'PL/pgSQL checking if a row exists
I'm writing a function in PL/pgSQL, and I'm looking for the simplest way to check if a row exists.
Right now I'm SELECTing an integer
into a boolean
, which doesn't really work. I'm not experienced with PL/pgSQL enough yet to know the best way of doing this.
Here's part of my function:
DECLARE person_exists boolean;
BEGIN
person_exists := FALSE;
SELECT "person_id" INTO person_exists
FROM "people" p
WHERE p.person_id = my_person_id
LIMIT 1;
IF person_exists THEN
-- Do something
END IF;
END; $$ LANGUAGE plpgsql;
Update - I'm doing something like this for now:
DECLARE person_exists integer;
BEGIN
person_exists := 0;
SELECT count("person_id") INTO person_exists
FROM "people" p
WHERE p.person_id = my_person_id
LIMIT 1;
IF person_exists < 1 THEN
-- Do something
END IF;
Solution 1:[1]
Simpler, shorter, faster: EXISTS
.
IF EXISTS (SELECT FROM people p WHERE p.person_id = my_person_id) THEN
-- do something
END IF;
The query planner can stop at the first row found - as opposed to count()
, which scans all (qualifying) rows regardless. Makes a big difference with big tables. The difference is small for a condition on a unique column: only one row qualifies and there is an index to look it up quickly.
Only the existence of at least one qualifying row matters. The SELECT
list can be empty - in fact, that's shortest and cheapest. (Some other RDBMS don't allow an empty SELECT
list on principal.)
Improved with @a_horse_with_no_name's comments.
Solution 2:[2]
Use count(*)
declare
cnt integer;
begin
SELECT count(*) INTO cnt
FROM people
WHERE person_id = my_person_id;
IF cnt > 0 THEN
-- Do something
END IF;
Edit (for the downvoter who didn't read the statement and others who might be doing something similar)
The solution is only effective because there is a where clause on a column (and the name of the column suggests that its the primary key - so the where clause is highly effective)
Because of that where
clause there is no need to use a LIMIT or something else to test the presence of a row that is identified by its primary key. It is an effective way to test this.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
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Solution 1 | |
Solution 2 |