Dave [Hawk-Systems] wrote:
> (CASE WHEN (keywords ~* '.*MySearchString.*') THEN (substring_count(keywords
> from '.*MySearchString.*')*5) ELSE 0 END)
This question (essentially) came up twice in July (once here and once on
the SQL list -- author copied) and I don't recall seeing an answer
posted. I've been saving it for when I had the time and energy to work
out a solution. Not sure you're still in need of this, and I'm sure
there are more efficient ways to do this sort of thing (certainly
implementing it in C would help), but FWIW this seems to do what you wanted:
create or replace function substr_count(text, text) returns integer as '
declare
t1 alias for $1; -- source string
t2 alias for $2; -- search pattern
match integer := 0;
pos integer := 0;
p integer := 0;
px integer := 0;
len1 integer := 0;
len2 integer := 0;
begin
len1 := length(t1);
len2 := length(t2);
if len2 < 1 then
return 0; /* result for empty pattern */
end if;
/* no use in searching t1 past point where t2 will fit */
px := len1 - len2 + 1;
for p in 1..px loop
if substr(t1, p, len2) = t2 then
match := match + 1;
end if;
end loop;
return match;
end;
' language plpgsql immutable strict;
regression=# select substr_count('abcdefedcba', 'a');
substr_count
--------------
2
(1 row)
regression=# select substr_count('aaa', 'aa');
substr_count
--------------
2
(1 row)
regression=# select substr_count('aabaacaaadefabaa', 'aa');
substr_count
--------------
5
(1 row)
HTH,
Joe
