我收到以下php错误[重复]

Question

This question already has an answer here:

• “Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP 28 answers

This is the part of my code giving the problem.I tried to use isset but i think my syntax is incorrect. I still get the Notice: Undefined index: id.

if($_GET['id'] && $record['allusers'] != "Y"){
        $q2 = mysql_query("SELECT * from users WHERE alljobs = 'Y'");
        while($arr2 = mysql_fetch_assoc($q2)){
            $userjobs2[] = $arr2['id']; 
        }

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Answer 1

Your code is correct this is not an Error it is a Notice to get rid of this you can use isset($_GET['id'])

if(isset($_GET['id']) && $record['allusers'] != "Y"){
        $q2 = mysql_query("SELECT * from users WHERE alljobs = 'Y'");
        while($arr2 = mysql_fetch_assoc($q2)){
            $userjobs2[] = $arr2['id']; 
        }
}