Like operator

Question

Post reply

Like operator

mtz676

SSCrazy Eights

Points: 9217
More actions
May 4, 2009 at 12:33 pm

#384500

My dataset
SELECT description
FROM tab1
where description = '%@Description%'
also tried
SELECT description
FROM tab1
where description = '%" + @Description + '%'
also tried
SELECT description
FROM tab1
where description = '*@Description*'
I am trying to create a report with a filter on a text field for which I am using like operator . The user will pass a parameter based on the column values(description column) for this like operator.
But when I run this query it does not result any results.
Thanks

Viewing 4 posts - 1 through 3 (of 3 total)

You must be logged in to reply to this topic. Login to reply

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 1

Have you tried this:

SELECT description

FROM tab1

where description = '''%' + @Description + '%'''

Please note, that is three single quotes at the start and end.

LutzM SSC Guru Points: 107049 More actions · Answer 2

Lynn Pettis (5/4/2009)
Have you tried this:
SELECT description
FROM tab1
where description = '''%' + @Description + '%'''
Please note, that is three single quotes at the start and end.

You probably need to change the query from an EQUAL query to an LIKE query (the failure wasn't caused by the single quotes):

SELECT description

FROM tab1

where description LIKE '%' + @Description + '%'

Lutz
A pessimist is an optimist with experience.

How to get fast answers to your question[/url]
How to post performance related questions[/url]
Links for Tally Table [/url] , Cross Tabs [/url] and Dynamic Cross Tabs [/url], Delimited Split Function[/url]

Lynn Pettis SSC Guru Points: 442467 More actions · Answer 3

lmu92 (5/4/2009)
Lynn Pettis (5/4/2009)
Have you tried this:
SELECT description
FROM tab1
where description = '''%' + @Description + '%'''
Please note, that is three single quotes at the start and end.
You probably need to change the query from an EQUAL query to an LIKE query (the failure wasn't caused by the single quotes):
SELECT description
FROM tab1
where description LIKE '%' + @Description + '%'

Good catch, didn't see the forest for the trees.