PHP output into a table - help with layout

SteveB_NI · 9 Feb 2006 at 13:32

I have a PHP code which pulls out the pc status field from a database, and displays a Green square if its working, Amber if its awaiting attention and Red if its faulty.

The code at the minute displays 1 square below another - in separate table rows it appears. Is there any way of changing the layout so it presents the images in rows of 5 columns.

i.e.

R R R R R
G G R R G
A A A A A

etc. (R = Red, A = Amber, G = Green images).

Thanks.

Code:

<table>
<tr><th>Colour</th></tr>

<?php
$workstations = @mysql_query($select . $from . $where);
if (!$workstations) {
  echo '</table>';
  exit('<p>Error retrieving workstation data from database!<br />'.
      'Error: ' . mysql_error() . '</p>');
}

while ($computer = mysql_fetch_array($workstations)) {
  echo "<tr valign='top'>\n";
  $status = htmlspecialchars($computer['status']);
  
  if ($status=="WORKING") {
  echo "<td align=center><img src='green.jpg'></td>";
  }
  if ($status=="ATTENTION") {
  echo "<td align=center><img src='amber.jpg'></td>";
  }
  if ($status=="FAULTY") {
  echo "<td align=center><img src='red.jpg'></td>";
  }
  
}
?>

</table>

SteveB_NI · 9 Feb 2006 at 14:01

Beansprout said:
Also, you can reduce those 3 ifs to none by naming the images after the states, or naming the states after the images

The code supplied by Dj Jestar (thanks!) is just right for displaying the images in rows, and columns (just as I wanted

)

How do I go about implementing your idea BeanSprout?

SteveB_NI · 9 Feb 2006 at 14:20

Thanks very much - just what I needed

Good comments about the code too!

SteveB_NI · 10 Feb 2006 at 17:37

Quick followup to my question - I'm having difficulty displaying a variables data within the table cell.

I've changed the image ($status.jpg) to the table cell background, and wish to display the data stored by $stationid over it.

The code at the mo is:

Code:

	if(in_array($status,$statuses)){
		$str .= '<td width="70" height="70" align="center" background="' . $status. '.jpg", valign="center"> $stationid </td>';

Obviously this just outputs $stationid in the table cell. How do I get it to ACTUALLY output $stationid ?

Cheers