英文:
Splitting string by new line with a condition
问题
以下是翻译好的部分：
我正在尝试仅在字符串不在我的“动作块”中时才通过\n拆分字符串。
这里有一个文本示例：message\n [testing](hover: actions!\nnew line!) more\nmessage。我希望只在\n不在[](this \n should be ignored)内部时拆分它。我为此制作了一个正则表达式，您可以在这里查看：https://regex101.com/r/RpaQ2h/1/。在示例中，它似乎工作正常，因此我在Java中进行了实现：
final List<String> lines = new ArrayList<>();
final Matcher matcher = NEW_LINE_ACTION.matcher(message);
String rest = message;
int start = 0;
while (matcher.find()) {
if (matcher.group("action") != null) continue;
final String before = message.substring(start, matcher.start());
if (!before.isEmpty()) lines.add(before.trim());
start = matcher.end();
rest = message.substring(start);
}
if (!rest.isEmpty()) lines.add(rest.trim());
return lines;
这应该忽略任何在上述模式内部的\n，但是它从不匹配“action”组，似乎当它添加到Java中并且存在\n时，它从不匹配。我有点困惑，不明白为什么，因为它在regex101上运行得非常完美。
英文:
I am trying to split a String by \n only when it's not in my "action block".
Here is an example of a text message\n [testing](hover: actions!\nnew line!) more\nmessage I want to split when ever the \n is not inside the [](this \n should be ignored), I made a regex for it that you can see here https://regex101.com/r/RpaQ2h/1/ in the example it seems like it's working correctly so I followed up with an implementation in Java:
final List<String> lines = new ArrayList<>();
final Matcher matcher = NEW_LINE_ACTION.matcher(message);
String rest = message;
int start = 0;
while (matcher.find()) {
if (matcher.group("action") != null) continue;
final String before = message.substring(start, matcher.start());
if (!before.isEmpty()) lines.add(before.trim());
start = matcher.end();
rest = message.substring(start);
}
if (!rest.isEmpty()) lines.add(rest.trim());
return lines;
This should ignore any \n if they are inside the pattern showed above, however it never matches the "action" group, seems like when it is added to java and a \n is present it never matches it. I am a bit confused as to why, since it worked perfectly on the regex101.
答案1
得分: 2
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
final String regex = "(?<action>\\[[^\\]]*\\]\\([^)]*\\))|(?<break>\\\\n)";
final String string = "message\\n [testing test](hover: actions!\\nnew line!) more\\nmessage";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
final String result = matcher.replaceAll("$1");
System.out.println(result);
}
}
英文:
Instead of checking whether the group is action, you can simply use regex replacement with the group $1 (the first capture group).
I also changed your regex to (?<action>\[[^\]]*]\([^)]*\))|(?<break>\\n) as [^\]]* doesn't backtrack (.*? backtracks and causes more steps). I did the same with [^)]*.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Main {
public static void main(String[] args) {
final String regex = "(?<action>\\[[^\\]]*\\]\\([^)]*\\))|(?<break>\\\\n)";
final String string = "message\\n [testing test](hover: actions!\\nnew line!) more\\nmessage";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
final String result = matcher.replaceAll("$1");
System.out.println(result);
}
}
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