[Resolved] Image url in an HTML table

[vaavi8r]

I have an HTML table that displays data from my database.

One of the columns in the HTML table is called 'Image', where I have the url of an image appearing.

Instead of having the url show, I want to just have an icon show up in that block that will link to the image location. But if there is no data for that column, I want the HTML table to display an empty block.

Here is the section of the code for the HTML table:

Line in question is the last line in the table.

<table border="2" cellspacing="2" cellpadding="2">
<tr>
<th><font face="Arial, Helvetica, sans-serif">ID</font></th>
<th><font face="Arial, Helvetica, sans-serif">MACID</font></th>
<th><font face="Arial, Helvetica, sans-serif">Name</font></th>
<th><font face="Arial, Helvetica, sans-serif">Date</font></th>
<th><font face="Arial, Helvetica, sans-serif">Type</font></th>
<th><font face="Arial, Helvetica, sans-serif">Departure</font></th>
<th><font face="Arial, Helvetica, sans-serif">Destination</font></th>
<th><font face="Arial, Helvetica, sans-serif">Time</font></th>
<th><font face="Arial, Helvetica, sans-serif">Charter #</font></th>
<th><font face="Arial, Helvetica, sans-serif">Charter Pay</font></th>
<th><font face="Arial, Helvetica, sans-serif">Comment</font></th>
<th><font face="Arial, Helvetica, sans-serif">Image</font></th>

</tr>

<?
$i=0;
while ($i < $num) {
$id=mysql_result($result,$i,"id");
$macid=mysql_result($result,$i,"macid");
$date=mysql_result($result,$i,"date");
$name=mysql_result($result,$i,"name");
$type=mysql_result($result,$i,"type");
$departure=mysql_result($result,$i,"departure");
$destination=mysql_result($result,$i,"destination");
$hours=mysql_result($result,$i,"hours");
$charterno=mysql_result($result,$i,"charterno");
$charterdoll=mysql_result($result,$i,"charterdoll");
$comment=mysql_result($result,$i,"comment");
$image=mysql_result($result,$i,"image");
$datePosted = "$date"; 
$formatted = date("M-d-Y", strtotime($datePosted));
$departure = strtoupper($departure);
$destination = strtoupper($destination);
?>

<tr>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$id"; ?></a></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$macid"; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$name"; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$date) { echo "&nbsp;"; } else { echo $formatted; } ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$type) { echo "&nbsp;"; } else { echo $type; } ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$departure) { echo "&nbsp;"; } else { echo $departure; } ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$destination) { echo "&nbsp;"; } else { echo $destination; } ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? echo "$hours"; ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$charterno) { echo "&nbsp;"; } else { echo $charterno; }?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$charterdoll) { echo "&nbsp;"; } else { echo $charterdoll; } ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$comment) { echo "&nbsp;"; } else { echo $comment; } ?></font></td>
<td><font face="Arial, Helvetica, sans-serif"><? if (!$image) { echo "&nbsp;"; } else ?><a href="<?{ echo $image; } ?>"><? echo "$image";?></a></font></td>
</tr>

<?
++$i;
} 
echo "</table>";

?>

Any ideas how to go about this?

[BuzzLY]

Ouch, my eyes. You seriously need to learn CSS, associative arrays, and creative looping. You could reduce that code significantly.

As for your URL...

replace this:
<? echo "$image";?>

with this:
<?php echo "<img src=\"$image\">";?>

(get in the habit of using <?php instead of short tags like <?. It's not that much extra work, and will make your code more portable and robust.)

[vaavi8r]

OK, maybe I didnt explain it well.

What I want is for the column to appear empty if there is no image to display,

and to display an icon (not the actual image) that will link you to the url to display the image itself.

Thanks for the help!

[bigswifty]

add a new var...

$holder = "<a href=\"$image\"><img src=\"icon.gif\"></a>";

<?php echo "$holder"; ?></td>

in your table after your else statement

icon.gif is your icon, so what you have is a new var that creates a link to your image via a linked generic image (icon)

the parsed result should be

<td><a href=\"path to your image.jpg\"><img src=\"icon.gif\"></a></td>

with the code cleaned up, you get this in your script....

<td><font face="Arial, Helvetica, sans-serif">
<? if (!$image) {
echo "&nbsp;"
} else {
echo "$holder"
} ?> </font></td>

this may have some syntax errors (missing ";") in it since Im not used to including snipplets in html files (I always use pure php files), you realy might want to consider making the page php rather then including multiple php statements in a html file.

for that added effect you could place a "no image available" image rather then "blank" for the no image condition

2nd var
$noimage = "http://www.yourdomain/noimage.jpg";

(or the proper path to the noimage.jpg file)

then you do this..

<td><font face="Arial, Helvetica, sans-serif">
<? if (!$image) {
echo "$noimage"
} else {
echo "$holder"
} ?> </font></td>

I had lots of fun with this on my image-laden search pages.

[vaavi8r]

OK, this works.......thanks to everyone for their input:

if ($image == ""){
$imagecell = "&nbsp";
} else {
$imagecell = "<a href='$image'><img src='http://www.margaritaair.com/flightlog/plane.gif' width='52' height='32' border='0' alt='Click to view'></a>";
}


THEN: 

<td><font face="Arial, Helvetica, sans-serif"><? echo $imagecell; ?></font></td>

Thanks to all who made suggestions....as always, you guys make this much easier!!