Java 正则表达式日期字符串

Question

我需要帮助创建一个正则表达式来解析以下字符串:

09-22-11 12:58:40       SEVERE       ...ractBlobAodCommand:104           -   IllegalStateException: version:1316719189017 not found in recent history                             Dump: /data1/aafghani/dev/devamir/logs/dumps/22i125840.dump

对我来说最困难的部分是解析日期。我不是真正的 Java 正则表达式专家 - 感谢您的帮助。

Answer 1

The question is a bit misleading as it implies the need to parse the date into java.util.Date object or similar. The real question is how to split up the input data into the desired fields:

• date
• level
• location name & line
• exception name & message
• dump file

这是一种使用正则表达式的解决方案。

String pattern = "^(\\d{2}-\\d{2}-\\d{2} \\d{2}:\\d{2}:\\d{2})" // date
    + "[ ]+(SEVERE|WARNING|INFO|CONFIG|FINE|FINER|FINEST)" // level
    + "[ ]+([^:]+):(\\d+)" // location name, location line
    + "[ ]+-[ ]+([^:]+): (.*?)" // exception name, exception message
    + "[ ]+Dump: ([a-zA-Z0-9\\./]+)" // dump
    + "$";

Pattern regex = Pattern.compile(pattern);
String input = "09-22-11 12:58:40       SEVERE       ...ractBlobAodCommand:104           -   IllegalStateException: version:1316719189017 not found in recent history                             Dump: /data1/aafghani/dev/devamir/logs/dumps/22i125840.dump";
Matcher m = regex.matcher(input);
assertTrue(m.matches());
assertSame(7, m.groupCount());
for (int i = 1; i <= m.groupCount(); i++) {
  System.out.format("[%d] \"%s\"%n", i, m.group(i));
}

输出

[1] "09-22-11 12:58:40"
[2] "SEVERE"
[3] "...ractBlobAodCommand"
[4] "104"
[5] "IllegalStateException"
[6] "version:1316719189017 not found in recent history"
[7] "/data1/aafghani/dev/devamir/logs/dumps/22i125840.dump"