Reputation: 170846
Is there any shorter way to look for multiple matches:
SELECT *
from table
WHERE column LIKE "AAA%"
OR column LIKE "BBB%"
OR column LIKE "CCC%"
This questions applies to PostgreSQL 9.1, but if there is a generic solution it would be even better.
Upvotes: 85
Views: 254477
Reputation: 11
if you want use not like to skip values use this query example:
select * from table
where not field like any(array['%aaa%','%bbb%','%ccc%','%ddd%'])
Upvotes: 1
Reputation: 3916
Following query helped me. Instead of using LIKE
, you can use ~*
.
select id, name from hosts where name ~* 'julia|lena|jack';
Upvotes: 6
Reputation: 191
You can use regular expression operator (~), separated by (|) as described in Pattern Matching
select column_a from table where column_a ~* 'aaa|bbb|ccc'
Upvotes: 15
Reputation: 1
You might be able to use IN, if you don't actually need wildcards.
SELECT * from table WHERE column IN ('AAA', 'BBB', 'CCC')
Upvotes: -6
Reputation: 325141
Use LIKE ANY(ARRAY['AAA%', 'BBB%', 'CCC%'])
as per this cool trick @maniek showed earlier today.
Upvotes: 50
Reputation: 125544
Using array or set comparisons:
create table t (str text);
insert into t values ('AAA'), ('BBB'), ('DDD999YYY'), ('DDD099YYY');
select str from t
where str like any ('{"AAA%", "BBB%", "CCC%"}');
select str from t
where str like any (values('AAA%'), ('BBB%'), ('CCC%'));
It is also possible to do an AND
which would not be easy with a regex if it were to match any order:
select str from t
where str like all ('{"%999%", "DDD%"}');
select str from t
where str like all (values('%999%'), ('DDD%'));
Upvotes: 36
Reputation: 3451
Perhaps using SIMILAR TO
would work ?
SELECT * from table WHERE column SIMILAR TO '(AAA|BBB|CCC)%';
Upvotes: 97