Issue
I'm trying to use a function to print the type of files and if the file is a directory it will call the function recursively with the list of files from the directory.
however when sending via ls
command the list of the files inside the directory $(file -i filename)
gives me "cannot open `a1' (no such file or directory)" even though the files are there and he got them from ls.
my testing directory is:
a1 - a directory file, gives (no such file or directory)
bunzip2test.txt - a text file, gives (no such file or directory for some reason)
dir1.tar.gz - a compressed file, gives the correct info
t.txt.gz - a compressed file, gives the correct info
zipping.zip - a zip, gives the correct info
my code is:
#! /bin/bash
function main()
{
f=0
printing $@
}
function printing()
{
cFile=($@)
echo "printing all files:"
for i in ${cFile[@]}
do
echo $i "and it's type is: $(file -i $i)"
done
if ((f == 0))
then
f=1
printing $(cd testing && ls)
fi
}
main $*
output:
testing and it's type is: testing: inode/directory; charset=binary
printing all files:
a1 and it's type is: a1: cannot open `a1' (No such file or directory)
bunzip2test.txt and it's type is: bunzip2test.txt: cannot open `bunzip2test.txt' (No such file or directory)
dir1.tar.gz and it's type is: dir1.tar.gz: application/gzip; charset=binary
t.txt.gz and it's type is: t.txt.gz: application/gzip; charset=binary
zipping.zip and it's type is: zipping.zip: application/zip; charset=binary
running cd testing && file -i $(ls)
from the terminal it does work as predicted and identify all of the files correctly
Solution
every time you call printing()
, you call/execute cd testing
... that is why only the first time the command works, and then it does not.. pull the cd testing
before/outside the function OR add the folder name within $(file -i $i)
as $(file -i testing/$i)
Answered By - Ron