Zepo. Posted December 23, 2007 Share Posted December 23, 2007 $apps = mysql_query("SELECT signupid,name.gamename,sn,email,pass,position,info,time,ip FROM apply"); while ($apps = mysql_fetch_array($apps)){ Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource. The loop works, its just that it gives that warning.... Quote Link to comment Share on other sites More sharing options...
redarrow Posted December 23, 2007 Share Posted December 23, 2007 wht this part mate is it a join name.gamename Quote Link to comment Share on other sites More sharing options...
Zepo. Posted December 23, 2007 Author Share Posted December 23, 2007 $apps = mysql_query("SELECT signupid,name,gamename,sn,email,pass,position,info,time,ip FROM apply"); while ($apps = mysql_fetch_array($apps)){ Sorry. Quote Link to comment Share on other sites More sharing options...
MadTechie Posted December 23, 2007 Share Posted December 23, 2007 try this <?php $apps = mysql_query("SELECT signupid,name,gamename,sn,email,pass,position,info,time,ip FROM apply") or die(mysql_error()); while ($apps = mysql_fetch_array($apps)){ ?> whats the error ? Quote Link to comment Share on other sites More sharing options...
redarrow Posted December 23, 2007 Share Posted December 23, 2007 add this please sorry i got beat lol <?php $apps = mysql_query("SELECT signupid,name,gamename,sn,email,pass,position,info,time,ip FROM apply")or die(mysql_error()); // if a error show it..... echo $apps; // if the error dosent show cheek query while ($apps = mysql_fetch_assoc($apps)){ ?> Quote Link to comment Share on other sites More sharing options...
Zepo. Posted December 23, 2007 Author Share Posted December 23, 2007 It doesnt die, and it echos Resource id #9 Quote Link to comment Share on other sites More sharing options...
redarrow Posted December 23, 2007 Share Posted December 23, 2007 try <?php $apps = mysql_query("SELECT `signupid`,`name`,`gamename`,`sn`,`email`,`pass`,`position`,`info`,`time`,`ip` FROM `apply`")or die(mysql_error()); // if a error show it..... echo $apps; // if the error dosent show cheek query while ($apps = mysql_fetch_assoc($apps)){ ?> Quote Link to comment Share on other sites More sharing options...
Zepo. Posted December 23, 2007 Author Share Posted December 23, 2007 No difference. The weird part is that its actually working, just coming up with that error though. Quote Link to comment Share on other sites More sharing options...
redarrow Posted December 23, 2007 Share Posted December 23, 2007 try assoc not array Quote Link to comment Share on other sites More sharing options...
Zepo. Posted December 23, 2007 Author Share Posted December 23, 2007 I did, nothing. Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource Quote Link to comment Share on other sites More sharing options...
Zepo. Posted December 23, 2007 Author Share Posted December 23, 2007 Bump. Quote Link to comment Share on other sites More sharing options...
MadTechie Posted December 23, 2007 Share Posted December 23, 2007 lol, i should of noticed.. try changing to <?php $mySQL = mysql_query("SELECT signupid,name,gamename,sn,email,pass,position,info,time,ip FROM apply")or die(mysql_error()); while ($apps = mysql_fetch_assoc($mySQL)){ ?> Quote Link to comment Share on other sites More sharing options...
Zepo. Posted December 23, 2007 Author Share Posted December 23, 2007 Yup worked, that makes since to me now, defining a query to the same thing thats an array. Gotcha, thanks. Quote Link to comment Share on other sites More sharing options...
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