Variable within path to file in bash script

Question

I've got a file, that contains about 20k strings. Each string in that file should be an argument for a command within bash script.

At first my script generates a file with this strings.

find /home/alotoffolders -type f -name "*.mp4" | grep some_greps > ~/tests/find_smth 

After that trying to set a variable, that will be a part of argument

filename=$(cut -d/ -f11 ~/tests/find_smth) 

Then I'm trying to read all file line by line and use every string as an input argument (seems like it works), but for output - there is some issue.

 for i in `cat ~/tests/find_smth`; do ./other_script -input $i -output /home/folder1/folder2/$filename; done 

Script can't see the static path before $filename.

Where is the problem?

Answer 1

• NEVER usefor i in $(cat ...) ... or similar. It loops over words, not lines. And even if it would, filenames are allowed to contain newlines.
• Use shell parameter expansion or basename command instead of cut to get the file name.
• Instead of using grep on the results of find, you might use e.g. find's -regex or -iregex option.

---

Use find -exec with sh -c:

find /home/alotoffolders -type f -iregex 'some_pattern.mp4' \ -exec sh -c ' /path/to/other_script -input "$1" -output "/home/folder1/folder2/${1##*/}" ' find-sh {} \;