Обсуждение: LIKE problem
Hi Everybody!
I'm asking you, because recently I have some problems with the LIKE so I test with an easy and simple query but doesn't work:
SELECT *
FROM employee
WHERE id like 'h%';
The error says:
"ERROR: operator does not exist: integer ~~ unknown
LINE 5: WHERE id like 'h%';
^
HINT: No operator matches the given name and argument type(s). You might need to add explicit type casts.
********** Error **********
ERROR: operator does not exist: integer ~~ unknown
SQL state: 42883
Hint: No operator matches the given name and argument type(s). You might need to add explicit type casts.
Character: 34"
I don't know why because I did before more difficult querys and works well. What is the problem?
Thanks a lot
JP Cook
2009/7/6 Juan Pablo Cook <juampick@gmail.com>: > Hi Everybody! > I'm asking you, because recently I have some problems with the LIKE so I > test with an easy and simple query but doesn't work: > SELECT * > FROM employee > WHERE id like 'h%'; > The error says: > "ERROR: operator does not exist: integer ~~ unknown SELECT * FROM employee WHERE name like 'h%'; --<<< there are not LIKE operator for integer values. you cannot use LIKE on integer columns regards Pavel > > LINE 5: WHERE id like 'h%'; > > ^ > > HINT: No operator matches the given name and argument type(s). You might > need to add explicit type casts. > > ********** Error ********** > > ERROR: operator does not exist: integer ~~ unknown > > SQL state: 42883 > > Hint: No operator matches the given name and argument type(s). You might > need to add explicit type casts. > > Character: 34" > > I don't know why because I did before more difficult querys and works well. > What is the problem? > > Thanks a lot > > JP Cook
In response to Juan Pablo Cook :
> Hi Everybody!
> I'm asking you, because recently I have some problems with the LIKE so I test
> with an easy and simple query but doesn't work:
>
> SELECT *
> FROM employee
> WHERE id like 'h%';
>
> The error says:
> "ERROR: operator does not exist: integer ~~ unknown
>
> LINE 5: WHERE id like 'h%';
>
> ^
>
> HINT: No operator matches the given name and argument type(s). You might need
> to add explicit type casts.
You can't compare a INT with a TEXT:
test=*# select '1'::int4 like '1'::text;
ERROR: operator does not exist: integer ~~ text
LINE 1: select '1'::int4 like '1'::text;
^
HINT: No operator matches the given name and argument type(s). You
might need to add explicit type casts.
But you can CAST the INT into a TEXT:
test=# select '1'::int4::text like '1%'::text;
?column?
----------
t
(1 row)
Okay?
But in your case,you have an INT-column (id) and you compare that with
'h%'. You can CAST the id-column, but all INT::TEXT don't match 'h%'.
--
Andreas Kretschmer
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