How to remove everything from path after a word using sed in bash/ shell script?

Question

Where the word is repeated and only wanted to be removed from specific word location

Lets say my path is - /opt/xyz/config/config.xml

Solution I want after using sed is: /opt/xyz/config/

how can this be obtained?

I am sick of using {sed 's/config.*//'} >> This actually removes both config words such as it looks

/opt/xyz/

I have tried using this in multiple ways

>  sed 's/config.*//'

Only when the filename matches the previous directory? Or are you looking to always remove the filename part of the path? It's not clear what your rule is. Your `sed` fails because you are replacing the match of 'config' and anything after with nothing. — Gary_W, Jan 11 '23 at 22:18
See: [How to remove end folder name from a path in Linux script?](https://stackoverflow.com/q/29329093/3776858) — Cyrus, Jan 12 '23 at 00:25

score 2 · Accepted Answer · answered Jan 11 '23 at 22:04

2

Maybe with something like this?

sed 's/[^/]*$//'

But if the filepath is in a shell variable then you might as well use:

mydir=${myfilepath%/*}/

answered Jan 11 '23 at 22:04

Fravadona

score 1 · Answer 2 · answered Jan 11 '23 at 22:39

1

Another potential solution, depending on your use-case, is the dirname bash function, e.g.

dirname /opt/xyz/config/config.xml
/opt/xyz/config

answered Jan 11 '23 at 22:39

jared_mamrot

score 0 · Answer 3 · answered Jan 11 '23 at 23:14

0

Using BASH parameter expansion:

p="/opt/xyz/config/config.xml"; echo "${p%/*}"
/opt/xyz/config

answered Jan 11 '23 at 23:14

j_b

3 Answers3