NYCPHP Meetup

[nycphp-talk] image display

Michael Southwell michael.southwell at nyphp.com
Sun Aug 26 20:15:27 EDT 2007


At 06:15 PM 8/26/2007, you wrote:
>could someone please look at this code and please tell me what is wrong.
>it is suppose to display a 4 column table with an indiffenate number of rows.
>Instead it displays one image in row , then 3 images in row 2.  Then 
>one image in row 3 and e images in row 4, and so on.  thanks.

If I understand correctly what you are trying to do, the simple way 
to do this is to end a row and start a new row in your table every 
time you have finished displaying four rows from the database, like 
this, using your $num variable:
<table>
<tr>
// loop through result
// display stuff
if ( $num % 4 == 0 ) echo '</tr><tr>';
}
</tr>
</table>

>Here is the code:
>
>require_once('connect/mysql_connect.php');
>
>$query="SELECT item_id, modelnum, name, upload, size, description, 
>price, gender FROM clothing ORDER BY item_id ASC";
>   $result=@mysql_query($query);
>     $num=@mysql_num_rows($result);
>$col = 0;
>echo '<table border="0" width="90%" cellspacing="3" cellpadding="3" 
>align="center">';
>
>if($num>0) {
>while($row=mysql_fetch_array($result)) {
>if($col == 0)
>echo '<tr>';
>
>
>echo '<td align="center" width="20%"><font 
>color="teal">cAclOm'.$row['modelnum'].'</font>
><span style="padding-left:5%">'.$row['name'].'</span><br /><br />
><img src="/coles/admin/new_images/'.$row['upload'].'" width="120" 
>height="140" /><br />
>'.$row['description'].'<span style="padding-left:5%">'.$row['price'].'</span>
></td>';
>
>if($col == 0)
>
>echo '</tr>';
>
>$col++;
>$col = ($col == 4) ? 0 : $col;
>}
>}
>echo '</table>';
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Michael Southwell, Vice President for Education
New York PHP
http://www.nyphp.com/training - In-depth PHP Training Courses





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