Just use the split function. It returns a list, so you can keep the first element:

>>> s1.split(':')
['Username', ' How are you today?']
>>> s1.split(':')[0]
'Username'

You are looking for str.rsplit(), with a limit:

print x.rsplit('-', 1)[0]

.rsplit() searches for the splitting string from the end of input string, and the second argument limits how many times it'll split to just once.

Another option is to use str.rpartition(), which will only ever split just once:

print x.rpartition('-')[0]

For splitting just once, str.rpartition() is the faster method as well; if you need to split more than once you can only use str.rsplit().

Demo:

>>> x = 'http://test.com/lalala-134'
>>> print x.rsplit('-', 1)[0]
http://test.com/lalala
>>> 'something-with-a-lot-of-dashes'.rsplit('-', 1)[0]
'something-with-a-lot-of'

and the same with str.rpartition()

>>> print x.rpartition('-')[0]
http://test.com/lalala
>>> 'something-with-a-lot-of-dashes'.rpartition('-')[0]
'something-with-a-lot-of'

Well, to answer the immediate question:

>>> s = "http://www.domain.com/?s=some&two=20"

The rfind method returns the index of right-most substring:

>>> s.rfind("&")
29

You can take all elements up to a given index with the slicing operator:

>>> "foobar"[:4]
'foob'

Putting the two together:

>>> s[:s.rfind("&")]
'http://www.domain.com/?s=some'

If you are dealing with URLs in particular, you might want to use built-in libraries that deal with URLs. If, for example, you wanted to remove two from the above query string:

First, parse the URL as a whole:

>>> import urlparse, urllib
>>> parse_result = urlparse.urlsplit("http://www.domain.com/?s=some&two=20")
>>> parse_result
SplitResult(scheme='http', netloc='www.domain.com', path='/', query='s=some&two=20', fragment='')

Take out just the query string:

>>> query_s = parse_result.query
>>> query_s
's=some&two=20'

Turn it into a dict:

>>> query_d = urlparse.parse_qs(parse_result.query)
>>> query_d
{'s': ['some'], 'two': ['20']}
>>> query_d['s']
['some']
>>> query_d['two']
['20']

Remove the 'two' key from the dict:

>>> del query_d['two']
>>> query_d
{'s': ['some']}

Put it back into a query string:

>>> new_query_s = urllib.urlencode(query_d, True)
>>> new_query_s
's=some'

And now stitch the URL back together:

>>> result = urlparse.urlunsplit((
    parse_result.scheme, parse_result.netloc,
    parse_result.path, new_query_s, parse_result.fragment))
>>> result
'http://www.domain.com/?s=some'

The benefit of this is that you have more control over the URL. Like, if you always wanted to remove the two argument, even if it was put earlier in the query string ("two=20&s=some"), this would still do the right thing. It might be overkill depending on what you want to do.

There are a few ways. Here are a few off the top of my head:

>>> s = "abcd//efgh"

>>> s.find("/")
4
>>> s[:s.find("/")]
'abcd'

>>> s.split("/")
['abcd', '', 'efgh']
>>> s.split("/", maxsplit=1)
['abcd', '/efgh']
>>> s.split("/", maxsplit=1)[0]
'abcd'

>>> import re
>>> re.sub("/.*$", "", s)
'abcd'

The last is overkill here and I wouldn't use it, but regexs are often appropriate for doing search & replace operations. Either of the first two would work pretty well. The first depends on the search string appearing though. Otherwise, s.find will return -1 and then s[:-1] will lop off the last character:

>>> s = "abcdef"
>>> s[:s.find("/")]
'abcde'