Hi!
I need to create startswith function which returns true if char(n) database
column starts with some characters which
may can contain space at end.
Spaces should treated like other characters.
Using sample data below
startswith( test, 'A')
startswith( test, 'A ')
StartsWith(test, rpad('A',19) )
shuld return true
but
startswith( test, RPAD( 'A', 20)) should return false since there is extra
space in end of check string
Database contains test column which has char(20) type column and this cannot
changed.
I tried code below but it returns false.
How to fix this so that it returns true?
Using Postgres starting from 9.1
Andrus.
CREATE or replace FUNCTION public.likeescape( str text )
--
https://stackoverflow.com/questions/10153440/how-to-escape-string-while-matching-pattern-in-postgresql
RETURNS text AS $$
SELECT replace(replace(replace($1,'^','^^'),'%','^%'),'_','^_') ;
$$ LANGUAGE sql IMMUTABLE;
CREATE or replace FUNCTION public.StartWith( cstr text, algusosa text )
RETURNS bool AS $$
SELECT $2 is null or $1 like likeescape($2) ||'%' ESCAPE '^' ;
$$ LANGUAGE sql IMMUTABLE;
create temp table test ( test char(20) ) on commit drop;
insert into test values ('A' );
select StartWith(test, 'A ' ) from test
posted also in
https://stackoverflow.com/questions/51206529/how-to-create-startswith-function-for-charn-type-with-ending-space
