float) (((double) correct) / Math.max(u.length(), v.length()));
}
Test:
String a = "This is the first string.";
String b = "this is not 1st string!";
// for exact string comparison, use .equals
boolean exact = a.equals(b);
// For similarity check, there are libraries for this
// Here I'll try a simple example I wrote
float similarity = simple_similarity(a,b);
The == operator check if the two references point to the same object or not. .equals() check for the actual string content (value).
Note that the .equals() method belongs to class Object (super class of all classes). You need to override it as per you class requirement, but for String it is already implemented, and it checks whether two strings have the same value or not.
Case 1
String s1 = "Stack Overflow";
String s2 = "Stack Overflow";
s1 == s2; //true
s1.equals(s2); //true
Reason: String literals created without null are stored in the String pool in the permgen area of heap. So both s1 and s2 point to same object in the pool.
Case 2
String s1 = new String("Stack Overflow");
String s2 = new String("Stack Overflow");
s1 == s2; //false
s1.equals(s2); //true
Reason: If you create a String object using the new keyword a separate space is allocated to it on the heap.
== compares the reference value of objects whereas the equals() method present in the java.lang.String class compares the contents of the String object (to another object).
I think that when you define a String you define an object. So you need to use .equals(). When you use primitive data types you use == but with String (and any object) you must use .equals().
If the equals() method is present in the java.lang.Object class, and it is expected to check for the equivalence of the state of objects! That means, the contents of the objects. Whereas the == operator is expected to check the actual object instances are same or not.
Example
Consider two different reference variables, str1 and str2:
str1 = new String("abc");
str2 = new String("abc");
If you use the equals()
System.out.println((str1.equals(str2))?"TRUE":"FALSE");
You will get the output as TRUE if you use ==.
System.out.println((str1==str2) ? "TRUE" : "FALSE");
Now you will get the FALSE as output, because both str1 and str2 are pointing to two different objects even though both of them share the same string content. It is because of new String() a new object is created every time.
Operator == is always meant for object reference comparison, whereas the String class .equals() method is overridden for content comparison:
String s1 = new String("abc");
String s2 = new String("abc");
System.out.println(s1 == s2); // It prints false (reference comparison)
System.out.println(s1.equals(s2)); // It prints true (content comparison)
All objects are guaranteed to have a .equals() method since Object contains a method, .equals(), that returns a boolean. It is the subclass' job to override this method if a further defining definition is required. Without it (i.e. using ==) only memory addresses are checked between two objects for equality. String overrides this .equals() method and instead of using the memory address it returns the comparison of strings at the character level for equality.
A key note is that strings are stored in one lump pool so once a string is created it is forever stored in a program at the same address. Strings do not change, they are immutable. This is why it is a bad idea to use regular string concatenation if you have a serious of amount of string processing to do. Instead you would use the StringBuilder classes provided. Remember the pointers to this string can change and if you were interested to see if two pointers were the same == would be a fine way to go. Strings themselves do not.
You can also use the compareTo() method to compare two Strings. If the compareTo result is 0, then the two strings are equal, otherwise the strings being compared are not equal.
The == compares the references and does not compare the actual strings. If you did create every string using new String(somestring).intern() then you can use the == operator to compare two strings, otherwise equals() or compareTo methods can only be used.
In Java, when the == operator is used to compare 2 objects, it checks to see if the objects refer to the same place in memory. In other words, it checks to see if the 2 object names are basically references to the same memory location.
The Java String class actually overrides the default equals() implementation in the Object class – and it overrides the method so that it checks only the values of the strings, not their locations in memory.
This means that if you call the equals() method to compare 2 String objects, then as long as the actual sequence of characters is equal, both objects are considered equal.
The == operator checks if the two strings are exactly the same object.
The .equals() method check if the two strings have the same value.
Related
When comparing the output of json.getString() to a string of same value i manually inputted into a json file, it's a false comparison.
println("-");
println(jsonScene.getString("value"));
println(jsonScene.getString("value") == "-");
Outputs
-
-
false
The same value is printed but when compared, it's false but should be true.
