正则表达式，用于捕获特定数字

Question

从下面的日志中，我如何单独grep'951792'值

2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes
2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e
2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10   

我尝试了Java拆分/子字符串操作。 但是代码行很高。 使用正则表达式如何获取“ 951792 ”值

输出将是

951792
951895
951934 
075041

Answer 1

在这里，我们可能想要简单地使用右边界[mod靠近我们想要的数字，并在我们的第一个捕获组中收集数字，也许与此类似：

([0-9]+)\s\[m 

如果我们愿意，我们可以添加更多边界，例如：

(.+?)([0-9]+)\s\[m.+

演示

测试

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

final String regex = "(.+?)([0-9]+)\\s\\[m.+";
final String string = "2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes\n"
     + "2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e\n"
     + "2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10   \n";
final String subst = "\\2";

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

// The substituted value will be contained in the result variable
final String result = matcher.replaceAll(subst);

System.out.println("Substitution result: " + result);

演示版

 const regex = /(.+?)([0-9]+)\\s\\[m.+/gm; const str = `2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes 2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e 2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10 `; const subst = `$2`; // The substituted value will be contained in the result variable const result = str.replace(regex, subst); console.log('Substitution result: ', result); 

正则表达式

如果不需要此表达式，则可以在regex101.com中对其进行修改或更改。

RegEx电路

jex.im可视化正则表达式：

Answer 2

您可以尝试以下正则表达式：

(?<=[0-9]{6}-[0-9]{2}:[0-9]{2}:[0-9]{2}\.)[0-9]+

别忘了两次逃脱. （ \\\\. ），将其添加到您的Java代码中。

输入：

2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes
2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e
2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10 

火柴：

951792
951895
951934 

演示1

对于使用前瞻性和后瞻性的更严格的正则表达式，请使用：

(?<=[0-9]\]:\s[0-9]{6}-[0-9]{2}:[0-9]{2}:[0-9]{2}\.)[0-9]+(?=\s\[mod=REC)

演示2

Java代码示例：

String input = "2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes\n" + 
                "2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e\n" + 
                "2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10   ";
List<String> matches = new ArrayList<String>();
Matcher m = Pattern.compile("(?<=[0-9]{6}-[0-9]{2}:[0-9]{2}:[0-9]{2}\\.)[0-9]+")
.matcher(input);
while (m.find()) {
    matches.add(m.group());
}
System.out.println(matches);

代码输出：

[951792, 951895, 951934]

Answer 3

//逐行迭代循环。

String line = "2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes";
      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(1) ); //This would give 951792
              }else {
         System.out.println("NO MATCH");
      }

在此处获取正则表达式参考： https : //regex101.com/r/8F0D4w/1