So I have a in my Postgresql:
TAG_TABLE
==========================
id tag_name
--------------------------
1 aaa
2 bbb
3 ccc
To simplify my problem, What I want to do is SELECT 'id' from TAG_TABLE when a string "aaaaaaaa" contains the 'tag_name'. So ideally, it should only return "1", which is the ID for tag name 'aaa'
This is what I am doing so far:
SELECT id FROM TAG_TABLE WHERE 'aaaaaaaaaaa' LIKE '%tag_name%'
But obviously, this does not work, since the postgres thinks that '%tag_name%' means a pattern containing the substring 'tag_name' instead of the actual data value under that column.
How do I pass the tag_name to the pattern??
This question is related to
sql
postgresql
A proper way to search for a substring is to use position
function instead of like
expression, which requires escaping %
, _
and an escape character (\
by default):
SELECT id FROM TAG_TABLE WHERE position(tag_name in 'aaaaaaaaaaa')>0;
I personally prefer the simpler syntax of the ~ operator.
SELECT id FROM TAG_TABLE WHERE 'aaaaaaaa' ~ tag_name;
Worth reading through Difference between LIKE and ~ in Postgres to understand the difference. `
In addition to the solution with 'aaaaaaaa' LIKE '%' || tag_name || '%'
there
are position
(reversed order of args) and strpos
.
SELECT id FROM TAG_TABLE WHERE strpos('aaaaaaaa', tag_name) > 0
Besides what is more efficient (LIKE looks less efficient, but an index might change things), there is a very minor issue with LIKE: tag_name of course should not contain %
and especially _
(single char wildcard), to give no false positives.
SELECT id FROM TAG_TABLE WHERE 'aaaaaaaa' LIKE '%' || "tag_name" || '%';
tag_name
should be in quotation otherwise it will give error as tag_name doest not exist
Source: Stackoverflow.com