Remember in Processing (Java) you need to compare Strings using equals():
Compares two strings to see if they are the same. This method is necessary because it's not possible to compare strings using the equality operator (==). Returns true if the strings are the same and false if they are not.
For example:
println(jsonScene.getString("value").equals("-"));
Check the .charCodeAt(0) for both strings
.trim() both of them before the comparison
Check that typeof jsonScene.getString("value") === "string" (it could be an object with custom toString() defined)
When practicing with Octave I created a variable with the name my_name = ["Andrew"] and upon asking Octave to interpret whether it was a string it outputted a '0'. Again when using the typeinfo(my_name) I got ans = string. Why am I getting this sort of output?
octave:47> my_name = ["Andrew"]
my_name = Andrew
octave:48> isstring(my_name)
ans = 0
octave:49> typeinfo(my_name)
ans = string
According to the documentation (emphasis mine):
isstring (s)
Return true if s is a string array.
A string array is a data type that stores strings (row vectors of characters) at each element in the array. It is distinct from character arrays which are N-dimensional arrays where each element is a single 1x1 character. It is also distinct from cell arrays of strings which store strings at each element, but use cell indexing ‘{}’ to access elements rather than string arrays which use ordinary array indexing ‘()’.
Programming Note: Octave does not yet implement string arrays so this function will always return false.
That is, isstring will always return false (or 0), no matter what the input is.
You should use ischar to determine if the input is a character array (==string).
I was wondering why this is valid go code:
func FindUserInfo(id string) (Info, bool) {
it, present := all[id]
return it, present
}
but this isn't
func FindUserInfo(id string) (Info, bool) {
return all[id]
}
is there a way to avoid the temporary variables?
To elaborate on my comment, the Effective Go mentions that the multi-value assignment from accessing a map key is called the "comma ok" pattern.
Sometimes you need to distinguish a missing entry from a zero value. Is there an entry for "UTC" or is that the empty string because it's not in the map at all? You can discriminate with a form of multiple assignment.
var seconds int
var ok bool
seconds, ok = timeZone[tz]
For obvious reasons this is called the “comma ok” idiom. In this example, if tz is present, seconds will be set appropriately and ok will be true; if not, seconds will be set to zero and ok will be false.
Playground demonstrating this
We can see that this differs from calling a regular function where the compiler would tell you that something is wrong:
package main
import "fmt"
func multiValueReturn() (int, int) {
return 0, 0
}
func main() {
fmt.Println(multiValueReturn)
asgn1, _ := multiValueReturn()
asgn2 := multiValueReturn()
}
On the playground this will output
# command-line-arguments
/tmp/sandbox592492597/main.go:14: multiple-value multiValueReturn() in single-value context
This gives us a hint that it may be something the compiler is doing. Searching the source code for "commaOk" gives us a few places to look, including types.unpack
At the time of writing this it this the method's godoc reads:
// unpack takes a getter get and a number of operands n. If n == 1, unpack
// calls the incoming getter for the first operand. If that operand is
// invalid, unpack returns (nil, 0, false). Otherwise, if that operand is a
// function call, or a comma-ok expression and allowCommaOk is set, the result
// is a new getter and operand count providing access to the function results,
// or comma-ok values, respectively. The third result value reports if it
// is indeed the comma-ok case. In all other cases, the incoming getter and
// operand count are returned unchanged, and the third result value is false.
//
// In other words, if there's exactly one operand that - after type-checking
// by calling get - stands for multiple operands, the resulting getter provides
// access to those operands instead.
//
// If the returned getter is called at most once for a given operand index i
// (including i == 0), that operand is guaranteed to cause only one call of
// the incoming getter with that i.
//
The key bits of this being that this method appears to determine whether or not something is actually a "comma ok" case.
