CODEWITHSUNDEEP
phpvariables
user1213994
user1213994

Reputation: 13

Unidentified variable

Will someone PLEASE explain to me why I get an error saying that $new_cid is unidentified? It only happens when I use .= to append a value to the existing variable.

    $dereg_course_student= mysql_query("SELECT * FROM course_student");
$new_cid;
while($row=mysql_fetch_assoc($dereg_course_student)){
        $cid=explode(".",$row['cid']);
        foreach($cid as $cids){
            if($cids==$_GET['cid']){
                unset($cids);}
            if(isset($cids)){
                $new_cid=$new_cid.".{$cids}";}
            } mysql_query("UPDATE course_student SET cid={$new_cid} WHERE sno={row['sno']}");
    }

Upvotes: 0

Views: 158

Answers (3)

INaTrouble
INaTrouble

Reputation: 1

it should be in php $new_cid = "";

Upvotes: 0

Csak Zoli
Csak Zoli

Reputation: 428

Because you are using $new_cid; at the beginning and you aren't specifying what type of variable it should be. In php you don't have to declare variables you can just use $new_cid in the loop whit no pre-defined value.

    $dereg_course_student= mysql_query("SELECT * FROM course_student");
 $new_cid='';
while($row=mysql_fetch_assoc($dereg_course_student)){
        $cid=explode(".",$row['cid']);
        foreach($cid as $cids){
            if($cids==$_GET['cid']){
                unset($cids);}
            if(isset($cids)){
                $new_cid.=$cids;}
            } mysql_query("UPDATE course_student SET cid=$new_cid WHERE sno=row['sno']");
    }

Try it like this.

Upvotes: 1

SamHuckaby
SamHuckaby

Reputation: 1162

I don't know that it would actually cause a problem, but I would try setting $new_cid to an empty string, before you try concat'ing with it. Like so:

$new_cid = "";

put that where you have "$new_cid;" right now.

Upvotes: 0

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