Is it possible to load a image using:
<img src="image.php?image_id=1">
?
for image.php:
$image_id = $_GET['image_id'];
echo "image".$image_id.".png";
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Don't you mean src? I just corrected it.Shamim Hafiz - MSFT– Shamim Hafiz - MSFT2011-05-17 08:56:37 +00:00Commented May 17, 2011 at 8:56
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answer to your question is yesLawrence Cherone– Lawrence Cherone2011-05-17 08:57:33 +00:00Commented May 17, 2011 at 8:57
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Are you sure you want to use server's resources to read then echo an image? I wouldn't be so sure about that...Bogdan Constantinescu– Bogdan Constantinescu2011-05-17 09:34:47 +00:00Commented May 17, 2011 at 9:34
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6 Answers
you must return a valid image, use file_get_contents or readfile for get the conent of image then output to browser
header("Content-Type:image/png");
$image_id = $_GET['image_id'];
if(is_file($file = "image".$image_id.".png") || is_file($file = "no_image.png"))
readfile($file);
3 Comments
Quentin
Don't serve up images with a
text/html content-type.
Lawrence Cherone
output still needs the headers
Yoshi
Or, if this is not to hide the true image location, or if the image is not created on the fly, one could easily use the
Location header to redirect to the image. (If it's accessable through http)If you have the png contents in the database, follow the other answers. If all you need is the id from the database, simply <img src="image<?=$image_id?>.png">
Comments
No, this is not possible that way. You would have to write the contents of the image file to the output stream instead.
Comments
Is it possible to load a image use:
<img src="image.php?image_id=1">
Yes. URLs are URLs. It is content-type that determines content, not any characters in the URL itself.
$image_id = $_GET['image_id'];echo "image".$image_id.".png";
Not like that. You either have to read the image file in and then output it with a suitable content-type, or redirect (with a location header) to a static image.
Comments
You have to change the header content-type and print the contents of the image file. Here is an example for your image.php file:
$image_id = $_GET['image_id'];
$filename = "image".$image_id.".png";
header('Content-Type: image/png');
print file_get_contents($filename);
Comments
one liner....
echo (file_exists($_GET['image_id'])) ? header("Content-Type:image/png").readfile($_GET['image_id']).die() : false;