I've been trying to get a table row with this query:
SELECT * FROM `table` WHERE `field` LIKE "%\u0435\u0442\u043e\u0442%"
Field itself:
Field
--------------------------------------------------------------------
\u0435\u0442\u043e\u0442 \u0442\u0435\u043a\u0441\u0442 \u043d\u0430
Although I can't seem to get it working properly.
I've already tried experimenting with the backslash character:
LIKE "%\\u0435\\u0442\\u043e\\u0442%"
LIKE "%\\\\u0435\\\\u0442\\\\u043e\\\\u0442%"
But none of them seems to work, as well.
I'd appreciate if someone could give a hint as to what I'm doing wrong.
Thanks in advance!
EDIT
Problem solved.
Solution: even after correcting the syntax of the query, it didn't return any results. After making the field BINARY the query started working.
As documented under String Comparison Functions:
Note
Because MySQL uses C escape syntax in strings (for example, “\n” to represent a newline character), you must double any “\” that you use in LIKE strings. For example, to search for “\n”, specify it as “\\n”. To search for “\”, specify it as “\\\\”; this is because the backslashes are stripped once by the parser and again when the pattern match is made, leaving a single backslash to be matched against.
Therefore:
SELECT * FROM `table` WHERE `field` LIKE '%\\\\u0435\\\\u0442\\\\u043e\\\\u0442%'
See it on sqlfiddle.
it can be useful for those who use PHP, and it works for me
$where[] = 'organizer_info LIKE(CONCAT("%", :organizer, "%"))';
$bind['organizer'] = str_replace('"', '', quotemeta(json_encode($orgNameString)));
Related
I have a small mysql database with a column which has format of a field as following:
x_1_1,
x_1_2,
x_1_2,
x_2_1,
x_2_12,
x_3_1,
x_3_2,
x_3_11,
I want to extra the data where it matches last '_1'. So if I run a query on above sample dataset, it would return
x_1_1,
x_2_1,
x_3_1,
This should not return x_2_12 or x_3_11.
I tried like '%_1' but it returns x_2_12 and x_3_11 as well.
Thank you!
A simple method is the right() function:
select t.*
from t
where right(field, 2) = '_1';
You can use like but you need to escape the _:
where field like '%$_1' escape '$'
Or use regular expressions:
where field regexp '_1$'
The underscore character has special significance in a LIKE clause. It acts as a wildcard and represent one single character. So you would have to escape it with a backslash:
LIKE '%\_1'
RIGHT does the job too, but it requires that you provide the proper length for the string being sought and is thus less flexible.
Duh, I found the answer.
Use RIGHT (col_name, 2) = '_1'
Thank you!
if I have a column with some values that starts with "%" like this:
[ID]-----[VALUES]
1--------Amount
2--------Percentage
3--------%Amount
4--------%Percentage
how can I have only these two rows with a "select" query?:
[ID]-----[VALUES]
3--------%Amount
4--------%Percentage
I tried these queries but them don't work:
select * from TABLE where VALUES like '[%]%'
select * from TABLE where VALUES like '\%%'
I know that in Java, C and other languages, the backspace \ let you use a jolly character as a normal one like:
var s = "I called him and he sad: \"Hi, there!\"";
There is a similar character/function that do it in SQL?
All answers will be appreciated, thank you for reading the question!
Your query
select * from TABLE where VALUES like '\%%'
should work. The reason it doesn't is because you may have NO_BACKSLASH_ESCAPES enabled which would treat \ as a literal character.
A way to avoid it is using LIKE BINARY
select * from TABLE where VALUES like binary '%'
or with an escape character (can be any character you choose) specification.
select * from TABLE where VALUES like '~%%' escape '~'
try this :
select * from TABLE where VALUES like '%[%]%'
There is an ESCAPE option on LIKE:
select *
from TABLE
where VALUES like '$%%' escape '$';
Anything following the escape character is treated as a regular character. However, the default is backslash (see here), so the version with backslash should do what you want.
Of course, you could also use a regular expression (although that has no hope of using an index).
Note: escape is part of the answer standard so it should work in any database.
You're right that you'll need an escape character for this. In SQL you have to define the escape character.
SELECT * FROM TABLE where VALUES like ESCAPE '!';
I'm pretty sure you can use whatever character you want.
Here's a link to a microsoft explanation that goes into more detail.
Microsoft explanation
MySQL Explanation
I have table like this
I want get those record which content Unit Separator
I have try many things but not getting result.I try with char(31) and 0x1f and many other ways but not getting desired result.This is my query which i try
SELECT * FROM `submissions_answers` WHERE `question_id`=90 AND `answer` like '%0x1f%'
How can i do this? Please help me..
Problem
The expression you tried won't work because answer LIKE '%0x1f%' is looking for a string with literally '0x1f' as part of it - it doesn't get converted to an ASCII code.
Solutions
Some alternatives to this part of the expression that ought to work are:-
answer LIKE CONCAT('%', 0x1F, '%')
answer REGEXP 0x1F
INSTR(answer, 0x1F) > 0
Further consideration
If none of these work then there may be a further possibility. Are you sure the character seen in the strings is actually 0x1F? I only ask because the first thing I tried was to paste in ␟ but it turns out MySQL see this as a decimal character code of 226 rather than 31. Not sure which client you are using but if the 0x1F character is in the string, it might not actually appear in the output.
