Pagination Help - Please
Posted: Mon Sep 15, 2008 3:48 pm
I am having trouble setting up pagination on my site. I can connect and query the database even return results but the next page link at the bottom of the results will not work. Right now the limit of results per page is 10 and if the search returns more than 10 results the next links will not work.
See my code below
Also here is a link to a query so you can get a better idea about what I am trying to do.
http://www.fashiontells.com/members/sta ... ch.php?q=O
Please Help!!! Thanks in advance.
See my code below
Code: Select all
<?php
// Get the search variable from URL
$var = @$_GET['q'] ;
$trimmed = trim($var); //trim whitespace from the stored variable
// rows to return
$limit=10;
// check for an empty string and display a message.
if ($trimmed == "")
{
echo "<p>Please enter a search...</p>";
exit;
}
// check for a search parameter
if (!isset($var))
{
echo "<p>We dont seem to have a search parameter!</p>";
exit;
}
include "../config.php";
mysql_connect($server, $db_user, $db_pass) or die (mysql_error());
//specify database ** EDIT REQUIRED HERE **
mysql_select_db("$database") or die("Unable to select database"); //select which database we're using
// Build SQL Query
$query = "select * from $table where username like \"%$trimmed%\"
order by id"; // EDIT HERE and specify your table and field names for the SQL query
$numresults=mysql_query($query);
$numrows=mysql_num_rows($numresults);
// If we have no results, offer a google search as an alternative
if ($numrows == 0)
{
echo "<h4>Results</h4>";
echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>";
// google
echo "<p><a href=\"http://www.google.com/search?q="
. $trimmed . "\" target=\"_blank\" title=\"Look up
" . $trimmed . " on Google\">Click here[/url] to try the
search on google</p>";
}
// next determine if s has been passed to script, if not use 0
if (empty($s)) {
$s=0;
}
// get results
$query .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");
// display what the person searched for
echo "<p>You searched for: "" . $var . ""</p>";
// begin to show results set
echo "Results";
$count = 1 + $s ;
// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
$title = $row["username"];
echo "$count.) <a href='http://www.fashiontells.com/members/$title/$title.php'>$title[/url]
" ;
$count++ ;
}
$currPage = (($s/$limit) + 1);
//break before paging
echo "
";
// next we need to do the links to other results
if ($s>=1) { // bypass PREV link if s is 0
$prevs=($s-$limit);
print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><<
Prev 10[/url]  ";
}
// calculate number of pages needing links
$pages=intval($numrows/$limit);
// $pages now contains int of pages needed unless there is a remainder from division
if ($numrows%$limit) {
// has remainder so add one page
$pages++;
}
// check to see if last page
if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {
// not last page so give NEXT link
$news=$s+$limit;
echo " <a href=\"http://www.fashiontells.com/members/sta ... ews&q=$var\">Next 10 >>[/url]";
}
$a = $s + ($limit) ;
if ($a > $numrows) { $a = $numrows ; }
$b = $s + 1 ;
echo "<p>Showing results $b to $a of $numrows</p>";
?>Also here is a link to a query so you can get a better idea about what I am trying to do.
http://www.fashiontells.com/members/sta ... ch.php?q=O
Please Help!!! Thanks in advance.