Digging into that method tells us that it will check to see if the mode of the operands is indexing a map or if the mode is set to commaok (where this is defined does give us many hints on when it's used, but searching the source for assignments to commaok we can see it's used when getting a value from a channel and type assertions). Remember the bolded bit for later!
if x0.mode == mapindex || x0.mode == commaok {
// comma-ok value
if allowCommaOk {
a := [2]Type{x0.typ, Typ[UntypedBool]}
return func(x *operand, i int) {
x.mode = value
x.expr = x0.expr
x.typ = a[i]
}, 2, true
}
x0.mode = value
}
allowCommaOk is a parameter to the function. Checking out where unpack is called in that file we can see that all callers pass false as an argument. Searching the rest of the repository leads us to assignments.go in the Checker.initVars() method.
l := len(lhs)
get, r, commaOk := unpack(func(x *operand, i int) { check.expr(x, rhs[i]) }, len(rhs), l == 2 && !returnPos.IsValid())
Since it seems that we can only use the "comma ok" pattern to get two return values when doing a multi-value assignment this seems like the right place to look! In the above code the length of the left hand side is checked, and when unpack is called the allowCommaOk parameter is the result of l == 2 && !returnPos.IsValid(). The !returnPos.IsValid() is somewhat confusing here as that would mean that the position has no file or line information associated with it, but we'll just ignore that.
Further down in that method we've got:
var x operand
if commaOk {
var a [2]Type
for i := range a {
get(&x, i)
a[i] = check.initVar(lhs[i], &x, returnPos.IsValid())
}
check.recordCommaOkTypes(rhs[0], a)
return
}
So what does all of this tell us?
Since the unpack method takes an allowCommaOk parameter that's hardcoded to false everywhere except in assignment.go's Checker.initVars() method, we can probably assume that you will only ever get two values when doing an assignment and have two variables on the left-hand side.
The unpack method will determine whether or not you actually do get an ok value in return by checking if you are indexing a slice, grabbing a value from a channel, or doing a type assertion
Since you can only get the ok value when doing an assignment it looks like in your specific case you will always need to use variables
You may save a couple of key strokes by using named returns:
func FindUserInfo(id string) (i Info, ok bool) {
i, ok = all[id]
return
}
But apart from that, I don't think what you want is possible.
Simply put: the reason why your second example isn't valid Go code is because the language specification says so. ;)
Indexing a map only yields a secondary value in an assignment to two variables. Return statement is not an assignment.
An index expression on a map a of type map[K]V used in an assignment or initialization of the special form
v, ok = a[x]
v, ok := a[x]
var v, ok = a[x]
yields an additional untyped boolean value. The value of ok is true if the key x is present in the map, and false otherwise.
Furthermore, indexing a map is not a "single call to a multi-valued function", which is one of the three ways to return values from a function (the second one, the other two not being relevant here):
There are three ways to return values from a function with a result type:
The return value or values may be explicitly listed in the "return" statement. Each expression must be single-valued and assignable to the corresponding element of the function's result type.
The expression list in the "return" statement may be a single call to a multi-valued function. The effect is as if each value returned from that function were assigned to a temporary variable with the type of the respective value, followed by a "return" statement listing these variables, at which point the rules of the previous case apply.
The expression list may be empty if the function's result type specifies names for its result parameters. The result parameters act as ordinary local variables and the function may assign values to them as necessary. The "return" statement returns the values of these variables.
As for your actual question: the only way to avoid temporary variables would be using non-temporary variables, but usually that would be quite unwise - and probably not much of an optimization even when safe.
So, why doesn't the language specification allow this kind of special use of map indexing (or type assertion or channel receive, both of which can also utilize the "comma ok" idiom) in return statements? That's a good question. My guess: to keep the language specification simple.
I'm no Go expert but I believe you are getting compile time error when you are trying to return the array i.e. return all[id]. The reason could be because the functions return type is specially mentioned as (Info, bool) and when you are doing return all[id] it can't map the return type of all[id] to (Info, bool).
However the solution mentioned above, the variables being returned i and ok are the same that are mentioned in the return type of the function (i Info, ok bool) and hence the compiler knows what it's returning as opposed to just doing (i Info, ok bool).
By default, maps in golang return a single value when accessing a key
https://blog.golang.org/go-maps-in-action
Hence, return all[id] won't compile for a function that expects 2 return values.
