Re: Undefined variable problem

Hello

This is the code: $m = 1; while ($line = mysql_fetch_assoc($rs)) { //dumping into an array

     foreach ($line as $field => $value) {
     $tmp = $field.$m;

$$tmp.= $value; /* ********This is the error line 117******* */

     }
     $m++;

} This is basically so that i can use the array via variables like so: if($tcl2==1){echo $tcl2;} instead of: if($arrayname['tcl']['1']==1){echo $arrayname['tcl']['1'];} and so on....esp since i have 38 fields. I can of course just shut off error reporting...but thats not the best solution right? and i still wouldnt know whats causing these notices...

Simply define $$tmp -variable before adding stuff in it. [code] $m = 1; while ($line = mysql_fetch_assoc($rs)) { //dumping into an array

     foreach ($line as $field => $value) {
        $tmp = $field.$m;
        $$tmp = "";
	 $$tmp .= $value;      }
     $m++;

} [/code] No it should whine about undefined vars. Cheers, Joona -- Joona Kulmala <joona@fiveam.org> PHP Finland