view python/open.py @ 785:205dd903e4c8

add some mediocre workflow
author Jeff Hammel <k0scist@gmail.com>
date Mon, 12 Sep 2016 14:11:49 -0700
parents 2025368488ee
children
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def load(resource):
    """
    open a file or URL for reading.  If the passed resource string is not a URL,
    or begins with 'file://', return a ``file``.  Otherwise, return the
    result of urllib2.urlopen()
    """

    # handle file URLs separately due to python stdlib limitations
    if resource.startswith('file://'):
        resource = resource[len('file://'):]

    if not is_url(resource):
        # if no scheme is given, it is a file path
        return file(resource)

    return urllib2.urlopen(resource)