这个PHP通知是什么意思？

Question

<?php
    $username="xxx";
    $password="xxx";
    $database="mobile_app";

    mysql_connect('localhost',$username,$password);

    @mysql_select_db($database) or die( "Unable to select database");

    foreach (array('courseid','roomChosen') as $varname) {
        $$varname = (isset($_POST[$varname])) ? $_POST[$varname] : '';
    }

    if (isset($_POST['prequestion'])) {
        $roomquery = "
                     SELECT Room
                     FROM Room
                     WHERE
                     (Room = '".mysql_real_escape_string($roomChosen)."')
                     ";

        $roomnum = mysql_num_rows($roomresult = mysql_query($roomquery));
        mysql_close();

        if($roomnum ==0){
            $msg = "This Room is Invalid '$roomChosen'";
        } 
        else {
            $msg = "This Room is Valid '$roomChosen'";
        }
    }
    $d = array("msg" => $msg);
    echo json_encode($d);
?>

In this line of code near the bottom:

$d = array("msg" => $msg);

I am getting this notice:

Notice: Undefined variable: msg in /u08877587/Mobile_app/room2.php on line 46 {"msg":null}

How do I fix this notice?

Answer 1

You should place $d = array("msg" =>$msg) inside if(isset($_POST['prequestion'])) { }

I.e:

if (isset($_POST['prequestion'])) 
{
    if($roomnum ==0)
    {
        $msg = "This Room is Invalid '$roomChosen'";
    } 
    else 
    {            
        $msg = "This Room is Valid '$roomChosen'";
    }


   $d = array("msg" => $msg);
   echo json_encode($d);    
}