I need to send a GET request to my page pic.php, and I want to get a real picture in return.

For now I implemented this idea like this:

<?php
if ((isset($_GET['pic']))&&(isset($_GET['key']))){
$pic = $_GET['pic'];
$pic=stripslashes($pic);

header('Location: /content/'.$pic);

}
?>

But it's not really what I want - it redirects to image directly. What I want is to keep the same URL, but get a needed file depending on what values were submitted. What is the best way to do that? thx.

I need to send a GET request to my page pic.php, and I want to get a real picture in return.

For now I implemented this idea like this:

<?php
if ((isset($_GET['pic']))&&(isset($_GET['key']))){
$pic = $_GET['pic'];
$pic=stripslashes($pic);

header('Location: /content/'.$pic);

}
?>

But it's not really what I want - it redirects to image directly. What I want is to keep the same URL, but get a needed file depending on what values were submitted. What is the best way to do that? thx.

• php
• javascript
• html
• css

• 2 stackoverflow./questions/7069112/… – miki Commented Mar 16, 2012 at 17:49
• try file_get_contents('/content/' . $pic); – Ascherer Commented Mar 16, 2012 at 17:54
• I'm not 100% on this, but I think file uploads have to be POST requests. simples reference. w3schools./php/php_file_upload.asp – laymanje Commented Mar 19, 2012 at 18:35

4 Answers 4

This example code snippet should do what you ask. I've also included code to only strip slashes if magic quotes is enabled on the server. This will make your code more portable, and patible with future versions of PHP. I also added use of getimagesize() to detect the MIME type so that you output the proper headers for the image, and do not have to assume it is of a specific type.

<?php
if(isset($_GET['pic']))
{
    //Only strip slashes if magic quotes is enabled.
    $pic = (get_magic_quotes_gpc()) ? stripslashes($_GET['pic']) : $_GET['pic'];

    //Change this to the correct path for your file on the server.
    $pic = '/your/path/to/real/image/location/'.$pic;

    //This will get info about the image, including the mime type.
    //The function is called getimagesize(), which is misleading 
    //because it does much more than that.
    $size = getimagesize($pic);

    //Now that you know the mime type, include it in the header.
    header('Content-type: '.$size['mime']);

    //Read the image and send it directly to the output.
    readfile($pic);
}
?>

I can see you doing this in two ways:

1) Return the URL to the image, and print out an image tag:

print '<img src=' . $img_url . ' />';

2) Alternatively, you could just pull the data for the image, and display it. For instance, set the header appropriately, and then just print the image data.

header("content-type: image/png");
print $img_data;

This assumes that you have the image data stored in a string $img_data. This method will also prevent you from displaying other things on the page. You can only display the image.

You can load the image, send the image headers, and display the image as such:

header('Content-Type: image/jpeg');
readfile('/path/to/content/pic.jpg');

Obviously the headers would depend on the filetype, but that's easy to make dynamic.

Not sure if I understand what you're after, but guessing that you want to load the picture in an img tag?

If I'm right you just do:

<img src=http://www.domain./pic.php?"<?php echo image here ?>" />

Basically you just make the source of the image the webpage you get directed to where the image is.