url builder for python

Question

I know about urllib and urlparse, but I want to make sure I wouldn't be reinventing the wheel.

My problem is that I am going to be fetching a bunch of urls from the same domain via the urllib library. I basically want to be able to generate urls to use (as strings) with different paths and query params. I was hoping that something might have a syntax like:

url_builder = UrlBuilder("some.domain.com")
# should give me "http://some.domain.com/blah?foo=bar
url_i_need_to_hit = url_builder.withPath("blah").withParams("foo=bar")  # maybe a ".build()" after this

Basically I want to be able to store defaults that get passed to urlparse.urlunsplit instead of constantly clouding up the code by passing in the whole tuple every time.

Does something like this exist? Do people agree it's worth throwing together?

Answer 1

Are you proposing an extension to http://docs.python.org/library/urlparse.html#urlparse.urlunparse that would substitute into the 6-item tuple?

Are you talking about something like this?

def myUnparse( someTuple, scheme=None, netloc=None, path=None, etc. ):
    parts = list( someTuple )
    if scheme is not None: parts[0] = scheme
    if netloc is not None: parts[1]= netloc
    if path is not None: parts[2]= path
    etc.
    return urlunparse( parts )

Is that what you're proposing?

This?

class URLBuilder( object ):
    def __init__( self, base ):
        self.parts = list( urlparse(base) )
    def __call__( self, scheme=None, netloc=None, path=None, etc. ):
        if scheme is not None: self.parts[0] = scheme
        if netloc is not None: self.parts[1]= netloc
        if path is not None: self.parts[2]= path
        etc.
        return urlunparse( self.parts )

bldr= URLBuilder( someURL )
print bldr( scheme="ftp" )

Something like that?

Answer 2

You might want consider having a look at furl because it might be an answer to your needs.