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+'%'
The SQL LIKE Operator The LIKE operator is used in a WHERE clause to search for a specified pattern in a column. There are two wildcards often used in conjunction with the LIKE operator: The percent sign (%) represents zero, one, or multiple characters. The underscore sign (_) represents one, single character.
You can find duplicates by grouping rows, using the COUNT aggregate function, and specifying a HAVING clause with which to filter rows.
To find the duplicate Names in the table, we have to follow these steps: Defining the criteria: At first, you need to define the criteria for finding the duplicate Names. You might want to search in a single column or more than that. Write the query: Then simply write the query to find the duplicate Names.
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.
You can use ESCAPE
:
WHERE columnName LIKE '%\%%' ESCAPE '\'
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