Demo
Some tests demonstrating the points above: SQL Fiddle demo
You can use:
SELECT * FROM submissions_answers WHERE question_id=90 AND instr(answer,char(31))>0
The keyword here being the INSTR MySQL function, which you can read about here. This function returns the position of the first occurrence of substring (char(31)) in the string (answer).
Yet another way...
SELECT * FROM `submissions_answers`
WHERE `question_id`=90
AND HEX(`answer`) REGEXP '^(..)*1F'
Explanation of the regexp:
^ - start matching at the beginning (of answer)
(..)* -- match any number (*) of 2-byte things (..)
then match 1F, the hex for US.
You could convert the answer column into a HEX value, and then look for values containing that hex string.
SELECT * FROM `submissions_answers`
WHERE HEX(`answer`) LIKE '%E2909F%'
Consider the following regex
#(.*\..*){2,}
Expected behaviour:
a#b doesnt match
a#b.c doesnt match
a#b.c.d matches
a#b.c.d.e matches
and so on
Testing in regexpal it works as expected.
Using it it in a mysql select doesn't work as expected. Query:
SELECT * FROM `users` where mail regexp '#(.*\..*){2,}'
is returning lines like
foo#example.com
that should not match the given regex. Why?
I think the answer to your question is here.
Because MySQL uses the C escape syntax in strings (for example, “\n”
to represent the newline character), you must double any “\” that you
use in your REGEXP strings.
MYSQL Reference
Because your middle dot wasn't properly escaped it was treated as just another wildcard and in the end your expression was effectively collapsed to #.{2,} or #..+
#anubhava's answer is probably a better substitute for what you tried to do though I would note #dasblinkenlight's comment about using the character class [.] which will make it easy to drop in a regex you've already tested in at RegexPal.
You can use:
SELECT * FROM `users` where mail REGEXP '([^.]*\\.){2}'
to enforce at least 2 dots in mail column.
I would match two dots in MySQL using like:
where col like '%#.%.%'
The problem with your code is that .* (match-everything dot) matches dot '.' character. Replacing it with [^.]* fixes the problem:
SELECT *
FROM `users`
where mail regexp '#([^.]*[.]){2,}'
Note the use of [.] in place of the equivalent \.. This syntax makes it easier to embed the regex into programming languages that use backslash as escape character in their string literals.
Demo.
I am having the following problem:
I have a table T which has a column Name with names. The names have the following structure:
A\\B\C
You can create on yourself like this:
create table T ( Name varchar(10));
insert into T values ('A\\\\B\\C');
select * from T;
Now if I do this:
select Name from T where Name = 'A\\B\C';
That doesn't work, I need to escape the \ (backslash):
select Name from T where Name = 'A\\\\B\\C';
Fine.
But how do I do this automatically to a string Name?
Something like the following won't do it:
select replace('A\\B\C', '\\', '\\\\');
I get: A\\\BC
Any suggestions?
Many thanks in advance.
You have to use "verbatim string".After using that string your Replace function will
look like this
Replace(#"\", #"\\")
I hope it will help for you.
The literal A\\B\C must be coded as A\\\\A\\C, and the parameters of replace() need escaping too:
select 'A\\\\B\\C', replace('A\\\\B\\C', '\\', '\\\\');
output (see this running on SQLFiddle):
A\\B\C A\\\\B\\C
So there is little point in using replace. These two statements are equivalent:
select Name from T where Name = replace('A\\\\B\\C', '\\', '\\\\');
select Name from T where Name = 'A\\\\B\\C';
Usage of regular expression will solve your problem.
This below query will solve the given example.
1) S\\D\B
select * from T where Name REGEXP '[A-Z]\\\\\\\\[A-Z]\\\\[A-Z]$';
if incase the given example might have more then one char
2) D\\B\ACCC
select * from T where Name REGEXP '[A-Z]{1,5}\\\\\\\\[A-Z]{1,5}\\\\[A-Z]{1,5}$';
note: i have used 5 as the max occurrence of char considering the field size is 10 as its mentioned in the create table query.
We can still generalize it.If this still has not met your expectation feel free to ask for my help.
You're confusing what's IN the database with how you represent that data in SQL statements. When a string in the database contains a special character like \, you have to type \\ to represent that character, because \ is a special character in SQL syntax. You have to do this in INSERT statements, but you also have to do it in the parameters to the REPLACE function. There are never actually any double slashes in the data, they're just part of the UI.
Why do you think you need to double the slashes in the SQL expression? If you're typing queries, you should just double the slashes in your command line. If you're generating the query in a programming language, the best solution is to use prepared statements; the API will take care of proper encoding (prepared statements usually use a binary interface, which deals with the raw data). If, for some reason, you need to perform queries by constructing strings, the language should hopefully provide a function to escape the string. For instance, in PHP you would use mysqli_real_escape_string.
But you can't do it by SQL itself -- if you try to feed the non-escaped string to SQL, data is lost and it can't reconstruct it.
You could use LIKE:
SELECT NAME FROM T WHERE NAME LIKE '%\\\\%';
Not exactly sure by what you mean but, this should work.
select replace('A\\B\C', '\', '\\');
It's basically going to replace \ whereever encountered with \\ :)
Is this what you wanted?