I need a command that will return the top level base directory for a specified path in bash.
I have an approach that works, but seems ugly:
echo "/go/src/github.myco.com/viper-ace/psn-router" | cut -d "/" -f 2 | xargs printf "/%s"
It seems there is a better way, however all the alternatives I've seen seem worse.
Thanks for any suggestions!
One option is using awk:
echo "/go/src/github.myco.com/viper-ace/psn-router" |
awk -F/ '{print FS $2}'
/go
As a native-bash approach forking no subshells and invoking no other programs (thus, written to minimize overhead), which works correctly in corner cases including directories with newlines:
topdir() {
local re='^(/+[^/]+)'
[[ $1 =~ $re ]] && printf '%s\n' "${BASH_REMATCH[1]}"
}
Like most other solutions here, invocation will then look something like outvar=$(topdir "$path").
To minimize overhead even further, you could pass in the destination variable name rather than capturing stdout:
topdir() {
local re='^(/+[^/]+)'
[[ $1 =~ $re ]] && printf -v "$2" '%s' "${BASH_REMATCH[1]}"
}
...used as: topdir "$path" outvar, after which "$outvar" will expand to the result.
Here's a sed possibility. Still ugly. Handles things like ////////home/path/to/dir. Still blows up on newlines.
$ echo "////home/path/to/dir" | sed 's!/*\([^/]*\).*!\1!g'
/home
Newlines breaking it:
$ cd 'testing '$'\n''this'
$ pwd
/home/path/testing
this
$ pwd | sed 's!/*\([^/]*\).*!/\1!g'
/home
/this
If you know your directories will be rather normally named, your and anubhava's solutions certainly seem to be more readable.
not sure better but with sed
$ echo "/go/src/github.myco.com/viper-ace/psn-router" | sed -E 's_(/[^/]+).*_\1_'
/go
This is bash, sed and tr in a function :
#!/bin/bash
function topdir(){
dir=$( echo "$1" | tr '\n' '_' )
echo "$dir" | sed -e 's#^\(/[^/]*\)\(.*\)$#\1#g'
}
topdir '/go/src/github.com/somedude/someapp'
topdir '/home/somedude'
topdir '/with spaces/more here/app.js'
topdir '/with newline'$'\n''before/somedir/somefile.txt'
Regards!
