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Home/ Questions/Q 3440306
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Editorial Team
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Editorial Team
Asked: May 18, 20262026-05-18T08:25:00+00:00 2026-05-18T08:25:00+00:00

I need to uniquely identify and store some URLs. The problem is that sometimes

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I need to uniquely identify and store some URLs. The problem is that sometimes they come containing “..” like http://somedomain.com/foo/bar/../../some/url which basically is http://somedomain.com/some/url if I’m not wrong.

Is there a Python function or a tricky way to resolve this URLs ?

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  1. Editorial Team
    Editorial Team
    2026-05-18T08:25:00+00:00Added an answer on May 18, 2026 at 8:25 am

    There’s a simple solution using urllib.parse.urljoin:

    >>> from urllib.parse import urljoin
    >>> urljoin('http://www.example.com/foo/bar/../../baz/bux/', '.')
    'http://www.example.com/baz/bux/'
    

    However, if there is no trailing slash (the last component is a file, not a directory), the last component will be removed.

    This fix uses the urlparse function to extract the path, then use (the posixpath version of) os.path to normalize the components. Compensate for a mysterious issue with trailing slashes, then join the URL back together. The following is doctestable:

    from urllib.parse import urlparse
    import posixpath
    
    def resolve_components(url):
        """
        >>> resolve_components('http://www.example.com/foo/bar/../../baz/bux/')
        'http://www.example.com/baz/bux/'
        >>> resolve_components('http://www.example.com/some/path/../file.ext')
        'http://www.example.com/some/file.ext'
        """
        parsed = urlparse(url)
        new_path = posixpath.normpath(parsed.path)
        if parsed.path.endswith('/'):
            # Compensate for issue1707768
            new_path += '/'
        cleaned = parsed._replace(path=new_path)
        return cleaned.geturl()
    
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