[Tutor] How can a CGI program get the URL that called it?

[Terry Carroll]

Ah, I discovered what my problem was...

On Fri, 10 Jan 2014, Alan Gauld wrote:

> its calling your file. You should know where your file is?

My problem was that, I know where the file is in the host's file system, 
and relative to my CGI program.  I do not have a URL to that file.

> If you want to create a png file and display it to the user then you 
> just store the file somewhere in your web site and create an html file 
> that has an img tag referencing that location.

Right; I am producing HTML output (using the print command, not as a 
file), with an img tag.  The img tag has a src attribute, which must 
provide the URL of the referenced image file.

But I do not have that URL.  I know where that file is in the file system, 
and relative to my CGI program.  But I do not have a URL to the file.  My 
thinking was that, if I have the URL to my program, it's pretty trivial to 
construct the URL to the file.

And here's where my real problem was:  I had tried specifying just a 
relative path in the src tag, and that did not work consistently; 
specifically, I found that it worked in Chrome, but not Firefox.

As it turns out, since I was testing on a Windows box, os.path.relpath was 
(reasonably) using a '\' as the separator character (surprisingly, so does 
posixpath.relpath).  By simply adding:

    relative_path = relative_path.replace('\\', '/')

It uses the '/' required in a URL (even a relative-path URL) and works. 
Chrome was forgiving of the '\'; other browsers not so much.

It was not until I posted the <img> tag into a draft of this reply that I 
noticed the '\' characters.

I swear I looked at this for hours without noticing this before.