java - Java Regex调整

谁能帮我在哪里做错了？

我的示例文字：

{[|Name:A|Class:1|Sex:Male|][|Name:B|Class:2|Sex:Female|][|Name:C|Class:3|Sex:Male|]}

预期产量：

|Name:A|Class:1|Sex:Male|
Name:A
Class:1
Sex:Male
|Name:B|Class:2|Sex:Female|
Name:B
Class:2
Sex:Female
|Name:C|Class:3|Sex:Male|
Name:C
Class:3
Sex:Male

电流输出

|Name:A|Class:1|Sex:Male|
Name:A
Sex:Male
|Name:B|Class:2|Sex:Female|
Name:B
Sex:Female
|Name:C|Class:3|Sex:Male|
Name:C
Sex:Male

我的程序：

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Regex {

    public static void main(String[] args) {

        String example = "{[|Name:A|Class:1|Sex:Male|][|Name:B|Class:2|Sex:Female|][|Name:C|Class:3|Sex:Male|]}";

        Pattern curlyBraces = Pattern.compile("\\[(.*?)\\]");

        Matcher m = curlyBraces.matcher(example);
        while (m.find()) {
            System.out.println(m.group(1));
            String element = m.group(1);
            Pattern pipe = Pattern.compile("\\|(.*?)\\|");
            Matcher mPipe = pipe.matcher(element);
            while (mPipe.find()) {
                System.out.println(mPipe.group(1));
            }
        }
    }
}

最佳答案

您的问题是"\\|(.*?)\\|"仅在行中匹配|Name:A|和|Sex:Male|

|Name:A|Class:1|Sex:Male|

因为正则表达式会消耗它匹配的字符，因此|和Name:A之间的Class:1只能匹配一次。

使用lookaround assertions来解决该问题-他们不使用匹配的文本：

        Pattern pipe = Pattern.compile("(?<=\\|).*?(?=\\|)");
        Matcher mPipe = pipe.matcher(element);
        while (mPipe.find()) {
            System.out.println(mPipe.group(0));
        }

如果您不希望为空值，则另一种可能性是匹配所有“非管道”字符：

        Pattern pipe = Pattern.compile("[^|]+");
        Matcher mPipe = pipe.matcher(element);
        while (mPipe.find()) {
            System.out.println(mPipe.group(0));
        }

关于java - Java Regex调整，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/35194704/