Java正则表达式，无法正确捕获最后一组

Question

[txt1] [txt2] [txt3] /some/long/path?params=1,2,3
[txt1] [txt2] [txt3] /path/
[txt1] [txt2] [txt3] /

我正在纠结于捕获最后一组--使用我的正则表达式，只有第一行匹配模式，而不是其他两行。如果我将正则表达式更改为^\[(.*？）\]\\[(.*？）\\\[(.*？）\](/.*)(\？*)？，则所有3行都匹配，但第一行捕获不成功（我只获得1组/some/long/path？params=1,2,3，而不是2组/some/long/path和？params=1,2,3)。

如何编写这个正则表达式，使所有行都有5个匹配组？

Answer 1

使您只剩一个.*为惰性，使最后一个捕获组为可选组，并附加$字符串结尾锚点：

^\[(.*?)] \[(.*?)] \[(.*?)] (/.*?)(\?.*)?$
                                ^       ^^

请参阅正则表达式演示

• (/.*？)组中的.*？应该是惰性的，因为我们需要允许用尽可能多的字符填充后续组
• (\？.*)？-必须是可选的，因为文本可以不存在
• $是必需的，因为前面的两个组是可选的，因此字符串末尾的文本可能不会匹配。这样，我们就需要regex引擎来获取行的其余部分。

Pattern pattern = Pattern.compile("^\\[(.*?)] \\[(.*?)] \\[(.*?)] (/.*?)(\\?.*)?$");
String[] ss = { "[txt1] [txt2] [txt3] /some/long/path?params=1,2,3", "[txt1] [txt2] [txt3] /path/", "[txt1] [txt2] [txt3] /"};
for (String s: ss) {
    Matcher matcher = pattern.matcher(s);
    while (matcher.find()){
        System.out.println("Next match for \"" + s + "\"" ); 
        System.out.println(matcher.group(1));
        System.out.println(matcher.group(2)); 
        System.out.println(matcher.group(3)); 
        System.out.println(matcher.group(4)); 
        System.out.println(matcher.group(5)); 
    } 
}

Next match for "[txt1] [txt2] [txt3] /some/long/path?params=1,2,3"
txt1
txt2
txt3
/some/long/path
?params=1,2,3
Next match for "[txt1] [txt2] [txt3] /path/"
txt1
txt2
txt3
/path/
null
Next match for "[txt1] [txt2] [txt3] /"
txt1
txt2
txt3
/
null