I want to save a true/false in my MySQL database. I'm saving 1/0 in an INT column to do this. When I select it, I get the 1 or 0, but I want it to return true/false to my PHP code, without having to rewrite the database.
Can I use another column type? Should I save it differently?
Update: My question is about not wanting to rewrite the returned value. I'm getting a lot of results from my database. Many of those are true/false, but some are 0s because the price is 0, so I don't want to universally rewrite all 1s and 0s. I also don't want to manually rewrite 10 columns.
To follow up my comment, here's a more detailed response which also covers the PHP side, although this probably belongs on StackOverflow.
I've always just used tinyint, although you can use bool/boolean which are synonyms for tinyint(1)
However as of MySQL 5.0.3 you can use the bit type:
As of MySQL 5.0.3, the BIT data type is used to store bit-field values. A type of BIT(M) enables storage of M-bit values. M can range from 1 to 64.
Next, assuming you have an active column, perhaps to store if a user is active, you could use PHP's automatic type conversion to handle this quite simply.
// Obviously you'd replace this with your database call
$results = [['active' => 1], ['active' => 0]];
foreach($results as $row) {
if ($row['active'] == true) {
echo "true\n";
}
else {
echo "false\n";
}
}
You don't strictly need to do anything.
PHP does not, and can not, use strongly typed variables. So, if you receive an (int) 1 from your query results, you can simply use this 1 as a boolean without rewriting or changing anything.
$intOne = (int) 1; //explicitly treat the variable as an integer
var_dump((bool) $intOne); //treat the variable as a boolean
When used in any boolean context, like if ($variable)... then any of these types will be considered to be false by PHP:
the boolean FALSE itself
the integer 0 (zero)
the float 0.0 (zero)
the empty string, and the string "0"
an array with zero elements
an object with zero member variables (PHP 4 only)
the special type NULL (including unset variables)
SimpleXML objects created from empty tags
... And, most importantly;
Every other value is considered TRUE (including any resource).
Source: PHP Manual > Booleans (english)
So while you can change the storage type of your column in mysql, this won't really change the way PHP handles the variable retrieved from your results at all.
Historically, I've always used a column of type TINYINT(1) to store boolean values in mysql, and as Tom Green points out, recent mysql versions provide a new BIT type, which might be appropriate. To the best of my knowledge, mysql does not currently have an actual boolean data type.
You could just as easily use a column of type VARCHAR(1), though, because PHP can and will use any value as a boolean, thanks to the glorious, majestic, and sometimes maddening, PHP Type Juggling.
If you're trying to use the values you're retrieving for boolean logic, just use the values you receive from mysql like booleans and it will work:
if ($valueFromResults) {
//The value was something like true
} else {
//The value was something like false
}
If you're trying to actually echo out the words "true" and "false", then you're probably best served by explicitly echoing the words out yourself, like this;
if ($valueFromResults) {
echo "true";
} else {
echo "false";
}
or, in my preferred shorthand;
echo ($valueFromResults) ? "true" : "false" ;
Update You mentioned in a comment that you want to pass the values through json_encode() and use them in javascript.
JavaScript treats any real value, like int 1, as true and any empty value, like int 0, or an empty string, as false. So if your json_encode() output gets used in actual JavaScript, the int values will still work as boolean values. So the integer values from your database should still work as they are.
Just check that your integer results are encoded as integers by PHP and not as strings - they should be encoded correctly by default - because "0" == true in javascript, but 0 == false.
For a boolean value (true/false), you should use the mySql type bit or tinyint(1).
$boolean = $mysql_data ? true : false;
At the beginning of a method I want to check if the method is called with these exact parameters before, and if so, return the result that was returned back then.
At first, with one parameter, I used a Dictionary, but now I need to check 3 parameters (a String, an Object and a boolean).
I tried making a custom Object like so:
var cacheKey:Object = { identifier:identifier, type:type, someBoolean:someBoolean };
//if key already exists, return it (not working)
if (resultCache[cacheKey]) return resultCache[cacheKey];
//else: create result ...
