Can anybody Help?

szcieder

[Moved this reply to its own thread -- Mod.]

Hi everyone, I'm a new member here, and I need a little help with my PHP coding, I saw your coding just now and I registered lately. I was hoping maybe someone could help me with something.

I'm trying to find a code for my site ([link deleted -- Mod.]) and I have alot of images.

So how can I make a my Picture Generator out of PHP?

I want something like...

[noparse]www.example.net/showpic.php?name=[/noparse]IMAGE

so when we try to view something like that it results to

[noparse]www.example.net/image.png[/noparse]

so when I want to have another name for it...

[noparse]www.example.net/showpic.php?name=[/noparse]IMAGE2

it also results to...
[noparse]www.example.net/image.png[/noparse]

and will apply to many images I want to put on the php file?

how can I do that?

Xager

First of all you can get values from the URL by typing $_GET['name']

Then you check if there is an image by the name of $_GET['name'] and if it is you set the content type with header("Content-Type: image/jpeg"); or the image type that it is and the print("/path/to/the/file.file");

szcieder

Xager wrote:
First of all you can get values from the URL by typing $_GET['name']

Then you check if there is an image by the name of $_GET['name'] and if it is you set the content type with header("Content-Type: image/jpeg"); or the image type that it is and the print("/path/to/the/file.file");

What if I want to access the image using a fake URL like...

www.example.net/showpic.php?name=IMAGE

and the real URL of the file is...

www.example.net/myphoto123.png

how can I do that?

Xager

szcieder wrote:
What if I want to access the image using a fake URL like...

www.example.net/showpic.php?name=IMAGE

and the real URL of the file is...

www.example.net/myphoto123.png

how can I do that?

Exactly the way i said (with some exceptions):

$file_name = $_GET['name'];
$file_path = "/var/www/bilder/";

if(file_exists($file_path. $file_name) == true && isset($_GET['name'])){
    $mime_type = mime_content_type($file_path . $file_name);
    header("Content-type: $mime_type");
    readfile($file_path. $file_name);
}
else{
    echo("The file does not exist!");
}

szcieder

thanks.