如何从Shell中的路径列表中删除文件名

Question

I would like to remove a file name only from the following configuration file.

Configuration File -- test.conf

knowledgebase/arun/test.rf
knowledgebase/arunraj/tester/test.drl
knowledgebase/arunraj2/arun/test/tester.drl

The above file should be read. And removed contents should went to another file called output.txt

Following are my try. It is not working to me at all. I am getting empty files only.

#!/bin/bash
file=test.conf
while IFS= read -r line
do
#       grep --exclude=*.drl line
#       awk 'BEGIN {getline line ; gsub("*.drl","", line) ; print line}'
#       awk '{ gsub("/",".drl",$NF); print line }' arun.conf
#       awk 'NF{NF--};1' line arun.conf
echo $line | rev | cut -d'/' -f 1 | rev >> output.txt
done < "$file"

Expected Output :

knowledgebase/arun
knowledgebase/arunraj/tester
knowledgebase/arunraj2/arun/test

Answer 1

There's the dirname command to make it easy and reliable:

#!/bin/bash
file=test.conf
while IFS= read -r line
do
    dirname "$line"
done < "$file" > output.txt

There are Bash shell parameter expansions that will work OK with the list of names given but won't work reliably for some names:

file=test.conf
while IFS= read -r line
do
    echo "${line%/*}"
done < "$file" > output.txt

There's sed to do the job — easily with the given set of names:

sed 's%/[^/]*$%%' test.conf > output.txt

It's harder if you have to deal with names like /plain.file (or plain.file — the same sorts of edge cases that trip up the shell expansion).

You could add Perl, Python, Awk variants to the list of ways of doing the job.

Answer 2

Using awk one liner you can do this:

awk 'BEGIN{FS=OFS="/"} {NF--} 1' test.conf

Output:

knowledgebase/arun
knowledgebase/arunraj/tester
knowledgebase/arunraj2/arun/test

Answer 3

You can get the path like this:

  path=${fullpath%/*}

It cuts away the string after the last /