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Displaying Images in a Table

Posted: Wed Feb 15, 2006 6:56 am
by tambourine
Hello, im new to php and i have failed to work out how to display images in a table that consist of 5 rows by 5 columns.

I think a IF loop is needed, but i couldn't seem to any to work :cry:

The variable im trying to displays is

Code: Select all

<?php echo $row_blobby['previewpath']; ?>
Im basically trying to get my images to display in a 5 * 5 table.

Hope, someone can help or guide me in the correct direction. Thanks! :D

Posted: Wed Feb 15, 2006 7:34 am
by AndrewBacca
this is untested but give it ago

Code: Select all

<?php
print "<table>";
for ($i = 0; $i <= 5; $i++) {
     if ($i % 5 == 0) {
         print "<tr>"; 
    }

     print "<td>" . $row_blobby['previewpath'] . "</td>";

     if ($i % 5 == 5) {
         print "</tr>";
    }
}
print "</table>";
?>
hope that helps/works

Posted: Wed Feb 15, 2006 8:21 am
by tambourine
Ok, this code works kinda....

Code: Select all

<?php
print "<table>"; 
for ($i = 0; $i <= 5; $i++) { 
     if ($i % 5 == 0) { 
         print "<tr>"; 
    } 

     print '<td><img src="http://'.$row_blobby['previewpath'].'" /></td>'; 

     if ($i % 5 == 5) { 
         print "</tr>"; 
    } 
} 
print "</table>"; 
?>
....it displays the same image 5 times accross and 1 time down

Exmaple: http://www.surgeryofsound.co.uk/mobile- ... all-uk.php

Any more suggestions?

Thank you!

Posted: Wed Feb 15, 2006 8:29 am
by AndrewBacca
I think this should work

Code: Select all

<?php
print "<table>"; 
for ($i = 0; $i <= 4; $i++) { 
     if ($i % 5 == 0) { 
         print "<tr>"; 
    } 

     print '<td><img src="http://'.$row_blobby['previewpath'].'" /></td>'; 

     if ($i % 5 == 5) { 
         print "</tr>"; 
    } 
} 
print "</table>"; 
?>