Using regular expressions for my error log in Java

Question

I'm trying to use regular expressions to parse text like this:

'''ErrorID:  951574305
Time:     Mon Apr 25 16:01:34 CEST 2011
URL:      /documents.do
HttpCode: null
Error:    class java.lang.NullPointerException: null'''

Where keywords ErrorID: , Time: , URL: are always the same and I need to search for them. How do I parse this text?

Seems overkill for regex... you could just split on newline then colon and trim whitespace. — Dan Breen
– Dan Breen, Commented Apr 25, 2011 at 18:24

Gabi Purcaru · Accepted Answer · 2011-04-25 18:30:17Z

1

import re
re.findall("ErrorID:\s+(.*)", text)
# ['951574305']
re.findall("Time:\s+(.*)", text)
# ['Mon Apr 25 16:01:34 CEST 2011']
re.findall("URL:\s+(.*)", text)
# ['/documents.do']

The regex works this way: it matches on ErrorID:(or other delimiter) plus some spaces, plus the rest of the string until the newline/end of string. Then it returns that "something" after the whitespace. Also, the result will be a list in which you will need the first item. There can be other strategies of finding what you need, but I found this the most appropriate.

edited Apr 25, 2011 at 18:30

answered Apr 25, 2011 at 18:24

Gabi Purcaru

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1 Comment

Natta Over a year ago

i need just matching regexp for this pattern , i don't care about values.

General Grievance · Accepted Answer · 2023-05-03 01:17:26Z

0

If your implementation supports named groups...

/ErrorID:\s+(?<ID>.*)\nTime:\s+(?<Time>.*)\nURL:\s+(?<URL>.*)/g

You can then reference them by name.

Otherwise by index

/ErrorID:\s+(.*)\nTime:\s+(.*)\nURL:\s+(.*)/g

$1 for ID, $2 for Time and $3 for URL.

edited May 3, 2023 at 1:17

General Grievance

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answered Apr 26, 2011 at 19:04

Paul Alexander

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Comments

user557597 · Accepted Answer · 2011-04-25 22:22:29Z

0

If you require all of these in the string and don't know where they are and can use lookahead assertions:

(?=[\S\s]*ErrorID:)(?=[\S\s]*Time:)(?=[\S\s]*URL:)

answered Apr 25, 2011 at 22:22

user557597

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