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corillo181

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under wher ein php info would i find that?

cuz i tthink is if i had gd? and my server do

i try this but it does not work.
[code]
<?php
function imageResize($width, $height, $target) {
if ($width > $height) {
$percentage = ($target / $width);
} else {
$percentage = ($target / $height);
}
$width = round($width * $percentage);
$height = round($height * $percentage);
return "width=\"$width\" height=\"$height\"";
}

include 'includes/db.php';
$query=mysql_query("SELECT path FROM tra_cotm_photo")or die("error: ".mysql_error());
$path=mysql_fetch_array($query);
$test = getimagesize("../cotm/457565197.jpg");
?>
<img src="../cotm/457565197.jpg" <?php imageResize($test[0],$test[1], 110); ?> />
[/code]
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If you have the GD library installed, it's clearly visible in phpinfo (it has its own section of information).

If you have GD installed and your code "doesn't work", there something not right wih your code. What is (not) happening with that script that makes you think there's a problem?
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tha tpage does nto resize with the function made.. so i think the imagesiz eit's not doign the work

and yes the gd is on..

plus i try using
[code]<img src="image/123.jpg" atl="<?php getimagesize("image/123.jpg");?>">
[/code]
to see if it would show me the image size but it doesn't..
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Sorry, try this:
[code]$size = getimagesize("image/123.jpg");
echo "<img src='image/123.jpg' ". $size[3]. ">";[/code]
If GD is working and if image/123.jpg exists (it's your example, so I assume it would), then $size[3] will display the width and height of the image - exactly what it says in the php manual.
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this worked..i got the hang of it.. thanx
[code]

$filename = 'images/123.jpg';
$size = getimagesize("images/123.jpg");
$width=$size[0];
$height=$size[1];
if($width >100 || $height >100){
$n_w=$width / 2;
$n_h=$height / 2;
echo '<img src="images/123.jpg" width="'.$n_w.'" height="'.$n_h.'" />';
}
?>


[/code]
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