Grep command to find a text after the matching pattern until the space

Question

I need a command that grep command to find a text after the matching pattern until the space.

I have a line in a file

shell="/mix/delta/LINUX"

and I need the output only as LINUX.

The below command give the output delta/LINUX,

grep -o 'delta/[^"]*'  file.txt

What I need is only LINUX. Please let me know what to change in this to get the expected output.

Answer 1

With grep that supports PCRE with -P flag:

grep -Po 'delta/\K[^"]*' file.txt

\K = Keep out

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Alternative, pipe the result to another command, e.g. cut or a second grep, e.g.

grep -o 'delta/[^"]*'  file.txt | cut -d/ -f2

Answer 2

You can use sed:

sed '/delta\//!d;s_.*/__;s/"//' file

• /delta\//!d does delete all lines that don't (the !) contain your pattern delta/, just in case there could be other lines in your file
• s_.*/__ substitutes everything upto the slash with nothing, so it gets removed
• s/"// finally removes the trailing double quote

Answer 3

sed -E 's|(.*=./.*/.*/)(.*).|\2|'

Match and print the second group in this case LINUX

Answer 4

You can use awk and tell it to use both / and " as field separators. e.g.

$ echo 'shell="/mix/delta/LINUX"' | awk -F'[/"]' '/^shell=/ {print $5}'
LINUX

LINUX is field 5 because awk counts even empty fields (i.e. the empty field between the first " and the first /, and the empty field after the final ")

If you don't know how long the pathname is going to be but know you're always going to be the last path element before the closing double-quote, you can use $(NF-1) to output the second-last field:

$ echo 'shell="/mix/delta/LINUX"' | awk -F'[/"]' '/^shell=/ {print $(NF-1)}'
LINUX