Posted by Robert on 05/31/07 01:47

Hi y'all,

The below gives resource error # 4. I've cut a lot of code out in an
attempt to just display adv_id twice. The first time works as expected.
The second time I get the error.

When echoing result, I should be looking for an array to be displayed?
If so, how?

Thanks.

Robert

*******************

<?php
include ("connect.php");
?>

<form action="<?php echo "display_details.php" ?>" method="post">
<p>Which record would you like to see?</p>
<P>Enter Record #<br>
<input type=text name="adv_id" size=5>
<input type=submit name="submit" value="View Record">
</form>

<?php
if (!empty($_POST))
{
$adv_id = $_POST['adv_id'];
echo "$adv_id";
$result=mysql_query("SELECT * FROM advertiser_tbl WHERE adv_id =
'adv_id'") or die

(mysql_error());
echo "$result";
}

?>

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