Android的常规前pression - 返回匹配的字符串 [英] Android regular expression - return matched string

问题描述

在我的Andr​​oid项目我有一个正规的前pression和一个字符串，其中我应该有匹配的前pression。问题是，我只找到了一个匹配（）方法，它返回boolean值。是不是有什么，仅返回匹配的字符串（例如，如果我的字符串到中午12点店，我要检查是否有此字符串时（在这个例子 - 12点），如果它是 - 返回它）？
先谢谢了。

In my Android project I have a regular expression and a string, in which I should have the matched expression. The problem is, I've only found a matches() method, which returns boolean. Is there something, which returns only the matched string (for instance, if my string is "go to the shop at 12pm", I want to check if there is a time in this string (in this example - "12pm"), if it is - return it) ? Thanks in advance.

推荐答案

您应该得到捕获你需要组。阅读这个。你会发现回答你的问题。结果
这是给你一个简单的例子。我想你会明白。

you should get capturing group you need. Read this. You'll find answer to your question.
Here is a simple example for you. I think you'll understand it.

Pattern p = Pattern.compile(".*?(\\d{2}(am|pm)).*");
Matcher m = p.matcher("go to the shop at 12pm");
if(m.matches())
   return m.group(1);

这将返回 12点结果
其实你可以得到你想要的东西有更好的办法。

This will return 12pm
Actually you can get what you want with better way.

Pattern p = Pattern.compile("\\d{2}(am|pm)");
Matcher m = p.matcher("go to the shop at 12pm");
if(m.find())
   return m.group(0);    //or you can write return m.group(); result will be the same.

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