What is wrong with this?

jamesm87

Warning: extract() [function.extract]: First argument should be an array in /home/expertsh/public_html/goldencoast/footer2.php on line 21

What does this mean?

here is the file

<?php

/* Header template */

echo "&nbsp;</td>
  <tr>
    <td width='33%'></td>
    <td width='92%' colspan='2' rowspan='2'>
</table>
<table border='0' cellpadding='0' cellspacing='0' width='100%'>
  <tr>
    <td width='5%' align='center'><a href='javascript:history.go(-1)'><img border='0' src='http://www.expertshelp.com/goldencoast/arrow.jpg'></a></td>
    <td width='50%'><b><font face='Lucida Console' color='#00AEEF'>Go Back</font></b></td>
  </tr>
</table>";
if (isset($staff_id_a))
	{
		$sqla="select message as mess from login where staff_id=('$staff_id_a')";
$resulta=mysql_query($sqla) or die("Error in mysql:<br><br> Sql = $sql <br><br>Log = ".mysql_error());
$rowa= mysql_fetch_array($resulta);
extract ($rowa);
	if($mess != 'n')
			{
				echo "<script>
	alert(\" $mess \")
	</script>";
		$sql="update login set message='n' where staff_id=('$staff_id')";
		$result=mysql_query($sql) or die("Error in mysql:<br><br> Sql = $sql <br><br>Log = ".mysql_error());
			}
	}
	?>

steadyguy

The error message means that your result set is not an array - your query is either faulty, or, more likely, is not returning an array (ie., you're only selecting one column). Something like this should work:

SELECT staff_id, message AS mess
FROM login
WHERE staff_id='$staff_id_a'

Alternatively, you could just skip the extract() step entirely:

$rowa= mysql_fetch_row($resulta);
if($rowa[0] != 'n') {
    echo '<script>alert("' . $rowa[0] . '")</script>';
    // etc ...
}