Extracting subset of a string using regex in linux

Question

I have a simple list like following

ss_Asd_n1_455_9_1
ss_Asd_n1_98_9_32
ss_Asd_n1_562_9_145
ss_Asd_n1_1_9_6

Using regex linux I want the list to trimmed like follwing

ss_Asd_n1_455_9
ss_Asd_n1_98_9
ss_Asd_n1_562_9 
ss_Asd_n1_1_9

I tried the following code but it did not print any output

grep '[a-z_]*[a-zA-Z]+(\_)[v0-9]+(\_)[0-9]+(\_)[0-9]'

Kindly guide me. Thanks in advance

Why grep if you need to modify the lines? Use sed -i.bak 's/$.*$_[^_]*$/\1/' file — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 24, 2017 at 9:46
Or sed -i.bak 's/_[0-9]*$//' file. Do you really need to check the format that thoroughly? — Wiktor Stribiżew
– Wiktor Stribiżew, Commented Nov 24, 2017 at 10:00

tripleee · Accepted Answer · 2017-11-24 10:24:50Z

1

Looks like you simply want sed 's/_[^_]*$//' filename

To edit the file in-place, use sed -i (or if you are on BSD / MacOS, sed -i '').

answered Nov 24, 2017 at 10:24

tripleee

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Comments

Calvin Taylor · Accepted Answer · 2017-11-24 10:19:43Z

0

using grep I'd use the perl regex syntax 'P' and match only the matching portion 'o'

echo ss_Asd_n1_455_9_1 | grep -oP '[a-z]*_(?:[a-z]|[A-Z])+_[a-z][0-9]+_[0-9]+_[0-9]+'
ss_Asd_n1_455_9

answered Nov 24, 2017 at 10:19

Calvin Taylor

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