Reputation: 131
I need to test if any part of a column value is in a given string, instead of whether the string is part of a column value. For instance:
This way, I can find if any of the rows in my table contains the string 'bricks' in column
:
SELECT column FROM table
WHERE column ILIKE '%bricks%';
But what I'm looking for, is to find out if any part of the sentence "The ships hung in the sky in much the same way that bricks don’t" is in any of the rows. Something like:
SELECT column FROM table
WHERE 'The ships hung in the sky in much the same way that bricks don’t' ILIKE '%' || column || '%';
So the row from the first example, where the column contains 'bricks', will show up as result.
I've looked through some suggestions here and some other forums but none of them worked.
Upvotes: 1
Views: 3525
Reputation: 656321
Your simple case can be solved with a simple query using the ANY
construct and ~*
:
SELECT *
FROM tbl
WHERE col ~* ANY (string_to_array('The ships hung in the sky ... bricks don’t', ' '));
~*
is the case insensitive regular expression match operator. I use that instead of ILIKE
so we can use original words in your string without the need to pad %
for ILIKE
. The result is the same - except for words containing special characters: %_\
for ILIKE
and !$()*+.:<=>?[\]^{|}-
for regular expression patterns. You may need to escape special characters either way to avoid surprises. Here is a function for regular expressions:
But I have nagging doubts that will be all you need. See my comment. I suspect you need Full Text Search with a matching dictionary for your natural language to provide useful word stemming ...
Related:
Upvotes: 3
Reputation: 36087
This query:
SELECT
regexp_split_to_table(
'The ships hung in the sky in much the same way that bricks don’t',
'\s' );
gives a following result:
| regexp_split_to_table |
|-----------------------|
| The |
| ships |
| hung |
| in |
| the |
| sky |
| in |
| much |
| the |
| same |
| way |
| that |
| bricks |
| don’t |
Now just do a semijoin against a result of this query to get desired results
SELECT * FROM table t
WHERE EXISTS (
SELECT * FROM (
SELECT
regexp_split_to_table(
'The ships hung in the sky in much the same way that bricks don’t',
'\s' ) x
) x
WHERE t.column LIKE '%'|| x.x || '%'
)
Upvotes: 1