display 2 or 3 images per row using php mysql

Question

hi i'm disaplyed images from mysql db table but it displays on by one means one row has one image. but i need 3 or 4 image per row. my coding is below. please give some idea.

<?php 
    include_once("config.php");
    $result=mysql_query("SELECT * FROM merchant");
    while($res=mysql_fetch_array($result))
    {
?>
<?php echo $res['description'];?></p>
    <img src="<?php echo $res['image'];?>" width="80" height="80"/>
<?php } ?>

Please make your code readable so people will be interested in looking at it. — Praveen Kumar Purushothaman
– Praveen Kumar Purushothaman, Commented Sep 5, 2012 at 11:22
@Farid He asked 5 questions already. Fluffeh, you are right, thanks for the link! — Dennis Hackethal
– Dennis Hackethal, Commented Sep 5, 2012 at 11:29

Muhammad Ummar · Accepted Answer · 2012-09-05 11:24:07Z

3

Do it in table like this, You might need to fix it a little bit, but it way how it will work

<table>

<?php 
        include_once("config.php");
        $result=mysql_query("SELECT * FROM merchant");
        $count = 0;
        while($res=mysql_fetch_array($result))
        {
            if($count==3) //three images per row
            {
               print "</tr>";
               $count = 0;
            }
            if($count==0)
               print "<tr>";
            print "<td>";
            ?>
                    <?php echo $res['description'];?></p>



                    <img src="<?php echo $res['image'];?>" width="80" height="80"/>



                <?php
            $count++;
            print "</td>";
        }
        if($count>0)
           print "</tr>";
        ?>

</table>

edited Sep 5, 2012 at 11:24

answered Sep 5, 2012 at 11:17

Muhammad Ummar

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Comments

Praveen Kumar Purushothaman · Accepted Answer · 2012-09-05 11:26:43Z

0

use a <table> to display.

<?php 
    include_once("config.php");
    $result=mysql_query("SELECT * FROM merchant");
    $count = 0;
    echo '<table>';
    while($res = mysql_fetch_array($result))
    {
        if($count % 2 == 0) echo '<tr>';
        ?>
    <td>
        <p><?php echo $res['description'];?></p>
        <img src="<?php echo $res['image']; ?>" width="80" height="80"/>
    </td>
<?php
    if($count % 2 == 0) echo '</tr>';
} ?>

answered Sep 5, 2012 at 11:26

Praveen Kumar Purushothaman

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