Create a link using a database field

[aphros]

Hi All,

I need to create a link to a picture based on a database field, i.e. if the 'Lot' field = 1, then the picture to be displayed = 1.jpg.

The closest I've been able to get is:
<img src="images/pixsmall/".<?php echo $row_rsCat0503['Lot']?>.".jpg" />

which doesn't work.

Any help greatly appreciated.

Regards,

Brendan

[ir4z0r]

<?PHP
  // query and stuff

echo "<img src=\"images/pixsmall/". $row_rsCat0503['Lot'] . ".jpg\" />";

?>

... I guess

[Moonglobe]

<img src="/images/pixsmall/<?=$rows_rsCat0503['lot']?>.jpg" />

edit: OR, if tehre are only two possibilities:

<?php if ($rows_rsCat0503['lot'] == 1) {
        $im=1;
} else {
        $im=2;
} ?>
<img src="/images/pixsmall/<?=$im;?>.jpg" />

[aphros]

Thanks for the responses, from which I was able to get the pictures working.

Now a related problem - if there is no picture associated with a particular lot number, I get a dead link icon where the picture would be if there was one.

Is there an 'if picture exists then display else nothing' syntax I can use to avoid this?

Thanks,

Brendan

[ir4z0r]

<?PHP
  // query and stuff

$image = "images/pixsmall/". $row_rsCat0503['Lot'] . ".jpg";
if (file_exists($image)) echo '<img src="' . $image . '" />';

?>

probably this way

[aphros]

Thank you for your valuable help. Site now working as expected.

Thanks & Regards,

Brendan