no results in the second query string!

chaza

I am trying to make a paginator script work.
It actually works ok in so far as it gives the correct amount of links, but when I go to the next page - ie next set of results on the same page - because the input boxes that define the search are on the same page and are not set , it's like I am searching again and 'cos nothing is set I am getting no results! (I think that is what is happening!)
Is there a way around this?

laserlight

because the input boxes that define the search are on the same page and are not set

um, set them?

I'm have a rough idea of what you are talking about, but it would help if you calmed down and explained a little more clearly, perhaps step by step.

chaza

I'm calm.

here is an example page of what is happening.

PS there are only five records in this table and I have set the page limit to one.

set weight to 20 plus to display all records

chaza

This must be something to do with the input boxes.
When I comment them out and define the sql search exactly, the paginator script works fine, but when I have even one input box the same is happening.

Any suggestions?

Pig

The form does us no good. Show us your code.

chaza

OK that is what happens and here is the stripped down code:

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "DTD/xhtml1-transitional.dtd">
<html>
<head>
<title>Paginator Test</title>
<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1" />
</head>
<body>
<?php

$weight = $_POST['weight'];

$database = "blah";
$table = "gallery";
$db = mysql_connect("*****", "***", "******") or die("couldnt connect");
mysql_select_db($database) or die("couldnt select database");

define('ROWS_PER_PAGE', 1);

if (!isset($page) || $page < 1) {
   $page = 1;
}

$sql = "SELECT COUNT(*) AS Total FROM gallery WHERE year='$year'";
$resultcount = mysql_query($sql , $db) or die("Error executing $sql<br />MySQL reported: ".mysql_error());
$rowcount = mysql_fetch_row($resultcount);
$numrows = $rowcount[0];
mysql_free_result($resultcount);

$numpages = ceil($numrows / ROWS_PER_PAGE);
if ($numpages < $page) {
   $page = $numpages;
}
$offset = ROWS_PER_PAGE * ($page - 1);

$query = "SELECT * FROM gallery WHERE year='$year'";
$query .= " LIMIT $offset, ".ROWS_PER_PAGE;
$result = mysql_query($query, $db) or die("Error executing $query<br />MySQL reported: ".mysql_error());

print "<table>";

	while ($data = mysql_fetch_object($result)) { 
	print "<tr>\n";  
	print "<td>";
	print "weight: ".$data->pounds." lb ".$data->ounces." oz";
	print "<br />";
	print "date caught: ".$data->date." ".$data->month." ".$data->year;
	print "<br />";
	print "caught by: ".$data->captor. '<br />'; 
	print "</td>\n";
	print "</tr>\n";

}
print	"</table>\n";

if ($page > 1) { ?>
<a href="<?php echo $PHP_SELF; ?>?page=<?php echo $page - 1; ?>">Previous</a>
<?php } 

for($i = 1; $i <= $numpages; $i++){

if(($page) == $i){
    echo "$i&nbsp;";
    } else {
        echo "<a href=\"".$_SERVER['PHP_SELF']."?page=$i\">$i</a>&nbsp;";
}
}
	if ($page < $numpages) { ?>
<a href="<?php echo $PHP_SELF; ?>?page=<?php echo $page + 1; ?>">Next</a>
<?php } ?>

<form action="<? echo $_SERVER['PHP_SELF']; ?>" method="post" id="search">
<select name="year">
<option value="">Year</option>
<option value="2002">2002</option>
<option value="2003">2003</option>
<option value="2004">2004</option>
</select><br />
<input type="submit" value="search" />
</form>
</body>
</html>

chaza

anyone?

I think I might need to save the search results somewhere and access them later somehow.