Splunk Search

How to create a regex that removes everything before the second underscore?

gbwilson
Path Finder

I'm trying to create a regex that removes everything before the second underscore in a string. The number of characters before the second underscore varies.

For example:

DR300_Corp_76
BELLOE_MX400_32
AB200_Corp_123

I only want the two or three digits after the second underscore (i.e. 76, 32, 123, etc.)

Tags (2)
0 Karma
1 Solution

somesoni2
Revered Legend

If the end of you string is always numbers, try like this

your current search 
| eval yourfield=replace(yourfield,".+_(\d+)$","\1")

if not, try this

your current search 
| eval yourfield=replace(yourfield,"^([^_]+_){2}(.+)$","\2")

View solution in original post

0 Karma

somesoni2
Revered Legend

If the end of you string is always numbers, try like this

your current search 
| eval yourfield=replace(yourfield,".+_(\d+)$","\1")

if not, try this

your current search 
| eval yourfield=replace(yourfield,"^([^_]+_){2}(.+)$","\2")
0 Karma

gbwilson
Path Finder

Thanks for your help! First one worked great.

0 Karma

gcusello
SplunkTrust
SplunkTrust

Hi gbwilson,
try this

(?<my_field>[^_]*_[^_]*)_.*

test it in https://regex101.com/r/YHRXW9/1
Bye.
Giuseppe

0 Karma
Get Updates on the Splunk Community!

New in Observability Cloud - Explicit Bucket Histograms

Splunk introduces native support for histograms as a metric data type within Observability Cloud with Explicit ...

Updated Team Landing Page in Splunk Observability

We’re making some changes to the team landing page in Splunk Observability, based on your feedback. The ...

New! Splunk Observability Search Enhancements for Splunk APM Services/Traces and ...

Regardless of where you are in Splunk Observability, you can search for relevant APM targets including service ...