Calling an image

johnnydog

I want to write a small script that calls an image based on the directory I'm in. I have the constants set up to know what directory I'm in already but I can't seem to get the code down for calling the image.

Can anyone help?

cordex

I guess I just don't understand your problem completely. Are you trying to call an image and trying to change the relative offset "../../" based on the depth? If so, use absolute positioning "http://www.myserver ..." and you won't have to worry about it. If you absolutely must have relative positioning, let me know and I'll cook up a parsing script for you.

Are you trying to use a different image depending on the directory? If so, you could have it call an image, say, "dirimage.jpg", with no offset. Then just change the image in each directory. But that's the dirty method.

Since you know the directory you're in (are you using environmental variables? that works best) I would make a table that stores what image is displayed depending on what directory you are in.

johnnydog

Hi! Thanks for responding. I'm kind of new to PHP and I haven't used it since for a few months (before I got layed off).

All of the images are in the same directory (1-title.gif, 2-title.gif. 3-title.gif ...)

I had done this once before by adding only one line of code to the existing code I had. I have the code that deciphers the directory and I want ONE image to be diplayed according to it's matching directory. (www.domain.com/1/ = 1-title.gif)

I hope I'm making sense here.

Anon

Hi,

You could try something like this:

// Path of directory with images
$dir_path = 'http://www.example.com/images/'

// When I tried it, $PATH_INFO returned
// '/1/index.htm', ie. These next few lines
// turn the paths into an array and then
// extract the second element (the first
// is what's behind the leading '/')
$path_array = explode('/', $PATH_INFO);
$img = $path_array[1];

// Spit out appropriate image url
echo $dir_path.$img.'-title.gif';

I haven't tested the script per se, but it should work.

Best Regards,
Bill Zeller

cordex

Nice script Bill.

Alternatively, since you say you know where you are in the directory structure (you have the "number" of the picture you want?) you could just do this:

$imgnam_st = $dirnumber_in . "-title.gif";

Bill's script, on the other hand, will allow you to name your pictures anything you damn well please as long as the name coincides with the directory structure you want to place it in.

Good luck.

Anon

You could also do this: have the script "read" the directory name, and choose an image depending on whether or not it matches a certain number. $PHP_SELF tells the script the name of the subdirectory it is in, as well as the filename of the script. Let's say you are at www.you.com/dir2/index.php.

// find out if the directory is "dir2"

$find = "dir2";

if(strstr($PHP_SELF,$find))

// if it is, set this image:

{ $img="dir2.jpg"; }

Since you're at www.you.com/dir2/index.php, $PHP_SELF="/dir2/index.php". the strstr() function looks at the string "/dir2/index.php" and checks if it contains the substring "dir2" (which you set as $find). Since it does contain the substring, $img is set to "dir2.jpg".

Hope that helps,

Steph