How echo a variable containing image

[mistercash60000]

Hello,

The problem is that I have the image in a variable, not the link but the imagefile self.
Because If I echo it i get 20 lines of strange code.

But when I place this in front of it:

header('Content-type: image/jpeg');

then I can view the image when I type

echo $img;

But the the problem is that it only shows the image and the text that I have in html under it doesn't show.
Do I have to announce after the image that know it is text?

Can somebody tell me how I can show the image?

I think this is something very simple

[onion2k]

2 scripts.

1 to generate your html that contains an img tag like "<img src='image.php'>".

1 to generate the image.

[mistercash60000]

Thanks for the help, it was indeed this way I had to do it
problem solved