//and save it in the cache
resultCache[cacheKey] = result;
But this doesn't work, because the seccond time the function is called, the new cacheKey is not the same object as the first, even though it's properties are the same.
So my question is: is there a datatype that will check the properties of the object used as key for a matching key?
And what else is my best option? Create a cache for the keys as well? :/
Note there are two aspects to the technical solution: equality comparison and indexing.
The Cliff Notes version:
It's easy to do custom equality comparison
In order to perform indexing, you need to know more than whether one object is equal to another -- you need to know which is object is "bigger" than the other.
If all of your properties are primitives you should squash them into a single string and use an Object to keep track of them (NOT a Dictionary).
If you need to compare some of the individual properties for reference equality you're going to have a write a function to determine which set of properties is bigger than the other, and then make your own collection class that uses the output of the comparison function to implement its own a binary search tree based indexing.
If the number of unique sets of arguments is in the several hundreds or less AND you do need reference comparison for your Object argument, just use an Array and the some method to do a naive comparison to all cached keys. Only you know how expensive your actual method is, so it's up to you to decide what lookup cost (which depends on the number of unique arguments provided to the function) is acceptable.
Equality comparison
To address equality comparison it is easy enough to write some code to compare objects for the values of their properties, rather than for reference equality. The following function enforces strict set comparison, so that both objects must contain exactly the same properties (no additional properties on either object allowed) with the same values:
public static propsEqual(obj1:Object, obj2:Object):Boolean {
for(key1:* in obj1) {
if(obj2[key1] === undefined)
return false;
if(obj2[key1] != obj2[key1])
return false;
}
for(key2:* in obj2)
if(obj1[key2] === undefined)
return false;
return true;
}
You could speed it up by eliminating the second for loop with the tradeoff that {A:1, B:2} will be deemed equal to {A:1, B:2, C:'An extra property'}.
Indexing
The problem with this in your case is that you lose the indexing that a Dictionary provides for reference equality or that an Object provides for string keys. You would have to compare each new set of function arguments to the entire list of previously seen arguments, such as using Array.some. I use the field currentArgs and the method to avoid generating a new closure every time.
private var cachedArgs:Array = [];
private var currentArgs:Object;
function yourMethod(stringArg:String, objArg:Object, boolArg:Boolean):* {
currentArgs = { stringArg:stringArg, objArg:objArg, boolArg:boolArg };
var iveSeenThisBefore:Boolean = cachedArgs.some(compareToCurrent);
if(!iveSeenThisBefore)
cachedArgs.push(currentArgs);
}
function compareToCurrent(obj:Object):Boolean {
return someUtil.propsEqual(obj, currentArgs);
}
This means comparison will be O(n) time, where n is the ever increasing number of unique sets of function arguments.
If all the arguments to your function are primitive, see the very similar question In AS3, where do you draw the line between Dictionary and ArrayCollection?. The title doesn't sound very similar but the solution in the accepted answer (yes I wrote it) addresses the exact same techinical issue -- using multiple primitive values as a single compound key. The basic gist in your case would be:
private var cachedArgs:Object = {};
function yourMethod(stringArg:String, objArg:Object, boolArg:Boolean):* {
var argKey:String = stringArg + objArg.toString() + (boolArg ? 'T' : 'F');
if(cachedArgs[argKey] === undefined)
cachedArgs[argKey] = _yourMethod(stringArg, objArg, boolArg);
return cachedArgs[argKey];
}
private function _yourMethod(stringArg:String, objArg:Object, boolArg:Boolean):* {
// Do stuff
return something;
}
If you really need to determine which reference is "bigger" than another (as the Dictionary does internally) you're going to have to wade into some ugly stuff, since Adobe has not yet provided any API to retrieve the "value" / "address" of a reference. The best thing I've found so far is this interesting hack: How can I get an instance's "memory location" in ActionScript?. Without doing a bunch of performance tests I don't know if using this hack to compare references will kill the advantages gained by binary search tree indexnig. Naturally it would depend on the number of keys.