So I need to write a bash script that copies all files to a specified directory however I need to rename the files to it's absolute path by replacing the / character with __.
For example if the file zad1.sh is in the directory /home/123456/ the file needs to be renamed to __home__123456__zad1.sh
Any ideas on how to do this?
To get the path of your file :
realpath <file> Replace in bash:
echo "${var//search/replace}" The first two slashes are for making global search. Using just / would only do one replacement.
So your code could be
path=$(realpath zad1.sh) path_replaced=${path//\//__} I believe this will accomplish what you are asking:
#!/bin/bash _from_dir=/path/to/files _to_dir=/path/to/dest for file in "${_from_dir}/"*; do nfile="$(sed 's#/#__#g' <<<"$file")" cp "$file" "${_to_dir}/$nfile" done Set the _from_dir variable to the path where your files are located and the _to_dir variable to the path where you want them copied.
The loop will go through each file in the _from_dir location. nfile will take the full path of the file and replace / with __. Then it will cp the file into the _to_dir path with a name representing the full path of it's origin.
The classic approach would be to use sed:
cp "${filename}" "$(realpath ${filename} | sed s:/:__:g)" The advantage is primarily portability across shells if you won't necessarily always be using bash.
Of course, it also lets you forego the script and just do it with find:
find /base/path -type f -exec cp \{\} `realpath \{\} | sed s:/:__:g` \; Find has sorting and filtering options you can use if you need them to only copy certain files.
edit: That find setup works on one of my systems, but not on the other one. Rather than sort out the difference, I just made it more portable:
find /base/path -type f | sed -e "p; s:/:__:g" | xargs -n2 cp 
find command will not work. The backticks will be evaluated before find and get you basically the equivalent of $PWD/{} with the slashes in $PWD replaced with double underscores. So, if you're in /home/user, this is equivalent to: find /base/path -exec cp {} __home__user__{} \; If you have a file f in /base/path, this will try to copy it to __home__user__/base/path/f. find -exec sh -c 'cp "$1" "$(realpath "$1" | sed s:/:__:g)"' sh {} \; should do xargs will have issues with names with whitespace, unless you use find -print0 | sed -z | xargs -0 (with at least GNU sed). Also any of those will make the copies to the current directory, which may or may not be what you want. /base/path and you just have regular files in there, no directories. However, even then you'd be copying to e.g. /base/path/__base__path__./f. That's not right. I have done by below method
ls -ltr /root/l.txt| awk '{print "cp" " " $NF,$NF}'| awk '{gsub("/","_",$NF);print $1,$2,"/root/p1/"$3}'| sh Here file /root/l.txt will be copied to /root/p1/ path but filename will be _root_l.txt Let me know for any confusion
Here /root/p1==> destination path 
some__really__important_file.txtwhen putting things back not going into/path/to/some/really/important_file.txt?__init__.py,__main__.py, etc.