[SOLVED] mysql select problem

[Lambneck]

Hello,

I have here, echoed as a link, the subject  of my submitted forms.

What I am trying to do is make it so each individual subject from a form submission becomes a link to a page displaying the rest of its form entry (ie. name, email, message).

 

The links page:

 

$result = mysql_query("SELECT col_4 FROM {$table} ORDER BY submission_date DESC");
if (!$result) {
    die("Query to show fields from table failed:".mysql_error());
}

while($row = mysql_fetch_array($result))
{

	echo '<a href="ResumeDisplay.php?id='.$row['id'].'">'.$row['col_4'].'</a>';
	echo "<br />";

}
mysql_free_result($result);
?>

 

and the display page:

 

$id = (int) $_GET['submission_id'];

$sql = "SELECT * FROM {$table} WHERE submission_id=$id, ";
$result = mysql_query($sql) or die(mysql_error());
if(mysql_num_rows($result) == 1){
$row = mysql_fetch_assoc($sql);
//print out information
}else{
echo 'That record ID does not exist!';
}
?>

 

 

the problem is with the code on the display page. I get the error message:

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

 

 

[toplay]

You have a comma at the end.  Change this:

 

$sql = "SELECT * FROM {$table} WHERE submission_id=$id, ";

 

 

to this:

 

$sql = "SELECT * FROM {$table} WHERE submission_id=$id";

 

 

 

[AndyB]

And please don't double post. Now you have two answers - both the same - and my time was wasted.