Posted by B Squared on 01/26/06 21:13

I'm trying to pass a filename (which is a jpeg image) to a php
function / file so that it will display. I know that its simple
to get PHP to display an image hardcoding in the filename. For
example, an href to this:

<?php
header("content-type:image/jpeg");
$filename = "image_file.jpg";
$im=ImageCreateFromJPEG($filename);
ImageJPEG($im);
?>

But when I have a filename I create dynamically in PHP and try to
call a php function with the filename, for example calling:

<?php
display_jpeg_file($image_path);
?>

<?php
function display_jpeg_file($filename) {
header("content-type:image/jpeg");
$im=ImageCreateFromJPEG($filename);
ImageJPEG($im);
}
?>

I get the "cannot modify header information" error. The error
message always refers to a php function early in the phmtl page.
So somehow it sees this function as a continuation of the existing
html document. How do I arrange the function call to my display
function so that it works? Or, if my approach is all wrong, what
is the correct approach? The important part is I have a filename
generated in PHP that I want to display.

Thanks in advance for any help.

B Squared

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Self esteem, n. An erroneous appraisement
-- Ambrose Bierce, The Devils Dictionary

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