Question

I have a url.

**"/deals/image/"+name**

name is a variable which gives the image name.

The view catering to this url is

def image(request,name = None):
    if name == None:
        name = "3gmjr0kme6_coffee-art.jpg"
    else:
        name = str(name)
    this_directory = settings.PROJECT_ROOT
    url = this_directory+"\\templates\\media\\images\\photos\\"
    full =url+name 
    image_data = open(full, "rb").read()
    return HttpResponse(image_data, mimetype="image/png")

The problem that I am facing is that it is unable to get this view because of the dot “.” i.e. “/deals/image/some_image.jpg” is not able to find the view. How can i accout for the “.”? or am i doing something wrong?

urls file is as follows

 url(r'^image/(\w+)$','image'),

Any help will be highly appreciated.

Answer 1

Change the URL configuration to something like this:

url(r'^image/([\w\.]+)$','image')

• The [...] part is a character set. It will match all characters inside.
• The \w is used to match normal characters (A to Z and a to z), digits and the underscore “_”.
• The \. matches a literal dot.
• The + means one or more of the previous expression.

When the \w and \. is put together inside a character set ([]), it will match all characters and digits, as well as underscore and the dot. Putting the + after means that there has to be at least one character in the set.