[RESOLVED] Image Garble

thompenn

The following code seems to work, but I can't get it to show in the browser. It keeps giving me raw data.

<?php

$filename = 'easy.jpg';
$percent = 0.5;
// Content type
header('Content-type: image/jpeg');
// Get new dimensions
list($width, $height) = getimagesize($filename);
$new_width = $width $percent;
$new_height = $height $percent;
// Resample
$image_p = imagecreatetruecolor($new_width, $new_height);
$image = imagecreatefromjpeg($filename);
imagecopyresampled($image_p, $image, 0, 0, 0, 0, $new_width, $new_height, $width, $height);
// Output
imagejpeg($image_p, null, 100);

bpat1434

To skip the filename argument in order to provide a quality argument just use an empty string ('').

imagejpeg($image_p, '', 100);

thompenn

Thank you very much for the replay... It was the code... I was trying to connect with an object and then called to this script but there was a problem between pulling the information dynamically and posting it into this file. Thank you for your assistance.

msandi

I'm a little rusty on my PHP. Can you please tell me what the @ is preceding a variable name? Thanks

When I wrap all the code in php and leave out the html, I only get the first rotated image, I don't understand why I don't get the last one. I even tried to insert a print line with a line break, but that did not work either.

Here's my code:

When all in php:

<?php

// Content type
header('Content-type: image/jpeg');

$pic1 = 'guitar-1_t.jpg';
$pic2 = 'guitar-7_t.jpg';

$path = 'guitar';

$img_res = imagecreatefromjpeg('images/'.$path.'/'.$pic1 . '');

$img_res2 = imagecreatefromjpeg('images/'.$path.'/'.$pic2 . '');

$pic1_90 = imagerotate($img_res,90,0);

$pic2_180 = imagerotate($img_res2,180,0);

imagejpeg($pic2_180, '', 100);
imagejpeg($pic1_90, '', 100);

Then if I add html code and try to get the image(s), I get nothing and garble or I get an error about headers already being sent.

<?php

$pic1 = 'guitar-1_t.jpg';
$pic2 = 'guitar-7_t.jpg';

$path = 'guitar';

// Content type
header('Content-type: image/jpeg');

$img_res = imagecreatefromjpeg('images/'.$path.'/'.$pic1 . '');

$img_res2 = imagecreatefromjpeg('images/'.$path.'/'.$pic2 . '');

$pic1_90 = imagerotate($img_res,90,0);

$pic2_180 = imagerotate($img_res2,180,0);

<?php imagejpeg($pic2_180, '', 100);?>
<br>

<?php imagejpeg($pic1_90, '', 100);?>
</body>
</html>

I've been trying to figure this out for a couple of days now, trying everything I can think of.

What I want is a random order of images which are located on the server and I have identified in a database that I am querying to create the randomness.

I am including other pages, one to login to the server and another for the queries.

I also tried to use this in my html to avoid having to use the header('Content-type: image/jpeg');

but that didn't work at all

.😕

bpat1434

The "@" symbol preceding a variable means to suppress any error messages which may show up because of that variable not existing. If you remove it, you will see any errors associated with it.

Usually it's considered bad practice to use the "@" symbol because it means you're hiding something instead of actually solving the problem; however, there are some instances where it's just easier to use it.

msandi

Thank you!