php mysql not returning desired integer, simple error

Question

I am having an issue with the following php/sql. I ran the sql in the mysql terminal and it worked perfectly. I also tried my connect statement alerting a string and it succesfully alert the string when connected to the db. There is some error within this code that is returning null when it should be returning two integers. Any suggestions?

$xmax = mysql_fetch_row(mysql_query('SELECT MAX(x) AS xmax FROM headerfooter WHERE site = "Brighter_Vista" AND location = 0'));
$xmax = $xmax['xmax'];
$ymax = mysql_fetch_row(mysql_query('SELECT MAX(y) AS ymax FROM headerfooter WHERE site = "Brighter_Vista" AND location = 0'));
$ymax = $ymax['ymax'];
echo '<script>alert("'.$xmax.$ymax.'")</script>';

Please use PDO rather than mysql_ functions, they both have major security flaws and are old... like deprecated old. — Jonathan
– Jonathan, Commented Nov 21, 2012 at 2:01
Try using print_r($xmax) after the mysql_fetch_row call to see what, if anything, is returned from the query. — Sean Powell
– Sean Powell, Commented Nov 21, 2012 at 2:03
just wondering: are you sure mysql supports double quotes for strings encapsulation? — Sebas
– Sebas, Commented Nov 21, 2012 at 2:03
Nothing is being returned in that print_r command. However, as I said, the query worked in the terminal and the db is definitely connected. — eatonphil
– eatonphil, Commented Nov 21, 2012 at 2:04

nickhar · Accepted Answer · 2012-11-21 02:15:36Z

1

We can't see the rest of your code, but, you can refine this into one query, doing something like:

  $connection = mysql_connect(
        $DB_hostname, 
        $DB_username, 
        $DB_password) or die(mysql_error());
  mysql_select_db($DB_name, $connection);

  $query = mysql_query("
      SELECT MAX(x) AS xmax, MAX(y) AS ymax 
      FROM headerfooter 
      WHERE site = 'Brighter_Vista' AND location = '0'
  ", $connection) or die(mysql_error());

  while ($row = mysql_fetch_array($query))
  {
     $xmax = $row['xmax'];
     $ymax = $row['ymax'];
  }

  echo '<script>alert("'.$xmax.$ymax.'")</script>';

However, if you have the time, ability and wish to learn, you should (as noted by others) look at and using either mysqli or PDO. Wise advice and in time you'll understand why.

answered Nov 21, 2012 at 2:15

nickhar

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Samuel Cook · Accepted Answer · 2012-11-21 02:04:40Z

1

The easist solution is to check if the sql statement is returning an error using:

mysql_query('SELECT MAX(y) AS ymax FROM headerfooter WHERE site = "Brighter_Vista" AND location = 0') or die('Error: '.mysql_error());

Side note: Don't use mysql_* functions anymore, that time has passed. Use either mysqli or PDO

answered Nov 21, 2012 at 2:04

Samuel Cook

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