asked on

array problems

I have the following code. I just want it to display $count.
[php]
<?php
       require_once 'classes/pg_class.php';
       extract($GLOBALS);
                  $db = new db(0);
// check if the username exists in the database
//$connect = $DB->PostgreDB();
$clean = pg_escape_string($_GET['username']);
$query = "SELECT count(username) as count FROM users WHERE username= '$clean'";
$data = $db->fetch($query);
extract($data);
var_dump($data);
/*if ($count = 0) {
      $_SESSION['output'] .= "A User already exists with that username. Please try another.<br />";
      $error = '1';
} else {
echo 'this is a good';
}*/
if ($count == '1') {
      echo 'this is good';
}
echo $count;
?>
[/php]
It generate the following
[ouput]
array(1) { [0]=> array(1) { ["count"]=> string(1) "1" } }
[/output]

Roonaan

Use
$data = $db->fetch($query);
$count = $data['count'];

-r-

casit

ASKER

Nope same exact thing. Nothing.

ASKER CERTIFIED SOLUTION

Roonaan

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holy007

Does this example help you out a bit?

<?php
$res = mysql_query("select * from blah") // a query that returns an empty set
$row = mysql_fetch_array($res); // get's 0 since there's no return
echo count($row); // echos 1 - since $row is not an array
echo $row[0]; // echos "", but casts $row as an array?
echo count($row); // echos 0 now
?>

With kind regards,

HoLy007

casit

ASKER

@Roonaan
ok var_export($data); shows the following
array ( 0 => array ( 'count' => '1', ), )

casit

ASKER

Thanks I got it.

Roonaan

Which solution did you use?
If you used the plain SELECT * FROM table along with $result->count / mysql_num_rows, then make sure to add a LIMIT 1 to the end of your query. You don't actually care how many rows there are returned, only if there are any.

-r-