Possible Duplicate:
How do I un-escape a backslash-escaped string in python?
I'm getting a response from an api that formats URLs like: http:\/\/domain.com\/. How do I make them normal URLs? I feel like there's a template tag that could be used.
UPDATE: I know I can use url.replace('\/', '/'), but I am wondering if there is something in the template I can do. I'm getting a dict back from the Instagram API and that contains the escaped slashes. I'd prefer to make this change in the template instead of having to parse the JSON in the view.