Home/ Questions/Q 1060505
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Asked: 2026-05-16T18:20:13+00:00

I’m using this piece of php code to create and rotate an image. And

  • 0

I’m using this piece of php code to create and rotate an image. And it works perfectly when i just call the img.php it lying in.

But if i try to include the img.php in anything else, or just include the snippet in another page (on same server) it just shows me a lot of questionmarks instead of an image.

My feeling is that the problem is related to the headers sent, but honestly i have no idea.

Any suggestions how to make this useful so it can be included in another page and still work?

<?php 
   header("Content-type: image/jpeg"); 
   // option one  
  // create a 250*77 image 
  $im = imagecreate(250, 77); 
  // option two  
  // creating a image from jpeg  
  $im = @imagecreatefromjpeg("images/test.jpg")  
  or die("Cannot Initialize new GD image stream"); 

  // white background and black text 
  $bg = imagecolorallocate($im, 255, 255, 255); 
  $textcolor = imagecolorallocate($im, 0, 0, 0); 

 //create the text
  $skrivetekst = date('d.m.Y');
    $skrivetekst2 = date('H:i:s');

 // write the string at the top left 
  imagestring($im, 5, 0, 0, $skrivetekst, $textcolor); 
  imagestring($im, 5, 0, 20, $skrivetekst2, $textcolor); 


 ///Rotate the image 
$rotate = imagerotate($im, 90, 0);
  // output the image 


// Output
imagejpeg($rotate);


  ImageDestroy ($rotate); 
?>

Regards Troels

Share

Leave an answer

You must login to add an answer.

Need An Account,

1 Answer

  1. Editorial Team

    This is the expected behavior, your script should response to the browser with an output of certain type, usual output is text/html, that contains html content. To output an image, you send the Content-type: image/jpeg header, followed by image binary content (via imagejpeg($rotate);).

    The errors showing in your case are probably due to you are trying to send the header after echoing html/text content. Once output is sent, you can’t send more headers.

    To output php-generated images within php-generated html, you need to split that in two pages, one to output the image, and another to output html, the html output will reference your image via the regular <img> tag, as @GSto mentioned.

    EDIT:

    page.php:

    <html>
 <head><title>HI!</title></head>
 <body><?php
  //here goes your html building logic
  /*If you are not using and PHP, make this page page.html*/
  ?>
  See my php-generated image: <br />
  <img src="img.php" />
  </body>
</html>

    img.php:

    <?php 
   header("Content-type: image/jpeg"); 
   // option one  
  // create a 250*77 image 
  $im = imagecreate(250, 77); 
  //rest of your image-generating code..

    You will access the first page in your browser, that will see the <img> tag, request the next file and get the content to show as that tag.

    • 0