Looping over grep results with Zsh - ssh

Question

I need to loop through a directory the first time I log into a new VM and set permissions on WSL shared folders. The SSH agent will not add private keys unless they have 0700 permissions set.
How do I use grep to get a list of file names and then loop through them?
I tried:
for file $(grep -lr "-----BEGIN OPENSSH PRIVATE KEY-----")
do
echo "Working on File: $file"
chmod 700 $file
done
The results were a mess. The grep command is being resolved into a single string of file names. Any mapped directories that have spaces in them (from the Win11 side of the house) messed things up further.

Answer 1

Suggesting to accomplish the task in one line:
chmod -v 700 $(grep -lr "-----BEGIN OPENSSH PRIVATE KEY-----" $PWD)
Explanation:
-v : Print verbose report on each changed file.
grep -lr "-----BEGIN OPENSSH PRIVATE KEY-----" $PWD : List all matched files to be changed, from all files under current directory.
Note: also try to replace $PWD with $HOME.

Answer 2

Here is my solution:
grep will output filenames with a null character after each item with the "-Z" flag.
Zsh can take that null delimiter and use it in an expanded string with the "0" flag.
The final result:
for file in ${(0o)"$(grep -lrZ "-----BEGIN OPENSSH PRIVATE KEY-----" *)"}
do
echo "Working on $file"
chmod 700 $file
done
A detailed explanation for posterity:
Zsh will execute anything inside () in a subshell. so the results of grep are in a nameless variable $().
Zsh does variable/parameter expansion with ${}. Anything inside gets expanded like hitting "Tab" at the command prompt.
We know that grep is going to return file paths with spaces in them. Therefore we can't use whitespace to expand our file list. We can use flags to make it happen with grep and Zsh using null characters as delimiters instead of spaces.
Zsh flags are specified immediately after the curly braces are opened up.
${(0o)$foo} will result in $foo being expanded into an array of items, splitting $foo on nulls because of the "0" and sorting them in ascending order because of the "o" flag.
We need the extra step of enclosing our grep results in double-quotes. ${(0o)"$foo"} We are after a string that contains both spaces and nulls. We are going to split the string on the nulls and preserve the spaces in the path names.
You could sort them in descending order with a capitalized "O" if you wanted to do it that way instead.
grep -lrZ will look for the string "-----BEGIN OPENSSH PRIVATE KEY-----" in all files found with the * wildcard.
The "r" is recursively look in sub-directories
The -l outputs just the filename
And -Z outputs a null separator at the end of each result.
So you end up with grep finding the files, Zsh expands the list and then loops through each of the items in the returned array.
As a side note in WSL, you have to turn on metadata in order to mount a windows folder and set Linux permissions. It is a nice solution to keep your ssh keys in a secure, backed-up location and use them on Linux images.

Answer 3

Iterate over all the files, but only operate on the ones that grep succeeds on. (You probably don't have that many files; don't worry so much over how many times grep has to run.)
for f in *; do
if grep -q "-----BEGIN OPENSSH PRIVATE KEY-----" "$f"; then
echo "Working on File: $f"
chmod 700 $f
fi
done
And because this is zsh, you can embed the call to grep in the glob:
is_key () {
grep -q "-----BEGIN OPENSSH PRIVATE KEY-----" "$REPLY"
}
for f in *(+is_key); do
echo "Working on File: $f"
chmod 700 "$f"
done
or even dispense with the explicit loop
chmod 700 *(+is_key)
(The shell function isn't actually necessary, but it does help with the readability of the glob qualifier.)