I am running a query on a column postal (type double).
SELECT * FROM `table` WHERE `postal` LIKE 'abcdef'; # returns 1 record
and the same query using = returns 100+ records.
SELECT * FROM `table` WHERE `postal` = 'abcdef'; # returns 107 record
What could be the reason?
You are using LIKE on a DOUBLE field, you should not do that.
LIKE is reserved for pattern matching on strings. Use = for numbers, or convert your digit to a string first using CONVERT and then apply the logic with LIKE.
= compares two values for identity.
LIKE is for pattern matching ie. that is, it matches a string value against a pattern string containing wild-card characters.
Refer here
LIKE will check and return similar values where as = will check for the exact value.
The following things affects the result (not the complete list!)
Implicit conversation
MySQL extension to standard SQL's LIKE operator
In each cases an implicit conversion occours: MySQL tries to convert the values to a common data type. In the first case case 'abcdef' will be converted to double which results to 0. This is why you get 107 records when comparing with equals (=).
SELECT * FROM `table` WHERE `postal` = 'abcdef'; # returns 107 record
You should get exactly the same result by running
SELECT * FROM `table` WHERE `postal` = 0;
In MySQL, LIKE is permitted on numeric expressions. (This is an extension to the standard SQL LIKE.)
This means that SELECT CASE WHEN 10 LIKE '1%' THEN 1 ELSE 0 END is allowed and results to 1 (matched)
To be honest, I'm not sure which double value could match with LIKE operator with the pattern 'abcdef'.
Related
I am new in Mysql and have a table where I'm going to select based on an integer column, the problem is that when I use an string on this column! I got no error but it gives me back all rows. for example:
SELECT * FROM `News` WHERE Cat='hello' order by id desc limit 20
It gives me 20 rows! what's wrong? did I do anything wrong or it's because of something else?
This is normal behavior for MySql, because in this expression:
Cat='hello'
what happens is an implicit conversion of the string literal 'hello' to INTEGER and as it is described in Type Conversion in Expression Evaluation the result of this conversion is 0, so the expression is equivalent to:
Cat=0
If you want to prevent this conversion you could instead convert the column Cat to string:
WHERE CONVERT(Cat, CHAR) = 'hello'
This way the comparison of Cat and 'hello' will be alphanumerical and will fail.
But if you pass a valid integer, then the correct result will be returned.
I'm doing some research in a MySQL database where some data is stored as XML. I already managed to find the string I was searching for:
select * from foo where Concat(foo) like '%bar%';
Now I'm trying to find only entries where "bar" appears 2 times. In the table where I'm searching "bar" always appears once so I want to find the entries with at least 2x "bar".
Can you give me some advice?
You should use the REGEXP method
select * from foo where Concat(foo) regexp '(bar).*(bar)';
Breakdown
()- First capturing group
bar - The expression to be captured
. - Matches any character
* - Matches zero or more of a character
(bar) - Second capturing group
https://regex101.com/r/wM3wX9/1
From the MySQL documentation
Performs a pattern match of a string expression expr against a pattern pat. The pattern can be an extended regular expression, the syntax for which is discussed later in this section. Returns 1 if expr matches pat; otherwise it returns 0. If either expr or pat is NULL, the result is NULL. RLIKE is a synonym for REGEXP, provided for mSQL compatibility.
I also created an SQL Fiddle for this.
http://sqlfiddle.com/#!9/49fd7/1/0
Here is a scalable solution, which can easily used on any repeated times, such as 1, 2, 10, 100 and etc.
-- match twice
select * from foo where (length(foo) - replace(foo,'bar',''))/length('bar') = 2;
-- match 3 repeated times
select * from foo where (length(foo) - replace(foo,'bar',''))/length('bar') = 3;
-- match 100 repeated times
select * from foo where (length(foo) - replace(foo,'bar',''))/length('bar') = 100;
Reference:
SQL function to get count of how many times string appears in column?
i'm working with mysql in a nodejs web app. I don't understand why when I ask for some id (key) it gives me more than 1 result.
When I:
SELECT * FROM products WHERE id = 1;
This happens, I get 3 results, but I only want 1:
1, 001 and 0000001.
I just want the info of one product (id: 1 in this example)
How can I fix this?
ID type is varchar(20)
If I use LIKE instead of = my result changes:
SELECT * FROM products WHERE id LIKE 0000001;
I get the info of id = 1 instead 0000001. Don't know why.
Thanks
The WHERE clause of your query contains a comparison of a literal numeric value with a string (column id).
When it needs to compare values of different type, MySQL uses several rules to convert one or both of the values to a common type.
Some of the type conversion rules are not intuitive. The last rule is the only one that matches a comparison of an integer number with a string:
In all other cases, the arguments are compared as floating-point (real) numbers.
When they are converted to floating-point (real) numbers, 1 (integer), '1', '0001' and '0000001' are all equal.
In order to get an exact match the literal value you put in the query must have the same type as the column id (i.e string). The query should be:
SELECT * FROM products WHERE id = '1'
The problem is that you are looking by a varchar type using an integer cast.
Try to add quotes to the id parameter:
SELECT * FROM products WHERE id = '1';
If you want to add integer ids with with leading zeros, I recommend you to use the zerofill option:
https://dev.mysql.com/doc/refman/5.5/en/numeric-type-attributes.html
If you want to use use alphanumeric values then keeps the ID type as varchar, but remember to enclose the search param into quotes.
Numbers in MySQL (and the real world) don't have leading zeros. Strings do.
So, you just need to make the comparison using the right type:
SELECT *
FROM products
WHERE id = '1';
What happens with your original query is that the id is converted to a number. And '1', '001' and '0000001' are all converted to the same integer -- 1. Hence, all three pass the filter.
