Member Avatar for Sarao

Hi,

I used the code

<?php
$con = mysql_connect("localhost","clickim_forum","thankyou1");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("clickim_forum", $con);

$result = mysql_query("SELECT userid FROM user ORDER BY userid DESC LIMIT 0,1");

echo "<table border='0'>";
echo "<tr>";
while(row = mysql_fetch_array($result))
  {
  echo "<td style='color:#FF0000 '>" . $row['userid'] . "</td>";
  } 
    {
  echo "<td>Users Online & counting. Are you one of them?</td>";
  }
  {
  echo "</tr>";
  }
echo "</table>";

mysql_close($con);
?>

to show the total registered members on a vBulletin forum. But Now I have upgraded my server and it stopped working and now its giving me error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/clickim/public_html/1.php on line 14

Please help.

  • 3 Contributors
  • 3 Replies
  • 158 Views
  • 9 Hours Discussion Span
  • Latest Post Latest Post by faroukmuhammad
Member Avatar for Insensus

Change

$result = mysql_query("SELECT userid FROM user ORDER BY userid DESC LIMIT 0,1");

to

$result = mysql_query("SELECT userid FROM user ORDER BY userid DESC LIMIT 0,1") or die(mysql_error());

and you should see an SQL error.
If you can't solve the error, just post it back here.

Member Avatar for Sarao

I get this error now

Parse error: syntax error, unexpected '=' in /home/clickim/public_html/1.php on line 14
Member Avatar for faroukmuhammad

Sarao, there was mistake (or lets say typographical error) you made in line 14. The variable row has no dollar sign...
change

while(row = mysql_fetch_array($result))

to

while($row = mysql_fetch_array($result))
Edited by faroukmuhammad because: n/a
Be a part of the DaniWeb community

We're a friendly, industry-focused community of developers, IT pros, digital marketers, and technology enthusiasts meeting, networking, learning, and sharing knowledge.