SQL 'LIKE' query using '%' where the search criteria contains '%'

Question

I have an SQL query as below.

Select * from table 
where name like \'%\' + search_criteria + \'%\' 

If search_criteria = \'abc\', it will r

Answer 1

You need to escape it: on many databases this is done by preceding it with backslash, \%.

So abc becomes abc\%.

Your programming language will have a database-specific function to do this for you. For example, PHP has mysql_escape_string() for the MySQL database.

Answer 2

May be this one help :)

DECLARE @SearchCriteria VARCHAR(25)
SET  @SearchCriteria = 'employee'
IF CHARINDEX('%', @SearchCriteria) = 0
BEGIN
SET @SearchCriteria = '%' + @SearchCriteria + '%'
END
SELECT * 
FROM Employee
WHERE Name LIKE @SearchCriteria

Answer 3

If you want a % symbol in search_criteria to be treated as a literal character rather than as a wildcard, escape it to [%]

... where name like '%' + replace(search_criteria, '%', '[%]') + '%'

Answer 4

Use an escape clause:

select *
  from (select '123abc456' AS result from dual
        union all
        select '123abc%456' AS result from dual
       )
  WHERE result LIKE '%abc\%%' escape '\'

Result

123abc%456

You can set your escape character to whatever you want. In this case, the default '\'. The escaped '\%' becomes a literal, the second '%' is not escaped, so again wild card.

See List of special characters for SQL LIKE clause

Answer 5

Escape the percent sign \% to make it part of your comparison value.

Answer 6

The easiest solution is to dispense with "like" altogether:

Select * 
from table
where charindex(search_criteria, name) > 0

I prefer charindex over like. Historically, it had better performance, but I'm not sure if it makes much of difference now.