CODEWITHSUNDEEP
linuxbashshellsed
user1987607
user1987607

Reputation: 2157

linux bash sed-command with variable

I have a variable in a linux bash ".sh" script

$data="test_1"

now I want to create a new variable ($name) that contains only the part of $data before the underscore, so 

$name="test"

I thought of doing this with sed

name=$(echo "$dataset" | sed 's/_.*//');

but this doesn't seem to work. What am I doing wrong?

Upvotes: 0

Views: 248

Answers (1)

riteshtch
riteshtch

Reputation: 8769

No need to call an external process(sed). Instead you can use shell's parameter substitution like this: 

$ data="test_1"

$ echo "${data%%_*}"
test

${var%%Pattern} Remove from $var the longest part of Pattern that matches the back end(from the right) of $var.

${var%Pattern} for removing shortest pattern

More info on parameter substitution can be found here.

You can store it in a variable like this: 

$ name="${data%%_*}"

$ echo "$name"
test

Upvotes: 5

Related Questions