PHP - Pagination

cyates · 15-11-07

Damn im absolutely shattered ...
less than 4 hours sleep in last 2 days

can anyone see what ive messed up here because my body and mind seems to have given up ..

PHP Code:


			
$limit = 2;
mysql_select_db($dbn, $dbc);
$cquery = "SELECT count(*) from datafeed";
$cdoresult = mysql_query($cquery, $dbc);
$cresult = mysql_fetch_assoc($cdoresult);
$totalrows = mysql_num_rows($cresult);
if(empty($page)){    // Checks if the $page variable is empty (not set) 
        $page = 1;      // If it is empty, we're on page 1 
    } 
    $limitvalue = $page * $limit - ($limit);  
    // Ex: (2 * 25) - 25 = 25 <- data starts at 25 
     
$query  = "SELECT * from datafeed WHERE type='Divan Bed' LIMIT $limitvalue, $limit";
$doresult = mysql_query($query, $dbc);
$result = mysql_fetch_assoc(doresult);
?>

Recordset Nav

PHP Code:


			
 <?php if($page != 1){  
        $pageprev = $page--;        
        echo("<a href=\"divan-beds.php?page=$pageprev\">PREV".$limit."</a> ");   
    }else 
        echo("PREV".$limit." ");  
        // If we're on page 1, PREV is not a link 
   if(($totalrows - ($limit * $page)) > 0){ 
        
        $pagenext   = $page++; 
            
        echo("<a href=\"divan-beds.php?page=$pagenext\">NEXT".$limit."</a>"); 
      
    }else{ 
        echo("NEXT".$limit);  

    } 

?>

any help appreciated.. really need my sleep

cyates · 15-11-07

forgot to mention ..

its adding teh correct num of products .. just a problem with prev & next

cyates · 15-11-07

also getting
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in C:\wamp\www\kingsizebeds\divan-beds.php on line 8

Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\wamp\www\kingsizebeds\divan-beds.php on line 18

PHP Code:


			
$limit = 6;

mysql_select_db($dbn, $dbc);

$cquery = "SELECT count(*) from datafeed";

$cdoresult = mysql_query($cquery, $dbc);

$cresult = mysql_fetch_assoc($cdoresult);

LINE 8 ->   $totalrows = mysql_num_rows($cresult);

if(empty($page)){    // Checks if the $page variable is empty (not set) 

        $page = 1;      // If it is empty, we're on page 1 

    } 



    $limitvalue = $page * $limit - ($limit);  

    // Ex: (2 * 25) - 25 = 25 <- data starts at 25 

     

$query  = "SELECT * from datafeed WHERE type='Divan Bed' LIMIT $limitvalue, $limit";

$doresult = mysql_query($query, $dbc);

LINE 18 -> $result = mysql_fetch_assoc(doresult);

MoonBug · 15-11-07

Quick scan says to me 'SELECT count' query wont return rows but the number of records in your table, or if it does perhaps this will be 1?

Also Line 18 is missing a '$' -> mysql_fetch_assoc($doresult)

cyates · 16-11-07

fixed the $ error but still nothing..

returning products but if i set limit to 2
it will only display 2 products and teh URLs for prev & next are inactive

any one any ideas ?

cyates · 16-11-07

and also if i add page=2 or page=3 etc to the url in teh address bar it only returns the same results as it does without $page being set

cyates · 16-11-07

also so people know where i got this code.. it was from here
PHP Help: Pagination: Easy as PREV 1 2 3 NEXT

ive ammeded it to match my db but everything else is teh same so dont see teh problem

drivetowin · 16-11-07

try adding

Code:

$page=$_GET['page'];

to the start of your code and then putting page=2, page=3 etc after your url.

cyates · 16-11-07

did nothing ...
still same as before

drivetowin · 16-11-07

can you put in a couple of lines after line 8

Code:

echo ("Totalrows = " . $totalrows);
echo ("Page = " . $page);

and see what they output.