Python - How to print a specific text

Question

def handler_users_answ(coze, res, type, source):
    if res:
        if res.getType() == 'result':
            aa=res.getQueryChildren()
            if aa:
                print 'workz1'
                for x in aa:
                    m=x.getAttr('jid')
                    if m:
                        print m

so this code returns me the values like this:

[email protected]/nickname1
[email protected]/nickname2

and so on, but i want it to print the value after the '/' only. like:

nickname1
nickname2

Thanks in advance.

Why do you use getter/setter methods? That's very unpythonic. If you need getter/setter logic, use properties. — ThiefMaster, Aug 08 '11 at 19:50

score 2 · Answer 1 · answered Aug 08 '11 at 19:49

2

You can use rpartition to get the part after the last \ in the string.

answered Aug 08 '11 at 19:49

Björn Pollex

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I just learned rpartition, always used rsplit(1)[1] for this, nice to know, +1, thanks. – utdemir Aug 08 '11 at 20:07
@utdemir: I think the difference is mostly one of style, although a quick `timeit` suggests that in certain cases `rpartition` is a bit faster. – Björn Pollex Aug 08 '11 at 20:14
rpartition is incorrect here, because this is a valid JID: `[email protected]/bar/baz`, and the resource portion is `bar/baz`, not `baz`. – Joe Hildebrand Aug 09 '11 at 02:56

score 1 · Answer 2 · answered Aug 08 '11 at 19:49

1

a = '[email protected]/nickname1'

b=a.split('/');
c=b[1];

answered Aug 08 '11 at 19:49

carlosdc

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This also fails on `[email protected]/nickname1/foo`. – Joe Hildebrand Aug 09 '11 at 03:00

score 1 · Answer 3 · answered Aug 08 '11 at 20:17

1

You can use rsplit which will do the splitting form the right:

a = '[email protected]/nickname1'
try:
    print a.rsplit('/')[1][1]
except IndexError:
    print "No username was found"

I think that this is efficient and readable. If you really need it to be fast you can use rfind:

a = '[email protected]/nickname1'
index = a.rfind('/')
if index != -1:
    print a[index+1:]
else:
    print "No username was found"

answered Aug 08 '11 at 20:17

Alex Plugaru

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Both of these fail on `[email protected]/nickname1/foo`. – Joe Hildebrand Aug 09 '11 at 02:59
Of course it fails. The problem was about [email protected]/nickname1 not [email protected]/nickname1/foo – Alex Plugaru Aug 09 '11 at 09:30
Right. The question was poorly worded. The *intent* of the question was "how do I parse an RFC 6122 Jabber ID?". – Joe Hildebrand Aug 09 '11 at 21:03

score 1 · Answer 4 · edited May 23 '17 at 10:34

1

To fully parse and validate the JID correctly, see this answer. There's a bunch of odd little edge cases that you might not expect.

edited May 23 '17 at 10:34

Community

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answered Aug 09 '11 at 02:59

Joe Hildebrand

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Python - How to print a specific text

4 Answers4