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SED Search Pattern and Replace with the Pattern

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# 1  
Old 03-14-2008
SED Search Pattern and Replace with the Pattern
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
# 2  
Old 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
# 3  
Old 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
# 4  
Old 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

# 5  
Old 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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