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Variable passed incorrectly

Posted: Fri Dec 09, 2005 8:01 am
by Catz
Hi All!

I have the following php code in a form:

Code: Select all

$Limit = 1;
$Page = $_REQUEST["Page"];
if(empty($Page)) {$Page = 1; }
	
$Start = $Page / $Limit - 1;
if($Start <= 1) {$Start = 0;}

$list = mysql_query("SELECT * FROM detail LIMIT $Start, $Limit") or die(mysql_error());
$num = mysql_num_rows($list);
$rows = $num / $Limit;
$next = $Page * $Limit;
...

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if($Page > 1) {
$p = $Page - 1;
?>

<input type="hidden" name="Page" value="<?php echo $p; ?>">
<input type="submit" name="Submit" value="Previous">

<?php
if($next < $rows) {
$f = $Page + 1;
?>

<input type="hidden" name="Page" value="<?php echo $f; ?>">
<input type="submit" name="Submit" value="Next"><?php }
The "next" button works fine, but the "previous" button acts weird. When pressed the first time, eg on page4, it will go to page 3. When pressed again on page3, it goes back to page 4....? I've added
<?php print $p; ?> and <?php print $Page ?> - $p outputs the correct page number : 2, but when submitted, $Page outputs "4" instead of "2"

Does any one know why this is?

Thnx!

d11wtq | Sorted out the

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 tags... NOTE: You only need one set of [php ][/ php] tags around your code.... <?php ?> inside you code is automatically highlighted ;)[/color]