Get string between two specific characters in unix

Question

I need to get string between 2nd # to .(dot) using unix command. For ex string is TEST#CV01#170403053938.csv output should be 170403053938

Can some one please advice how to get.

RavinderSingh13 · Accepted Answer · 2017-04-12 09:33:42Z

1

try:

A="TEST#CV01#170403053938.csv"
B=${A##*#}
echo ${B%%.*}

I am simply using parameter expansion to get the exact required values here.

answered Apr 12, 2017 at 9:33

RavinderSingh13

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Comments

VIPIN KUMAR · Accepted Answer · 2017-04-12 09:27:17Z

0

Try this -

$ echo "TEST#CV01#170403053938.csv"|egrep -o '[[:digit:]]{10,20}'
170403053938

OR

$ echo "TEST#CV01#170403053938.csv"|awk -F'[#.]' '{print $3}'
170403053938

answered Apr 12, 2017 at 9:27

VIPIN KUMAR

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J. Chomel · Accepted Answer · 2017-04-12 11:06:40Z

0

Could be done with sed (might be overkill), but for the sake of regex!

 echo "TEST#CV01#170403053938.csv"|\
    sed -e 's/\(.*#.*#\)\(.*\)\(\.csv\)/\2/'

answered Apr 12, 2017 at 11:06

J. Chomel

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