Question

The following script is showing only first page and link to next page is there but is not leading anywhere. Can someone help me?

$var = @$_GET['q'] ;
$trimmed = trim($var);
$limit = 10;

if ($trimmed == "")
{
    echo "<p>What are you looking for?...</p>";
    exit;
}

if (!isset($var)) 
{   
    echo "<p>We dont seem to have a search parameter!</p>";
    exit;
}

mysql_connect('xxx', 'yyy', 'zzz');
mysql_select_db('yyy') or die('Unable to select database');
$query = "select * from table  where NAME like '%$trimmed%' order by NAME";
$numresults = mysql_query($query);
$numrows = mysql_num_rows($numresults);

if ($numrows == 0)
{
    echo "<h4>Results</h4>";
    echo "<p>Sorry, your search: &quot;" . $trimmed . "&quot; returned zero results</p>";
    echo "<p><a href=\"http://www.google.com/search?q="
    . $trimmed . "\" target=\"_blank\" title=\"Look up
    " . $trimmed . " on Google\">Click here</a> to try the search on google</p>";
}

if (empty($s)) 
{
    $s = 0;
}

$query .= " limit $s,$limit";
$result = mysql_query($query) or die("Couldn't execute query");
echo "<p>You searched for: &quot;" . $var . "&quot;</p>";
echo "Results";
$count = 1 + $s;

while ($row= mysql_fetch_array($result)) 
{
    $title = $row["NAME"];
    echo "$count.-&nbsp;$title" ;
    $count++ ;
}

$currPage = (($s/$limit) + 1);
echo "<br />";

if ($s >= 1) 
{ 
    // bypass PREV link if s is 0
    $prevs = ($s - $limit);
    print "&nbsp;<a href=\"$PHP_SELF?s=$prevs&q=$var\">&lt;&lt;
    Prev 10</a>&nbsp;";
}

$pages = intval($numrows/$limit);

if ($numrows % $limit) 
{
    $pages++;
}  

if (!((($s+$limit)/$limit) == $pages) && $pages != 1) 
{
    $news = $s + $limit;
    print "<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
} 

$a = $s + ($limit);

if ($a > $numrows) 
{ 
    $a = $numrows; 
}

$b = $s + 1;
echo "<p>Showing results $b to $a of $numrows</p>";

Answer 1

In your code, $s gets reset every time the page is reloaded or the next page link is clicked. You should have $s = $_REQUEST['s'] at the beginning of your code.