How to get the HTML form to show once

[micmac]

Hi Guys,
Can anyone tell me where i am going wrong?
The HTML form is showing up however many times there is a 'confirm_cat' value in the database.

<?
$row = mysql_fetch_row(mysql_query("SELECT COUNT(*) FROM cat1 WHERE CatParent = $id"));
if($row[0] > 0){

$result = mysql_query("SELECT * FROM cat1 WHERE CatParent = $id AND active='yes' ORDER BY CatName");
while($row = mysql_fetch_array($result)){

    if($row['confirm_cat'] == 'nil'){
	?>
<form method="post" action="<?php echo $PHP_SELF?>">

HTML FORM CONTENTS HERE.....

</form>
	  <?
	  } else if($row['confirm_cat'] == 'yes'){
        echo "<center><TABLE width=\"400\"></center><TR><td>";
		echo "<p align=\"left\"><a href=\"add_url_free.php?id={$row['CatID']}&country=$country\">{$row['CatName']}</a></p>";
        echo "</TR><center></table></center>";	    

		}
		}
		}
		else {
		?>

  <form method="post" action="<?php echo $PHP_SELF?>">

HTML FORM CONTENTS HERE.....         

</form>

<?	
	}
	}
?>		

[assassin74]

Hi,

the form shows up once for each resulting row of you second query. To avoid this, move the form-tags outside the while-loop.

Like this:

untested

if($row[0] > 0){

echo "<form method='post' action='".$PHP_SELF?."'>";

$result = mysql_query("SELECT * FROM cat1 WHERE CatParent = $id AND active='yes' ORDER BY CatName");
while($row = mysql_fetch_array($result)){

    if($row['confirm_cat'] == 'nil'){                     
		echo "          
			// HTML FORM CONTENTS HERE.....
		";

    } else if($row['confirm_cat'] == 'yes'){
		echo "
			// HTML FORM CONTENTS HERE.....
        ";

     else { 
		/* and so on */ 
     }
 }

echo "</from>";