andrisp Posted May 3, 2005 Report Share 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 Link to comment Share on other sites More sharing options...
bubu Posted May 3, 2005 Report Share 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? Link to comment Share on other sites More sharing options...
andrisp Posted May 3, 2005 Author Report Share Posted May 3, 2005 ai.. fak.. galva galiigi vairs nestraadaa.. nu ja.. pats sajaucu vietaam un veel briinos.. paldies, ka paraadiiji manu stulbumu. :) Link to comment Share on other sites More sharing options...
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