你如何返回在 python 中用 re.search 找到的字符串？

Question

So this is my code:

class telnet(object):
    """conexiune"""
    def __init__(self):

        HOST = "route-views.routeviews.org"
        user = "rviews"
        password = ""

        tn = telnetlib.Telnet(HOST)

        tn.read_until("login: ", 5)
        tn.write(user + "\r\n")

        tn.read_until("Password: ", 5)
        tn.write(password + "\r\n")

        print tn.read_until(">", 10)
        tn.write("show ip route 192.0.2.1"+"\r\n")

        self.y = tn.read_until("free", 10)
        print self.y
        tn.write("exit"+ "\r\n")

        tn.close()

    def re(self):
        self.m = re.search(r' Known via "bgp \d{0,5}"', self.y)
        if self.m:
            print self.m.group(0)
        else:
            print False

What I need to do is return self.m instead of printing it. If I write 'return "This answer is: "+self.m', I get this error:

return "The answer is: " + self.m.group(0) TypeError: cannot concatenate 'str' and '_sre.SRE_Match' objects

If I use print it prints it, but I don't know how to do it with a return statement.

This is what it has to return:

Known via "bgp 6447"

from this telnet output: route-views> show ip route 192.0.2.1

Routing entry for 192.0.2.1/32

Known via "bgp 6447", distance 20, metric 0

Tag 19214, type external

Last update from 208.74.64.40 4w1d ago

Routing Descriptor Blocks:

• 208.74.64.40, from 208.74.64.40, 4w1d ago

  Route metric is 0, traffic share count is 1

  AS Hops 1

  Route tag 19214

  MPLS label: none

route-views>

I know that return is used for functions - that's why I added the '+'. Btw I'm a Python beginner. Any help would be appreciated.

Answer 1

I think in your 'return' statement you just missed out the '.group(0)'.

Try :

if self.m:
   return self.m.group(0)

Answer 2

From documentation

>>> m = re.match(r"(\w+) (\w+)", "Isaac Newton, physicist")
>>> m.group(0)       # The entire match
'Isaac Newton'
>>> m.group(1)       # The first parenthesized subgroup.
'Isaac'
>>> m.group(2)       # The second parenthesized subgroup.
'Newton'
>>> m.group(1, 2)    # Multiple arguments give us a tuple.
('Isaac', 'Newton')

So, try this

>>> import re
>>> y = ' Known via "bgp 54574"'
>>> m = re.search(r' Known via "bgp (\d{0,5})"', y)
>>> print m.group(1) if m else False
54574
>>> y = ' Known via "bg p 54574"'
>>> m = re.search(r' Known via "bgp (\d{0,5})"', y)
>>> print m.group(1) if m else False
False
>>>