Getting path from find command to variable

Question

So this is my shell script so far, it's very simple
I just want the user to give a directory, then it'll find it and give back the permissions and name of the files in the directory.

echo '#!/bin/bash' > ex1

echo 'echo 'Which is the directory to list?'' >> ex1

echo 'read directory' >> ex1

echo 'path=`find / -type d -name $directory`' >> ex1

echo '$(find $path -printf '%M %P\n')' >> ex1

I've done that so far but when I run it, it gets me an error:

./ex1: line 5: drwxr-xr-x: command not recognized

(ex1 is the name of the shell script and it's inside the folder I'm using to test the shell script)

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Answer 1

echo '$(find $path -printf '%M %P\n')' >> ex1

Get rid of the $(...). Just run the find command directly. You also should fix the inner single quotes, such as by switching them to double quotes.

echo 'find $path -printf "%M %P\n"' >> ex1

Getting path from find command to variable

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