Shell Programming and Scripting

Hi,

I have a file abc.txt with data like this

1 /test/
2 /test/file.txt
3 /data/
4 /data/file1.txt
5 /data/file2.txt

I want to take out every path from the file and check if its a directory or file.
I am trying it with cut with something like this but it doesnt work

while read chg_path
do
path=`cut -c5- $chg_path`
if [[ -d $path ]]
then
echo $path "directory"
else
echo $path "file"
fi
done < abc.txt

Please help me. Thanks in advance.

Assuming the line numbers are not part of the file, you are removing the directory name but leaving a leading slash. This is probably not what you want. (Why would you want to remove the first directory component anyway?) The trivial fix is to use cut -c6- but I suspect you could get a quite different script if you told us why you are doing this and what you are ultimately trying to achieve.

I guess easy way to solve this to find a / character on end of each line. if it is present I would like to treat the path as an input to another command in the check, otherwise ignore.
For example:
1 /test/ ignore
2 /test/file.txt take /test/file.txt to another command
3 /data/ ignore
4 /data/file1.txt take /data/file1.txt to another command
and so on..
Another command is same for all.
Path starts from 5th position in every line.

So the problem you are trying to figure out is how to skip the ones which end with a slash?

Code:

cut -c5- abc.txt |
while read path; do
  case $path in */) continue;; esac
  test -d "$path" && echo "$path" directory || echo "$path" not a directory
done

PWD=`pwd`
for VAR in `awk '{printf "%s/%s\n","'$PWD'",$2}' FILENAME`
do
test -d "$VAR" && echo "$VAR" is a DIRECTORY || echo "$VAR" is a FILE
done

Try this:

Code:

awk '{ print $2}'  filename | xargs ls -ld | awk ' { if (/^d/) print $NF " is a directory"; else print $NF " is a file"; }'

Thank you! Its solved