Doesn't work "SELECT COUNT(*) FROM..." in PHP script

Question

My function should return number of rows with email like '$email'. But whey return 0 all the time, although in database i have rows with email like I insert in variable '$email'. What could be the reason?

function checkMail($email){     
    $email = mysql_real_escape_string($email);
    $sql = "SELECT COUNT(*) FROM users WHERE email='$email'";       
    return mysql_query($sql);   
}

You say `like`... but you use `=`. Stop using mysql, the functions are deprecated. Use mysqli or PDO, and either pass a connection resource into the function or acquire it globally. — scrowler, Dec 19 '13 at 20:55
change the return to `return mysql_query($sql) or die("MySQL error for $sql:" . mysql_error());` It will show you an error message — beardhatcode, Dec 19 '13 at 20:55

score 0 · Answer 1 · answered Dec 19 '13 at 20:58

0

function checkMail($email){     
  $email = mysql_real_escape_string($email);
  $sql = "SELECT COUNT(*) FROM users WHERE email='$email'";       
  $resource=mysql_query($sql);
  $row=mysql_fetch_array($resource);
  return $row[0];   
}

answered Dec 19 '13 at 20:58

symcbean

47,736
6
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score 0 · Accepted Answer · edited May 23 '17 at 12:19

You aren't returning a result, you're returning a query resource:

function checkMail($email){     
    $email = mysql_real_escape_string($email);
    $sql = "SELECT COUNT(*) as emailCount FROM users WHERE email='$email'";       
    $query = mysql_query($sql) or die(mysql_error()); // show error if one happens
    return mysql_fetch_assoc($query);
}

This will return an associative array containing your results (if it succeeds), and you should be able to access your count by:

$res = checkMail('your@email.com');
$count = $res['emailCount'];

Side note:

mysql functions are deprecated, you should use mysqli or PDO syntax: https://stackoverflow.com/a/13944958/2812842

score -1 · Answer 3 · answered Dec 19 '13 at 20:59

-1

To fetch the count use:

mysql_query($sql)
$row = mysql_fetch_assoc($result);
return($row[0]);

The funny thing is that mysql_query return 0, which indicates query fail. Check the corresponding error message with:

echo mysql_error();

answered Dec 19 '13 at 20:59

vigour

45
1
8

Doesn't work "SELECT COUNT(*) FROM..." in PHP script

3 Answers3