PHP Breaks with second mysql_query

Question

Running a program which includes two queries. When I run either query alone it works but when both queries exist within the code it breaks.

$qry = "SELECT * FROM temp_user WHERE email='$email' AND pin='$pin'";
$result = mysql_query($qry);
$num_rows = mysql_num_rows($result);

$qry2 = "SELECT * FROM email WHERE email='$email'";
$result2 = mysql_query($qry2);
$num_rows2 = mysql_num_rows($result2);

How can I fix this?

EDIT:

For those asking what do I mean by it breaks, here is an image.

Error from log.

PHP Parse error: syntax error, unexpected '}' in /Users/philipkirkbride/Documents/apps/Today_test/confirm.php on line 30

Full code of page is

<?php
include 'connect.php';

$pin = $_GET['pin'];
$email = $_GET['email'];

$qry = "SELECT * FROM temp_user WHERE email='$email' AND pin='$pin'";
$result = mysql_query($qry);
$num_rows = mysql_num_rows($result);
if ($num_rows!=0){
    print "create user and delete temp";
    $sql = "INSERT INTO pin VALUES (DEFAULT, '$pin', '$email')";
    $result = mysqli_query($con,$sql);
    if ( false===$result ) {
      printf("error: %s\n", mysqli_error($con));
    } else {
        // Delete user from temp table
        $sql2 = "DELETE FROM temp_user WHERE email='$email' AND pin='$pin'";
        $delete = mysqli_query($con,$sql2);
        // Make query to see if user is new or existing
        $qry2 = "SELECT * FROM email WHERE email='$email'";
        $result2 = mysql_query($qry2);
        $num_rows2 = mysql_num_rows($result2);
        // Need to add a snippet to add a row to the email table, make sure to check user doesn't have an email already in table
        if($num_rows2==0){
            print "email doesn't exist, create new user."
            // $date = new DateTime;
            // $sql = "INSERT INTO email VALUES ('$email', '$date')";
            // $result = mysqli_query($con,$sql);
        }else{
            print "email exists already";
        }
    }
}else{
    print "Account request not found";
}
// End connection
mysqli_close($con);
?>

Answer 1

Try calling $result->close() before you call the second query, in order to release the connection. Your PHP script hasn't finished executing yet, and your previous result is holding that connection open. In general, it is good practice to explicitly clean up these kinds of resources rather than leaving it to the runtime.

EDIT: I believe @eis is correct that since this is the older (and deprecated, incidentally) mysql_ api, that the proper call would actually be mysql_free_result($result).

ANOTHER EDIT: I put this comment under @eis's answer, but I figure it bears bolder type, because @eis's answer was absolutely right and should not have been down-voted:

Just to elaborate redundantly on the whole buffered vs. unbuffered thing; the concept is that a buffered query will read all of the results back from the database into a buffer, then release the connection automatically (this will happen synchronously, so that the connection is free for another call immediately after the buffered call returns). An unbuffered query will essentially open the connection and hand back a reference to a database server cursor, from which you will then need to fetch records. As long as there is something left to fetch, the cursor is still open, and you won't be using the underlying connection for anything else. Or you can call mysql_free_result() and free the connection "early" (without reading all of the rows from the cursor first).

Answer 2

you should actually either 1) read the result that you retrieved or 2) free it (mysql_free_result)...

You have two distinct queries. If you don't read the result or close it in between, it might not let you run a new one.

(I don't see the point doing the queries if you don't actually need the actual items. Use SELECT COUNT(*) if you are not intrested in the actual items, and just about the count.)