How to remove a filename from the list of path in Shell - bash

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

You can get the path like this:
path=${fullpath%/*}
It cuts away the string after the last /

Answer 3

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