即使有重复，也要在一个字符之后获取所有内容。 蟒蛇

Question

I'm writing an IRC bot, and am parsing the message using a whole bunch of splits. The format for an IRC 'message' is:

:username!765a4aa2@gateway/web/freenode/ip.***.***.***.*** PRIVMSG #Channel :Message body

In particular is the one that get's the text of the message:

message = data.split(':')[2]

This get's everything after the last ':' But when the user sends a link over IRC, then the message will look like:

:username!765a4aa2@gateway/web/freenode/ip.***.***.***.*** PRIVMSG #Channel :http://web address.com/

and the code will only get the 'http' section of the message, as the rest is now in the third section of the split.

So how can I parse a message containing a link?

One way is to get the lenght of the first section, and remove it:

message = data[len(data.split(':')[1])+1:]

But I can't help feeling that there must be a better way. Is there one?

Answer 1

str.split()接受可选参数maxsplit 。

message = data.split(':', 2)

Answer 2

Set the amount of : you want to split:

data.split(':', 2)

And you'll have:

['',
'username!765a4aa2@gateway/web/freenode/ip.***.***.***.*** PRIVMSG #Channel ',
'http://web address.com/']