Bash: How do I check (and return) the results of a command filtered by file content-CodePudding

I executed a command on Linux to list all the files & subfiles (with specific format) in a folder.

This command is:

ls -R | grep -e "\.txt$" -e "\.py$"

In an other hand, I have some filenames stored in a file .txt (line by line).

I want to show the result of my previous command, but I want to filter the result using the file called filters.txt.

If the result is in the file, I keep it
Else, I do not keep it.

How can I do it, in bash, in only one line?

I suppose this is something like:

ls -R | grep -e "\.txt$" -e "\.py$" | grep filters.txt

An example of the file:

# filters.txt

README.txt
__init__.py

EDIT

I am trying to a file instead a list of argument because I get the error:

'/bin/grep: Argument list too long'

CodePudding user response：

You can use comm :

comm -12 <(ls -R | grep -e "\.txt$" -e "\.py$" ) <(cat filters.txt)

This will give you the intersection of the two lists.

EDIT It seems that ls is not great for this, maybe find Would be safer

CodePudding user response：

find . -type f | xargs grep $(sed ':a;N;$!ba;s/\n/\\|/g' filters.txt)

That is, for each of your files, take your filters.txt and replace all newlines with \| using sed and then grep for all the entries. Grep uses \| between items when grepping for more than one item. So the sed command transforms the filters.txt into such a list of items to be used by grep.

CodePudding user response：

grep -f filters.txt -r .

..where . is your current folder.

CodePudding user response：

'/bin/grep: Argument list too long'

I executed a command on Linux to list all the files & subfiles (with specific format) in a folder.

This command is:

ls -R | grep -e ".txt$" -e ".py$" In an other hand, I have some filenames stored in a file .txt (line by line).

I want to show the result of my previous command, but I want to filter the result using the file called filters.txt.