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Linux command to output from a string

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# 1  
Old 09-13-2012
Linux command to output from a string
Hi All,

I have a file with name
Quote:
"/iis_ft_dev_data1/custsys/publish/goldstd/WCC_CUST_GS_NON_TRAN_CTA_DLY.dat".
Is there a LINUX command that will help me to output the word after the 9th Underscore(_).
ie the output should be DLY in this case.

Can anybody pls help me.

Thanks much in advance,
Freddie
# 2  
Old 09-13-2012
Code:
echo "/iis_ft_dev_data1/custsys/publish/goldstd/WCC_CUST_GS_NON_TRAN_CTA_DLY.dat" | sed -n 's/.*_\(.*\)\.dat/\1/p'

DLY

This User Gave Thanks to in2nix4life For This Post:
dsfreddie
# 3  
Old 09-13-2012
Code:
echo ' /iis_ft_dev_data1/custsys/publish/goldstd/WCC_CUST_GS_NON_TRAN_CTA_DLY.dat' | awk -F'[_.]' '{print $(NF-1)}'

This User Gave Thanks to vgersh99 For This Post:
dsfreddie
# 4  
Old 09-13-2012
Thanks @vgersh99 & @in2nix4life, both worked fineSmilie

Wondering what we should do to get the last 2 words(before the .) in the output,

ie CTA_DLY

i tried modifying the same commands, but with no luck

Thanks again.
Freddie
# 5  
Old 09-13-2012
Code:
awk -F'[_.]' '{print $(NF-2), $(NF-1)}' OFS=_

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