How do I find ' % ' with the LIKE operator in SQL Server?

106,481

Solution 1

I would use

WHERE columnName LIKE '%[%]%'

SQL Server stores string summary statistics for use in estimating the number of rows that will match a LIKE clause. The cardinality estimates can be better and lead to a more appropriate plan when the square bracket syntax is used.

The response to this Connect Item states

We do not have support for precise cardinality estimation in the presence of user defined escape characters. So we probably get a poor estimate and a poor plan. We'll consider addressing this issue in a future release.

An example

CREATE TABLE T
(
X VARCHAR(50),
Y CHAR(2000) NULL
)

CREATE NONCLUSTERED INDEX IX ON T(X)

INSERT INTO T (X)
SELECT TOP (5) '10% off'
FROM master..spt_values
UNION ALL
SELECT  TOP (100000)  'blah'
FROM master..spt_values v1,  master..spt_values v2


SET STATISTICS IO ON;
SELECT *
FROM T 
WHERE X LIKE '%[%]%'

SELECT *
FROM T
WHERE X LIKE '%\%%' ESCAPE '\'

Shows 457 logical reads for the first query and 33,335 for the second.

Solution 2

You can use ESCAPE:

WHERE columnName LIKE '%\%%' ESCAPE '\'

Solution 3

Try this:

declare @var char(3)
set @var='[%]'
select Address from Accomodation where Address like '%'+@var+'%' 

You must use [] cancels the effect of wildcard, so you read % as a normal character, idem about character _

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106,481
Teju MB
Author by

Teju MB

A Professional:-)

Updated on April 17, 2020

Comments

  • Teju MB
    Teju MB about 4 years

    I have a column name Address which consists of some address which has '%' in between as:

    Address
    --------------------
    Aman Ja%lan%
    Stree% Ro%ad
    

    etc., etc.

    How I can write the LIKE operator to find that pattern?

    I tried:

    declare @var char(1)
    set @var='!%'
    select Address from Accomodation where Address like '%'+@var+'%'
    
  • Teju MB
    Teju MB over 10 years
    Thanks , actually i removed ESCAPE as i thought it may be for my reference.
  • gotqn
    gotqn over 10 years
    Using LIKE '%[%]%' should the make trick too.
  • gotqn
    gotqn over 10 years
    @491243 Actually, this is pretty good point. You are not able to escape ']' like this '[]]', but you can escape it using the the ESCAPE keyword and escape character. +1 for this and it is a pity there is no ESCAPE keyword for PATHINDEX too.
  • Martin Smith
    Martin Smith over 10 years
    @gotqn - Why do you say you can't escape it like that? SELECT 1 WHERE 'A[B' LIKE 'A[[]B' works fine. Using ESCAPE can give worse plans as my answer shows.
  • gotqn
    gotqn over 10 years
    @MartinSmith You are right. I was trying to say that you can not escape "close" bracket, but it seems that it is not special symbol and does not need to be escaped.
  • Teju MB
    Teju MB over 10 years
    Indeed martin, very lucid explanation, but what about the case when one has to search ']' ??? For example INSERT INTO T (X) SELECT TOP (5) '10[] off' FROM master..spt_values
  • Teju MB
    Teju MB over 10 years
    Indeed martin, very lucid explanation, but what about the case when one has to search ']' ??? For example INSERT INTO T (X) SELECT TOP (5) '10[] off' FROM master..spt_values For this when i try to search with [], it did not come SET STATISTICS IO ON; SELECT * FROM T WHERE X LIKE '%[]]%' But it worked with the below one SELECT * FROM T WHERE X LIKE '%\]%' ESCAPE '\'
  • Martin Smith
    Martin Smith over 10 years
    @TejuMB - That doesn't need escaping. Just use LIKE '%]%'. If you do LIKE '%[AB]]%' that means match a character in the set A,B then a ]. If you do LIKE '%[A]]%' that means match A]. If you do LIKE '%[]]%' then it means "match a single character in this empty set of characters followed by a ]. As the set is empty no character can ever match.