Find rows where text array contains value similar to input Find rows where text array contains value similar to input postgresql postgresql

Find rows where text array contains value similar to input


It's also important to understand that ANY is not an operator but an SQL construct that can only be used to the right of an operator. More:

The LIKE operator - or more precisely: expression, that is rewritten with to the ~~ operator in Postgres internally - expects the value to the left and the pattern to the right. There is no COMMUTATOR for this operator (like there is for the simple equality operator =) so Postgres cannot flip operands around.

Your attempt:

select * from someTable where '%someInput%' LIKE ANY(someColum);

has flipped left and right operand so '%someInput%' is the value and elements of the array column someColum are taken to be patterns (which is not what you want).

It would have to be ANY(someColum) LIKE '%someInput%' - except that's not possible with the ANY construct which is only allowed to the right of an operator. You are hitting a road block here.

Related:

You can normalize your relational design and save elements of the array in separate rows in a separate table. Barring that, unnest() is the solution, as you already found yourself. But while you are only interested in the existence of at least one matching element, an EXISTS subquery will be most efficient while avoiding duplicates in the result - Postgres can stop the search as soon as the first match is found:

SELECT *FROM   tblWHERE  EXISTS (    SELECT -- can be empty     FROM   unnest(someColum) elem    WHERE  elem LIKE '%someInput%'  );

You may want to escape special character in someInput. See:

Careful with the negation (NOT LIKE ALL (...)) when NULL can be involved:


My question was marked duplicate and linked to a question out of context by a careless mod. This question comes closest to what I asked so I leave my answer here. (I think it may help people for who unnest() would be a solution)

In my case a combination of DISTINCT and unnest() was the solution:

SELECT DISTINCT ON (id_) *FROM (  SELECT unnest(tags) tag, *  FROM someTable  ) xWHERE (tag like '%someInput%');

unnest(tags) expands the text array to a list of rows and DISTINCT ON (id_) removes the duplicates that result from the expansion, based on a unique id_ column.

Update

Another way to do this without DISTINCT within the WHERE clause would be:

SELECT *FROM someTable WHERE (  0 < (    SELECT COUNT(*)     FROM unnest(tags) AS tag    WHERE tag LIKE '%someInput%'  ));


An admittedly imperfect possibility might be to use ARRAY_TO_STRING, then use LIKE against the result. For example:

SELECT *FROM someTableWHERE ARRAY_TO_STRING(someColum, '||') LIKE '%someInput%';

This approach is potentially problematic, though, because someone could search over two array elements if they discover the joining character sequence. For example, an array of {'Hi','Mom'}, connected with || would return a result if the user had entered i||M in place of someInput. Instead, the expectation would probably be that there would be no result in that case since neither Hi nor Mom individually contain the i||M sequence of characters.