displaying table with images with PHP

I need to generate a table with php, that will display the images - names stored on database. It has to display 3 images in a row. The images are added 开发者_如何学JAVAto the database all the time, so I need that to be automatically generated, instead of hard coding the tables. I am not sure how do I do that? Please help!

You need to cycle the result recordset and print out the new row every 3rd element.

For example:

<table>
<tr>

<?php $i=0; foreach ($images as $image): ?>

  <td><?php echo $image['name'] ?> <img src="<?php echo $image['path'] ?>" /></td>
  <?php if(++$i%3==0): ?>
    </tr><tr>
  <?php endif ?>

<?php endforeach ?>

</tr>
</table>

suppose u get the all images name from database in an array

$img_array = array(
                   1=>'f.jpg',
                   2=>'s.jpg',
                   3=>'t.jpg',
                   4=>'f.jpg',
                   5=>'e.jpg'
            );

// now create dynamic table via php
<table border="1" cellpadding="2" cellspacing="2" width="100%">
<tr>
<?php 
    $i=0;
    foreach($img_array as $k){
    if($i%3==0) { ?> </tr><tr> <?php } ?>
     <td><img src="<?php echo $k?>" border="0"></td>
<?php $i++; } ?>
 </tr>
</table>

Note: please write full path of image in src before <?php echo $k?>

Iterate the image records, using modulo 3 to change to the next table row.

Something like:

echo '<table><tr>';
foreach ($images) {
    echo '<td>$image</td>';
    if ($i % 3 == 0) {
       echo '</tr><tr>';
    }
}
echo '</tr></table>';

A simple table like that would be like

<table>
<tr><td>1</td><td>2</td><td>3</td></tr>
<tr><td>1</td><td>2</td><td>3</td></tr>
</table>

To generate this automatically you need to store where you are in the table, first col, 2nd col or 3th col.

<?php
$pos = 1;

print "<table>"
for ($i=0; $i<=10;$i++)
{
  if ($pos==1)
  {
     print "<tr><td>1</td>";
     $pos=2;
   }
  else if ($pos==2)
  {
     print "<td>2</td>";
     $pos=3;
   }
  else if ($pos==3)
  {
     print "<td>3</td></tr>";
     $pos=1;
   }
 }
 if ($pos==2 || $pos==3)
    print "</tr>";

 print "</table>"

Keep in mind that if you use the options with $i%3 from the other comments, that your table will start end/or finish with an empty row. This would need additional checks. The $i%3 will give the remainder of the division of $i and 3. So when $i/3 == 0, means it is true on every third row.