How to do a field extraction on the date string of a Windows source path field?

[shariinPH]

Hi All

I have here a sample source path:

source="G:\\OperationData\\Ntr_RdngANI\\Work_Bill\\Dis_Rec\\03_12\\Fun_20150312_successful.txt"

I want to extact the 20150312

how can I do that?
I already tried

|rex field=source "G:\\OperationData\\Ntr_RdngANI\\Work_Bill\\Dis_Rec\\03_12\\Fun_(?\d+)_successful.txt"

but it doesn't work, I think its about the "\"

can someone help me with this?
Thanks!

[vincenteous]

Hi shariin,

What jeffland said about double backslashes and naming your capturing group was correct. Instead of using the full path as regex pattern, why don't you focus on the file name instead? Something like this:

Fun_(?<DateLog>\d{8})_.*

I've tested it and you can get what you wanted.

[masonmorales]

^Easiest solution

[shariinPH]

Yes! This works! Thanks!! 🙂

[jeffland · Accepted solution]

I believe there are two errors in your expression: a) you need to escape your backslashes, and b) you did not give your capture group a name. I can reccomend https://regex101.com/ to confirm your regular expressions.

In your case, your path contains double backslashes and you need to escape both of them, so your regex looks like | rex field=source "G:\\\\OperationData\\\\Ntr_RdngANI\\\\Work_Bill\\\\Dis_Rec\\\\03_12\\\\Fun_(?\d+)_successful.txt" - the capture in angular brackets being the name of your capture group, and \\\\ as a literal \\.

Hope this helps. And you should really give the page mentioned above a try!

[shariinPH]

Hello @jeffland .. This works too ! 🙂 thanks!! 🙂