Pagination with PHP and MySQL?

Question

I have 2 files. There are showReport.php and paging.php. The codes are fine, no errors, but it showing incorrect results. Can you help me for my final exams? Because I've tried to change it but still doesn't work. Please ignore for the variable name or table name, because I'm using Indonesian.

showReport.php

<?php 
            $query = mysql_query("select * from barang");
            $item_per_page = 2; // tambahan
            paging( $query,$item_per_page );
            $limitt = ($page - 1) * $item_per_page;

            $query_ = mysql_query("SELECT * FROM barang ORDER BY id_transaksi DESC LIMIT $limitt, $item_per_page"); // tmbh
            $sql_ = mysql_query($query_);       

            $i = 1;

            while ($data = mysql_fetch_array($query)) {
            ?>
            <tr class="<?php if ($i % 2 == 0) { echo "odd"; } else { echo "even"; } ?>">
                <td><?php echo $i; ?></td>
                <td>
                    <?php 
                    echo $data['tanggal']; 

                    //previlege admin
                    if ($_SESSION['role'] == 'operator') {
                    ?>
                        <div class="row-actions">
                        <a href="edit_barang.php?uid=<?php echo $data['id_transaksi'];?>">Update</a>
                         | <a href="delete_barang.php?uid=<?php echo $data['id_transaksi'];?>" onClick="return confirm('Delete this report?')" class="delete">Delete</a>




                    </div>
                    <?php } ?>
                </td>



                <td><?php echo $data['barang_in']; ?></td>
                <td><?php echo $data['bijih_out']; ?></td>
                <td><?php echo $data['htm_out']; ?></td>
                <td><?php echo $data['pth_out']; ?></td>
                <td><?php echo $data['bijih']; ?></td>
                <td><?php echo $data['kantong_htm']; ?></td>
                <td><?php echo $data['kantong_pth']; ?></td>
                <td><?php echo $data['note']; ?></td>
            </tr>



            <?php 
                $i++;
            } 
            ?>

paging.php

<?php
function paging($query,$item_per_page){
   $page          =  isset($_GET['page']) ? $_GET['page'] : 1 ; 
   if( ( $page < 1) && (empty( $page ) ) ){
      $page=1;
   }
   $sql         = mysql_query( $query_);
   $amount_data   = mysql_num_rows($query);
   $amount_page    = ceil( $amount_data/$item_per_page );
   if( $page>$amount_page ){
      $page=$amount_page;
   }
   $lanjut  = $page + 1;
   $sebelum = $page - 1;
   ?>
<style type="text/css">
.tengah {
    text-align: center;
}
</style>

<?php echo $page;?> dari <?php echo $amount_page;?><br />
   <a href="?page=1"></a><a href="?page=<?php echo $sebelum; ?>">sebelumnya</a>;
   ||
   <a href="?page=<?php echo $lanjut;?>">selanjutnya</a><a href="?page=<?php echo $jumlah_hal;?>"></a>
   Ke Halaman: <form action="" method="get"><input type="text" name="page"><input type="submit" value="Go"></form>
   <?php
}
?>

The number of pages are right, but it still showed all of data in one page. For example I have 4 data, I wanna show just 2 data for each page. So number of pages are 2 ( right ), but why my data won't to show 2 data for each page? Thank you in advance

Hope this can solve your problem.. stackoverflow.com/a/17159388/1358004 — Viral Shah
– Viral Shah, Commented Mar 12, 2014 at 13:03
Your paging query doesn't appear to return anything, nor does it use global variables or pass parameters by reference to update the values of passed parameters . Not sure what the value of $limitt will be when it gets to the query. — Kickstart
– Kickstart, Commented Mar 12, 2014 at 13:19
Visit [pagination using php and mysql][1]. <br> I answered here. [1]: stackoverflow.com/questions/12315920/… — user2552725
– user2552725, Commented Apr 18, 2014 at 4:45
visit: [pagination][1] here. I already answered for this question. [1]: stackoverflow.com/questions/12315920/… — user2552725
– user2552725, Commented Apr 18, 2014 at 4:46

Wesley De Keirsmaeker · Accepted Answer · 2014-03-12 13:09:57Z

0

Please put:

echo $query;

after:

$sql_ = mysql_query($query_);

in showReport.php.

Check if the query is created how you expect it to be. Also in showReport.php, I see you are using the variable '$page' while i don't see it was given a value first. Are both pages imported in another one?

answered Mar 12, 2014 at 13:09

Wesley De Keirsmaeker

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4 Comments

ulaaaan Over a year ago

yaaap, in showReport.php there is include('paging.php); I've tried, but the data still showing all of it hmm....

Wesley De Keirsmaeker Over a year ago

The echo command should show the query you are using. You can then try the query directly into your MySQL database to verify if it gives the correct data there.

ulaaaan Over a year ago

but how to make the data just showing 2 data per pages? the query is correct, but it confusing why the data won't show just 2 data, beside the item for page just set 2

Wesley De Keirsmaeker Over a year ago

In showReport.php your while will go trough all the records found with that query. If the query returns 4 records, 4 records will be shown while you only want to show 2. Check if the query only return 2 records instead of 4.

ulaaaan · Accepted Answer · 2014-03-12 14:03:26Z

For making limit the amount of data displayed

<?php 
$sqlCount = "select count(id_transaksi) from barang";  
$rsCount = mysql_fetch_array(mysql_query($sqlCount));  
$banyakData = $rsCount[0];  
$page = isset($_GET['page']) ? $_GET['page'] : 1;  
$limit = 2;  
$start_from = $limit * ($page - 1);  
$sql_limit = "select * from barang ORDER BY id_transaksi limit $start_from, $limit";  
$result = mysql_query($sql_limit);  ?>

For making number of pages / pagination

<?php
$numberofPages = ceil($amountData / $limit);  
echo 'Pages: ';  
for($i = 1; $i <= $numberofPages; $i++){  
 if($page != $i){  
 echo '[<a href="showReport.php?page='.$i.'">'.$i.'</a>] ';  
 }else{  
 echo "[$i] ";  
 }  
}  
?>

Based on my question, paging.php isn't any longer used. Hope can help anyone who gets problem like me before, cheers! ;)

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