Chyba php?

Jiri Chaloupka dev na b2bexpander.com
Pondělí Červen 4 21:55:47 CEST 2001


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Dne po  4. červen 2001 21:01 jste napsal(a):

>
> ----- Vypis chyby: -----
>
> Warning: 1 is not a valid MySQL-Link resource in kluby.php on line 53
>
> Warning: Supplied argument is not a valid MySQL result resource in
> kluby.php on line 54
>
> Warning: 1 is not a valid MySQL-Link resource in funkce.php on line 19
>
>
> ----- Odpovidajici radky: -----
>
> kluby.php : 53
> $vysl = mysql($databaze, "SELECT * FROM kluby WHERE id_skupiny = '$show'
> ORDER BY nazev DESC");

Nemel by jste radeji pouzivat mysql_db_query()?
U mysql_close() davate paranmetr s cislem connection (i kdyz je tusim 
nepovinny)? (mysql_close($con))

Chaloupka
- -- 
Jiri Chaloupka
B2BExpander.com
chaloupka na b2bexpander.com
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