Tema: Re: PHP paging problema...
Autorius: gopas
Data: 2010-04-07 07:20:07
tai echo $query, ir ziurek kas blogai.

"exoticdevil" <born4devil@gmail.com> wrote in message news:hpg6ft$be7$1@trimpas.omnitel.net...
> Turiu pasidares scriptuka paging, viskas ok veikia,naudoju kituose vietose 
> puslapyje,  bet dabar is db lenteles prireike paimti ne visus duomenis, o 
> tik kai kuriuos, ir kazko neveikia,tiksliau pusiau veikia, pirma puslapi 
> sugeneruoja, o paspaudus ant sekancio nieko nerodo, tarsi isvis paging'o 
> nebutu... nesuprantu kur klaida iveliau, su queriu kazkas ne taip 
> manau...visa vakara prasedejau, nebelaiko nervai jau, gal kas padesit..
> (sorry del netvarkos)
> 
> <?php
> 
> $cat = $_GET['cat'];
> 
> 
> include ('admin/sqlcon.php');
> 
> mysql_select_db("shop");
> 
> $rowsPerPage = 5;
> $pageNum = 1;
> 
> if(isset($_GET['page']))
> {
>    $pageNum = $_GET['page'];
> }
> 
> $offset = ($pageNum - 1) * $rowsPerPage;
> $query = " SELECT * FROM items WHERE itemsubcat = '$cat' LIMIT $offset, 
> $rowsPerPage";
> $result = mysql_query($query) or die('Error, query failed');
> 
> while($row = mysql_fetch_array($result))
> {
> 
>    echo $row["itemcode"]. "<br>";
> 
> 
> }
> 
> $query   = "SELECT COUNT(itemsubcat) AS numrows FROM items WHERE itemsubcat 
> = '$cat'";
> $result  = mysql_query($query) or die('Error, query failed2');
> $row     = mysql_fetch_array($result, MYSQL_ASSOC);
> $numrows = $row['numrows'];
> $maxPage = ceil($numrows/$rowsPerPage);
> $self = $_SERVER['PHP_SELF'];
> $nav  = '';
> 
> for($page = 1; $page <= $maxPage; $page++)
> {
>   if ($page == $pageNum)
>   {
>       $nav .= "[$page]";
>   }
>   else
>   {
>       $nav .= " <a href=\"$self?page=$page\">$page</a> ";
>   }
> }
> 
> if ($pageNum > 1)
> {
>    $page  = $pageNum - 1;
> }
> 
> if ($pageNum < $maxPage)
> {
>    $page = $pageNum + 1;
> }
> echo $nav ;
> 
> ?> 
>