问题描述
我在网上学习 Java 正则表达式教程时,对一个小程序感到困惑.
I was learning a Java regular expression tutorial online and got confused about one small program.
// String to be scanned to find the pattern.
String line = "This order was places for QT3000! OK?";
String pattern = "(.*)(\\d+)(.*)";
// Create a Pattern object
Pattern r = Pattern.compile(pattern);
// Now create matcher object.
Matcher m = r.matcher(line);
if (m.find( )) {
System.out.println("Found value: " + m.group(0) );
System.out.println("Found value: " + m.group(1) );
System.out.println("Found value: " + m.group(2) );
}
打印出来的结果是:
Found value: This order was places for QT3000! OK?
Found value: This order was places for QT300
Found value: 0
我不知道为什么 group(1) 会得到上述值?为什么它在'QT3000'的最后一个零之前停止?
I have no idea why the group(1) gets value the above value? Why it stops before the last zero of 'QT3000'?
非常感谢!
推荐答案
第一组 (.*)
(这是索引 1 - 索引 0 是整体正则表达式)是贪婪匹配.它尽可能多地捕捉,同时让整体表达仍然匹配.因此它可以占用字符串中的第二个 0
,只剩下 0
来匹配 (\\d+)
.如果您想要不同的行为,那么您应该阅读贪婪和非贪婪匹配,或者找到更合适的模式.
The first group of (.*)
(this is index 1 - index 0 is the overall regular expression) is a greedy match. It captures as much as it can while letting the overall expression still match. Thus it can take up to the second 0
in the string, leaving just 0
to match (\\d+)
. If you want different behaviour, then you should read up on greedy and non-greedy matches, or find a more appropriate pattern.
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