In a MySQL table i have a field, containing this value for a given record : "1908,2315,2316"
Here is my sql Query :
SELECT * FROM mytable WHERE 2316 IN (myfield)
I got 0 results!
I tried this :
SELECT * FROM mytable WHERE 2315 IN (myfield)
Still 0 results
And then i tried this :
SELECT * FROM mytable WHERE 1908 IN (myfield)
Surprisingly i obtained the record when searching with 1908! What should i do to also obtain the record when searching with 2315 and 2316 ? What am i missing ?
Thanks
You appear to be storing comma delimited values in a field. This is bad, bad, bad. You should be using a junction table, with one row per value.
But, sometimes you are stuck with data in a particular structure. If so, MySQL provides the find_in_set() functions.
SELECT *
FROM mytable
WHERE find_in_set(2316, myfield) > 0;
You can't use IN() over comma separated list of no.s its better to normalize your structure first for now you can use find_in_set to find results matching with comma separated string
SELECT * FROM mytable WHERE find_in_set('1908',myfield) > 0
This question has been asked and answered before, but I don't want to hunt for it; this question should be closed as a duplicate. But, to answer your question:
The commas in the string, the column value, are just characters. Those are part of the string. They aren't seen as "separators" between values in the SQL text. The way SQL sees it, the column contains a single value, not a "list" of values.
So, in your query, the IN (field) is equivalent to an equals comparison. It's equivalent to comparing to a string. For example:
... WHERE 2316 = '1908,2315,2316'
And those aren't equal, so the row isn't returned. The "surprisingly" finding of a match, in the case of:
... WHERE 1908 IN ('1908,2315,2316')
that's explained because that string is being evaluated in a numeric context. That is, the comparison returns true, because all of these also true:
... WHERE 1908 = '1908,2315,2316' + 0
... WHERE 1908 = '1908xyz' + 0
... WHERE 1908 = '1907qrs' + 1
(When evaluated in a numeric context, a string gets converted to numeric. It just happens that the string evaluates to a numeric value that equals the integer value it's being comparing to.)
You may be able to make use of the MySQL FIND_IN_SET function. For example:
... WHERE FIND_IN_SET(2316,'1908,2315,2316')
But, please seriously reconsider the design of storing comma separated list. I recommend Bill Karwin's "SQL Antipatterns" book...
http://www.amazon.com/SQL-Antipatterns-Programming-Pragmatic-Programmers/dp/1934356557
In mysql IN clause is utilized as
SELECT * FROM mytable WHERE column_name IN (set_of_values) ;
Mention column name instead of values
Please try
SELECT * FROM mytable WHERE LOCATE(CONCAT (',', 2316 ','), CONCAT (',',myfield,',' ) ) <>0
This case is similar to: S.O Question; mySQL returns all rows when field=0, and the Accepted answer was a very simple trick, to souround the ZERO with single quotes
FROM:
SELECT * FROM table WHERE email=0
TO:
SELECT * FROM table WHERE email='0'
However, my case is slightly different in that my Query is something like:
SELECT * FROM table WHERE email=(
SELECT my_column_value FROM myTable WHERE my_column_value=0 AND user_id =15 LIMIT 1 )
Which in a sense, becomes like simply saying: SELECT * FROM table WHERE email=0, but now with a Second Query.
PLEASE NOTE: It is a MUST that I use the SECOND QUERY.
When I tried: SELECT * FROM table WHERE email='( SELECT my_column_value FROM myTable WHERE my_column_value=0 LIMIT 1 )' (Notice the Single Quotes on the second query)
MySql SCREAMED Errors near '(.
How can this be achieved
Any Suggestion is highly honored
EDIT1: For a visual perspective of the Query
See the STEN_TB here: http://snag.gy/Rq8dq.jpg
Now, the main aim is to get the sten_h where rawscore_h = 0;
The CURRENT QUERY as a whole.
SELECT sten_h
FROM sten_tb
WHERE rawscore_h = (
SELECT `for_print_stens_rowscore`
FROM `for_print_stens_tb`
WHERE `for_print_stens_student_id` =3
AND `for_print_stens_factor_name` = 'Factor H' )
The result of the Second Query can be any number including ZERO.
Any number from >=1 Works and returns a single corresponding value from sten_h. Only =0 does not Work, it returns all rows
That's the issue.
CORRECT ANSWER OR SOLUTION FOR THIS
Just in case someone ends up in this paradox, the Accepted answer has it all.
SEE STEN_TB: http://snag.gy/Rq8dq.jpg
SEE The desired Query result here: http://snag.gy/wa4yA.jpg
I believe your issue is with implicit datatype conversions. You can make those datatype conversions explicit, to gain control.
(The "trick" with wrapping a literal 0 in single quotes, that makes the literal a string literal, rather than a numeric.)
In the more general case, you can use a CAST or CONVERT function to explicitly specify a datatype conversion. You can use an expression in place of a column name, wherever you need to...
For example, to get the value returned by my_column_value to match the datatype of the email column, assuming email is character type, something like:
... email = (SELECT CONVERT(my_column_value,CHAR(255)) FROM myTable WHERE ...
or, to get the a literal integer value to be a string value:
... FROM myTable WHERE my_column_value = CONVERT(0,CHAR(30)) ...
If email and my_column_value are just indicating true or false then they should almost certainly be both BIT NOT NULL or other two-value type that your schema uses for booleans. (Your ORM may use a particular one.) Casting is frequently a hack made necessary by a poor design.
If it should be a particular user then you shouldn't use LIMIT because tables are unordered and that doesn't return a particular user. Explain in your question what your query is supposed to return including exactly what you mean by "15th".
(Having all those similar columns is bad design: rawscore_a, sten_a, rawscore_b, sten_b,... . Use a table with two columns: rawscore, sten.)