SED Search Pattern and Replace with the Pattern

racbern · #1 (**permalink**) 03-14-2008

Hello All,

I have a string "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031", and I just want to extract LLSV1, but I dont get the expected result when using the sed command below.

# echo "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031" | awk '{print $4}' | sed 's/_$LLSV[a-zA-Z0-9]$_/\1/g'

Can you help me with the sed command how can I search a pattern and replace the whole word with the patter?

racbern

Franklin52 · #2 (**permalink**) 03-14-2008

Assuming the pattern is preceded by "R1H_" you can try this:

Code:

echo "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031" | sed 's/.*R1H_\(.*\)_.*/\1/'

Regards

jaduks · #3 (**permalink**) 03-15-2008

Code:

And assuming that, your interest pattern is always before the last field (_ dlimited),

$ echo "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031" | awk '{print $(NF-1)}' FS=_
LLSV1

//Jadu

ghostdog74 · #4 (**permalink**) 03-15-2008

from the sample code you provided, it seems that LLSV? is something nearly constant.

Code:

# echo "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031" | sed 's/.*\(LLSV[a-zA-Z0-9]\).*/\1/'
LLSV1

lijojoseph · #5 (**permalink**) 03-15-2008

If pattern is going to be the same then use this

echo "CP_STATUS OSSRC_R6_0_Shipment_R1H_CU AOM_901046 R1H_LLSV1_2008031" |cut -d ' ' -f4|cut -d _ -f2

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