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[SOLVED] Loop Error


Zepo.

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$apps = mysql_query("SELECT signupid,name.gamename,sn,email,pass,position,info,time,ip FROM apply");
while ($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....

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https://forums.phpfreaks.com/topic/82872-solved-loop-error/
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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)){
?>

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https://forums.phpfreaks.com/topic/82872-solved-loop-error/#findComment-421477
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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)){
?>

Link to comment
https://forums.phpfreaks.com/topic/82872-solved-loop-error/#findComment-421488
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