andrisp Posted May 3, 2005 Report Posted May 3, 2005 $izsaukums = mysql_fetch_array("SELECT * FROM jaunumi WHERE id = 1"); nu kur te ir iemesls lai buutu shitas errors ? Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
bubu Posted May 3, 2005 Report Posted May 3, 2005 myqsl_fetch_array() fjai kā parametrs jāpadod mysql_query fjas rezultāts. $q=mysql_query("SELECT * FROM jaunumi WHERE id = 1"); $izsaukums = mysql_fetch_array($q); Vai tad manuālī piemērus nedomāji skatīties?
andrisp Posted May 3, 2005 Author Report Posted May 3, 2005 ai.. fak.. galva galiigi vairs nestraadaa.. nu ja.. pats sajaucu vietaam un veel briinos.. paldies, ka paraadiiji manu stulbumu. :)
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