Question

I need help creating a regular expression that will parse the following string :

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

The most difficult part for me is parsing out the date. I’m not really an expert on Java regular expressions – any help is appreciated.

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

This is one solution using a regular expression.

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));
}

Output

[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"