extracting column values when a part of the column value is known

Question

Post reply

extracting column values when a part of the column value is known

t.mounika01

Mr or Mrs. 500

Points: 585
More actions
August 21, 2014 at 8:52 pm

#289363

hi
How can we extract a column values from database if we know a part of that column value.
For ex: if I have values for a column called statements like this:
(I am good Gal)
(you are a good boy) and so on..
and I want to extract all the statements that have a part called 'good'
what is the best way to query this?

Viewing 7 posts - 1 through 6 (of 6 total)

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

Eirikur Eiriksson SSC Guru Points: 182911 More actions · Answer 1

Quick thought, use the LIKE operator

😎

USE tempdb;

GO

DECLARE @test-2 TABLE

(

TEST_ID INT IDENTITY(1,1) NOT NULL PRIMARY KEY CLUSTERED

,TEST_STRING VARCHAR(50) NOT NULL

);

INSERT INTO @test-2(TEST_STRING)

VALUES

('I am a bad Gal')

,('I am a good Gal')

,('This is an old horse')

,('This is a good horse')

,('I am a good boy')

,('I am a naugthy boy');

SELECT

*

FROM @test-2 T

WHERE T.TEST_STRING LIKE '%good%';

Results

TEST_ID TEST_STRING

----------- ---------------------

2 I am a good Gal

4 This is a good horse

5 I am a good boy

t.mounika01 Mr or Mrs. 500 Points: 585 More actions · Answer 2

What to do if we want to replace the word 'good' with 'bad' in all those sentences?

I wrote it like this:

REPLACE(columnname,'good','bad')

Is there any other way to do this?

Ed Wagner SSC Guru Points: 287026 More actions · Answer 3

Using the REPLACE function is the simplest way to do it. Assuming there aren't any unusual requirements, that's exactly the way I'd do it.

Tally Tables - Performance Personified
String Splitting with True Performance
Best practices on how to ask questions

t.mounika01 Mr or Mrs. 500 Points: 585 More actions · Answer 4

t.mounika01

Mr or Mrs. 500

Points: 585

August 22, 2014 at 12:46 pm

#1740379

Thank you all for the help 🙂

Luis Cazares SSC Guru Points: 183706 More actions · Answer 5

Note that you might replace strings you don't want them to be replaced.

INSERT INTO @test-2(TEST_STRING)

VALUES

('I am a bad Gal')

,('I am a good Gal')

,('Oh my goodness! There''s no more goodwill')

,('He''s a patient with Osgood–Schlatter disease');

SELECT

*, REPLACE(TEST_STRING, 'good', 'bad')

FROM @test-2 T

WHERE T.TEST_STRING LIKE '%good%';

Luis C.
General Disclaimer:
Are you seriously taking the advice and code from someone from the internet without testing it? Do you at least understand it? Or can it easily kill your server?

How to post data/code on a forum to get the best help: Option 1 / Option 2

Eirikur Eiriksson SSC Guru Points: 182911 More actions · Answer 6

Luis Cazares (8/22/2014)
Note that you might replace strings you don't want them to be replaced.
INSERT INTO @test-2(TEST_STRING)
VALUES
('I am a bad Gal')
,('I am a good Gal')
,('Oh my goodness! There''s no more goodwill')
,('He''s a patient with Osgood–Schlatter disease');
SELECT
*, REPLACE(TEST_STRING, 'good', 'bad')
FROM @test-2 T
WHERE T.TEST_STRING LIKE '%good%';

DECLARE @STR VARCHAR(50) = 'That an absolute @£$%'

SELECT REPLACE(REPLACE(REPLACE(REPLACE(@STR,'e @','ly'),'£',' a '),'$','brilliant'),'%',' point!')

😎