用于捕获特定数字的正则表达式
RegEx for capturing particular digits
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 split/substring 操作。但是代码行很高。使用正则表达式如何获得“951792”值
输出将是
951792
951895
951934
075041
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]+
在添加 java 代码时，不要忘记对 . (\.) 进行双重转义。
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
匹配项：
951792
951895
951934
Demo 1
对于同时使用前瞻和后视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)
Demo 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]
//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
在这里，我们可能想简单地使用我们想要的数字旁边的右边界 [mod 并在我们的第一个捕获组中收集数字，可能与此类似：
([0-9]+)\s\[m
如果我们愿意，我们可以添加更多的边界，例如：
(.+?)([0-9]+)\s\[m.+
DEMO
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
const regex = /(.+?)([0-9]+)\s\[m.+/gm;
const str = `2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes
2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e
2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10
`;
const subst = ``;
// The substituted value will be contained in the result variable
const result = str.replace(regex, subst);
console.log('Substitution result: ', result);
正则表达式
如果不需要此表达式，可以在 regex101.com 中对其进行修改或更改。
正则表达式电路
jex.im 可视化正则表达式：
从下面的日志中，我如何单独 grep '951792' 值
2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes
2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e
2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10
我尝试了 java split/substring 操作。但是代码行很高。使用正则表达式如何获得“951792”值
输出将是
951792
951895
951934
075041
您可以尝试以下正则表达式：
(?<=[0-9]{6}-[0-9]{2}:[0-9]{2}:[0-9]{2}\.)[0-9]+
在添加 java 代码时，不要忘记对 . (\.) 进行双重转义。
输入：
2019 May 22 03:32:17.952296 france1v4 sh[4937]: 190522-03:32:17.951792 [mod=REC, lvl=INFO] [tid=26130] Recording A8602096210405800406L200218680503121519 size is 4145956224 bytes
2019 May 22 03:32:17.952387 france1v4 sh[4937]: 190522-03:32:17.951895 [mod=REC, lvl=INFO] [tid=26130] RecordingInfo = fffocap://0x401e
2019 May 22 03:32:17.952466 france1v4 sh[4937]: 190522-03:32:17.951934 [mod=REC, lvl=INFO] [tid=26130] recording_dvr_from_recording_info:physicalSegmentCount=10
匹配项：
951792
951895
951934
Demo 1
对于同时使用前瞻和后视的更严格的正则表达式，请使用：
(?<=[0-9]\]:\s[0-9]{6}-[0-9]{2}:[0-9]{2}:[0-9]{2}\.)[0-9]+(?=\s\[mod=REC)
Demo 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]
//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
在这里，我们可能想简单地使用我们想要的数字旁边的右边界 [mod 并在我们的第一个捕获组中收集数字，可能与此类似：
([0-9]+)\s\[m
如果我们愿意，我们可以添加更多的边界，例如：
(.+?)([0-9]+)\s\[m.+
DEMO
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
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 = ``;
// 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，可以在 regex101.com 中对其进行修改或更改。
正则表达式电路
jex.im 可视化正则表达式：