CODEWITHSUNDEEP
linuxshellunix
smisra3
smisra3

Reputation: 107

How to extract the value before the last underscore in shell script

I need to extract the value before the last underscore in shell script.

Example:

Input:   first_second_third_mmddyyy.csv
Output: first_second_third

Input: first_second_mmddyy.csv
Output: first_second

Upvotes: 1

Views: 2933

Answers (3)

glenn jackman
glenn jackman

Reputation: 246807

You can just use shell parameter expansion:

while read -r line; do echo "${line%_*}"; done < file
# ...........................^^^^^^^^^^

Upvotes: 1

Mustafa DOGRU
Mustafa DOGRU

Reputation: 4112

You can also use awk or cut as below;

awk -F '_' 'NF{NF-=1};1' file

echo "first_second_third_mmddyyy.csv" | rev | cut -d '_' -f2- | rev

Eg;

$ echo "first_second_third_mmddyyy.csv"  | awk -F '_' 'NF{NF-=1};1'
first second third

NF is a variable containing the number of fields in a line.

NF-=1 is subtracting one from the NF variable, to remove last field

Upvotes: 0

sat
sat

Reputation: 14949

You can use this sed:

sed 's/_[^_]*$//g' file

Test:

$ echo "first_second_third_mmddyyy.csv" | sed 's/_[^_]*$//g'
first_second_third

$ echo "first_second_mmddyy.csv" | sed 's/_[^_]*$//g'
first_second

Upvotes: 5

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