Question

Hey there i have this snippet:

$profile_pic = imagecreatefromjpeg("upload_Main/".$user[1].".jpg");

This function works (upload_Main: folder:777 ; files:644)

But this snippet:

$profile_pic = imagecreatefromjpeg("upload_Sub/".$user[1].".jpg");

Doesnt work (upload_Sub: folder:777 ; files:644).

They have exactly the same files and the same right´s, so why does the second one not work? Any idea´s? thanks and greetings!

Was it helpful?

Solution

imagecreatefromjpeg():

Returns an image resource identifier on success, FALSE on errors.

I had it return False on an image that was manually re-named to a JPG, when it was actually a PNG. I got an error saying:

'not a valid JPEG file' 

You should look to see what error you get when performing that function, That way you'll know exactly what's goind bad.

One way to check for the error is creating a PHP file next to the image with:

<?php
  echo imagecreatefromjpeg("path/to/image");
?>

run it in using 'php filename.php'
and looking in the console for the result

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