标记文件中的Regex get节

Question

我试图使用regex获取链接列表(https://raw.githubusercontent.com/Anything-Minecraft-Team/anything-minecraft/main/server/info/lists/plugins/anticheats.md)的每个部分。

我希望能像这样从列表中得到每一节

- [Anti Cheat](https://github.com/Paroxial/Anti-Cheat)  
Version: 1.8  
Rating: ???  
Discontinued

这样我就可以得到版本/评级信息，并知道它来自哪个链接。

我有一个(?:.*)(?:\n .*)*，它选择在上面/下面的文本之间有空格的文本(编辑，不，它不是真的哑)。但由于我的格式不起作用，是否有其他方法可以做到这一点，还是必须更改格式？

Answer 1

您可以使用3个捕获组获取链接和部分。

^- \[[^]\[]+]\((https?://[^\s()]+)\).*\R(Version:.*)\R(Rating:.*)\R(\S.+)$

• ^-匹配字符串开头的-
• 来自\[[^]\[]+]的[...]匹配
• \((https?://[^\s()]+)\)匹配(并捕获第1组中的url和匹配)
• .*\R匹配行的其余部分和换行符
• (Version:.*)捕获组2，匹配版本信息
• \R(Rating:.*)捕获组3，匹配评级信息
• \R(\S.+)$匹配换行符并捕获组4中的单个空格char和行的其余部分

Regex演示 x- Java演示

final String regex = "^- \\[[^]\\[]+]\\((https?://[^\\s()]+)\\).*\\R(Version:.*)\\R(Rating:.*)\\R(\\S.+)$";
final String string = "- [ABC Advanced Anticheat](https://www.spigotmc.org/resources/91606/) - Removed due to private reasons  \n"
+ "Version: 1.7 - 1.16  \n"
+ "Rating: 4  \n"
+ "Discontinued\n"
+ "- [AbdeslamNeverCheat](https://www.spigotmc.org/resources/61280)  \n"
+ "Version: 1.8  \n"
+ "Rating: 1  \n"
+ "Discontinued";

final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
final Matcher matcher = pattern.matcher(string);

while (matcher.find()) {
    System.out.println("Link: " + matcher.group(1));
    System.out.println("Version: " + matcher.group(2));
    System.out.println("Rating: " + matcher.group(3));
    System.out.println("Status: " + matcher.group(4));
}

输出

Link: https://www.spigotmc.org/resources/91606/
Version: Version: 1.7 - 1.16  
Rating: Rating: 4  
Status: Discontinued
Link: https://www.spigotmc.org/resources/61280
Version: Version: 1.8  
Rating: Rating: 1  
Status: Discontinued