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Java splitting string by custom regex match

开发者 https://www.devze.com 2022-12-25 13:36 出处:网络
I am completely new to regular expressions so I\'m looking for a bit of help here. I am compiling under JDK 1.5

I am completely new to regular expressions so I'm looking for a bit of help here.

I am compiling under JDK 1.5

Take this line as an example that I read from standard input:

ab:Some string po:bubblegum

What I would like to do is split by the two characters and colon. That is, once the line is split and put into a string array, these should be the terms:

ab:Some string
po:b开发者_StackOverflow中文版ubblegum

I have this regex right now:

String[] split = input.split("[..:]");

This splits at the semicolon; what I would like is for it to match two characters and a semicolon, but split at the space before that starts. Is this even possible?

Here is the output from the string array:

ab
Some String po
bubblegum

I've read about Pattern.compile() as well. Is this something I should be considering?


input.split(" (?=[A-Za-z]{2}:)")

The ?= creates a positive lookahead. This means the engine looks ahead to see if the next part matches, without consuming that part. If it does match, it splits on the space character. [A-Za-z] means a upper or lower-case letter, while {2} specifies we want two characters matching that class.


You asked about Pattern#compile(String pattern). You should consider using it if you are going to use the regex a lot since the aforementioned method compiles the regex into something that's fast to execute while using String#split(String regex) directly always recompiles the regex.

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