php gd script not outputting picture correctly

Question

This Code saves the image like it's suppose to, but instead of displaying it as a picture, what is displayed is a line of text inside the picture. help?

<?php

    imagecopymerge($image, $watermark, $dest_x, $dest_y, 0, 0, $watermark_width, $watermark_height, $opacity);


    // print image to screen
    header("content-type: image/jpeg");
    imagejpeg($image, "modified-images/".$codigo2."_modified_picture_status_".$status.".jpg");
    imagedestroy($image);  
    imagedestroy($watermark);

    ?>

Manuel Strausz · Accepted Answer · 2011-01-26 17:36:53Z

1

If you are using the second parameter of imagejpeg, the image will not be outputted to the browser but ONLY saved to the file. Try omitting the second parameter if you don't need to save it as a file, and it should output directly to the browser.

If you want to do both, try a print(file_get_contents($imagepath)) after your current block of code. $imagepath should obviously contain the path that you wrote the image to.

answered Jan 26, 2011 at 17:36

Manuel Strausz

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2 Comments

Tom Granot Over a year ago

You just solved a problem that took half of my day. THANKYOU!!

dqhendricks Over a year ago

why open and read the file you just created when you already have a resource for it in your code?

dqhendricks · Accepted Answer · 2011-01-26 17:53:25Z

1

make two lines:

// save image
imagejpeg($image, "modified-images/".$codigo2."_modified_picture_status_".$status.".jpg");
// output image
imagejpeg($image);

answered Jan 26, 2011 at 17:53

dqhendricks

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