Hi.
How do I show the content of an jpg-image with php?
Thanks.
Hi.
How do I show the content of an jpg-image with php?
Thanks.
echo '<img src="photo.jpg">';
If thats not what you want you need to give more information
Hi Materix
Do you mean the Exif information embedded within you’re image? If so try: http://pel.sourceforge.net/
I would like to do something like this:
<img src=“image.php”>
And in image.php I would like to output the content of a jpg-file (if it exists). Is this possible?
I still do not understand what this means:
output the content of a jpg-file
Do you want to display a jpg, create a file, show the EXIF data ???
Usually in PHP circles, the image is kept on the hard drive with a path, and a file name.
/var/www/website1/images/image1.jpg
… and the reference to it (ie its name, image1.jpg) is kept in a database table, usually.
You use PHP to pull that image address out of the database, and then echo it out onto the html page, as shown in the examples above.
<?php
// this could have come from your db table
$image = '/images/image1.jpg';
?>
A load of html
<div>
<img src="<?php echo $image?>" />
</div>
A load more html
If this is not what you mean, perhaps you can explain a little more clearly what you do mean.
Sorry for my bad description of the problem.
I have found, what I was looking for. Here it is:
<?php
header(‘Content-Type: image/jpeg’);
readfile(‘image.jpg’);
?>
So you have a page with:
<img src="image.php">
and image.php contains:
<?php
header('Content-Type: image/jpeg');
readfile('image.jpg');
?>