header()

Anon

I made one thus script:

<?
header("Content-type: image/jpeg");
$im=ImageCreate (50, 100) or die ("Cannot Initialize new GD image stream");
$background_color = ImageColorAllocate ($im, 255, 255, 255);
$text_color = ImageColorAllocate ($im, 233, 14, 91);
ImageString ($im, 1, 5, 5, "A Simple Text String", $text_color);
ImagePng ($im);
ImageDestroy($im);
?>

e occurs the following error:
Warning: Cannot add header information - headers already sent by (output started at c:\var\www/ricardo/counter/teste1.php:12) in c:\var\www/ricardo/counter/teste1.php on line 13
‰PNG IHDR2dþ&ñPLTEÿÿÿé[‘Ë£

what I must make?

thank

Anon

The Header() function must be the first function to print any output of any kind.

The error says that the error occurs at line 13, which should be your header line, but that is line 2 in the script. Do you have any html text before the script?

If you plan to use this function to include an image in your html, you can forget it, because you can only make an image if you use the header() statement, and you can only use the header statement if you don't print anything before the header statement.
Ergo, you can't print the "<img src=" tag before printing the image.
You will have to make a seperate script for the image.

Anon

thanks for your reply.
My test worked, but now I have another problem. The code is:

<?
header ("Content-type: image/jpeg");
$im = imagecreatefromjpeg("images/lesquerdo.jpg");
ImageJPEG($im);
ImageDestroy($im);
echo "<img src=\"$im\">";
?>

this message appears in the browser:
PHP -> General Failure
and then, the following message appears:
Internal Server Error

I will be so grateful if you could help me.

Anon

That can be anything.
Look at the webserver log to see if there is anything about what went wrong.

Anon

The error.log shows:

[Fri Aug 18 11:27:24 2000] [error] [client 127.0.0.1] Premature end of script headers: c:/php4/php.exe

thanks

Anon

Premature end of script headers

this means that the script did not send any header information at all.
Odd.. I'll look into that a litte later, as today is winding down....

Anyway, have a look at the PHP manual, because I think you are trying to do the impossible.
You start the page by telling the browser that this is an image, but lateron
you print HTML code:
echo "<img src=\"$im\">";

Also, you should print the image before you destroy it.

I am sure you already did this, but try one of the samples from the manual and see if those work.

Anon

In fact, when I cut out the line:
$im = ImageCreateFromJPEG("images/lesquerdo.jpg");
and replace it for other function, it works.
I'm sorry if I am boring you with my questions, ok?

Anon

As long as it works!