Pattern / Matcher group()在Java中获取子字符串? [英] Pattern/Matcher group() to obtain substring in Java?
问题描述
更新:感谢所有精彩的回复!我尝试了许多不同的正则表达式模式,但不明白为什么m.matches()没有做我认为它应该做的事情。当我切换到 m.find()时,以及调整正则表达式模式,我能够到达某个地方。
UPDATE: Thanks for all the great responses! I tried many different regex patterns but didn't understand why m.matches() was not doing what I think it should be doing. When I switched to m.find() instead, as well as adjusting the regex pattern, I was able to get somewhere.
我想匹配Java字符串中的模式,然后使用正则表达式提取匹配的部分(如Perl的$& operator)。
I'd like to match a pattern in a Java string and then extract the portion matched using a regex (like Perl's $& operator).
这是我的源字符串s: DTSTART; TZID = America / Mexico_City:20121125T153000
我想提取America / Mexico_City部分。
This is my source string "s": DTSTART;TZID=America/Mexico_City:20121125T153000
I want to extract the portion "America/Mexico_City".
我以为我可以使用Pattern和Matcher,然后使用m.group()提取但是它没有按照我的预期工作。我尝试使用不同的正则表达式字符串进行修改,而m.matches()上唯一似乎是。* TZID。*
这是没有意义的,因为它只是返回整个字符串。有人可以开导我吗?
I thought I could use Pattern and Matcher and then extract using m.group() but it's not working as I expected. I've tried monkeying with different regex strings and the only thing that seems to hit on m.matches() is ".*TZID.*"
which is pointless as it just returns the whole string. Could someone enlighten me?
Pattern p = Pattern.compile ("TZID*:"); // <- change to "TZID=([^:]*):"
Matcher m = p.matcher (s);
if (m.matches ()) // <- change to m.find()
Log.d (TAG, "looking at " + m.group ()); // <- change to m.group(1)
推荐答案
你使用 m.match()
尝试匹配整个字符串,如果你将使用 m.find()
,它将在里面搜索匹配,我也改进了你的正则表达式以使用零宽度后面的方式排除TZID前缀:
You use m.match()
that tries to match the whole string, if you will use m.find()
, it will search for the match inside, also I improved a bit your regexp to exclude TZID prefix using zero-width look behind:
Pattern p = Pattern.compile("(?<=TZID=)[^:]+"); //
Matcher m = p.matcher ("DTSTART;TZID=America/Mexico_City:20121125T153000");
if (m.find()) {
System.out.println(m.group());
}
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