$_POST Variable question.

[Josheeyy]

I have created a form which a user would enter an ID# , then this ID# would be used to make a query in another form. This part works. The proble is that more than one record contain this ID#, so I need to move to the next record, I can not because the query is expecting a value from the first form again.

This is the error I get when I try to move to the next record:
Notice: Undefined index: tagno in D:\PHPProject\UpdateSoftware.php on line 57

And this is my query:
mysql_select_db($database_PHpMysql, $PHpMysql);
$test = $HTTP_POST_VARS['tagno'];
$query_rsUpdateSoft = "SELECT * FROM soft WHERE soft.tagno ='".$test."'";
$query_limit_rsUpdateSoft = sprintf("%s LIMIT %d, %d", $query_rsUpdateSoft, $startRow_rsUpdateSoft, $maxRows_rsUpdateSoft);
$rsUpdateSoft = mysql_query($query_limit_rsUpdateSoft, $PHpMysql) or die(mysql_error());
$row_rsUpdateSoft = mysql_fetch_assoc($rsUpdateSoft);

PLEASSSEE Someone help!!

[megayoga]

Don't know if I undestood your problem, but I assume you need to pass on the tagno value.
If you have more than one result and have to move to next post, pass the tagno value in a hidden field:
<form action="somefile.php" method="post">
<input type="hidden" value="<?=$HTTP_POST_VARS['tagno']?>" name="tagno">
</form>

You can trigger this form either by a submit-button or javascript in a regular link.