从Bash脚本中的文件参数打印权限

Question

I'm having trouble reading the permissions of file arguments. I looks like it has something to do with hidden files but I'm not sure why.

Current Code:

#!/bin/bash

if [ $# = 0 ]
then
    echo "Usage ./checkPerm filename [filename2 ... filenameN]"
    exit 0
fi

for file in $@
do
    ls -l | grep $file | cut -f1 -d' '

    # Do Something

done

I can get the permissions for each input, but when a hidden file is run through through the loop it re-prints the permissions of all files.

-bash-4.1$ ll test*
-rw-r--r-- 1 user joe 0 Nov 11 19:07 test1
-r-xr-xr-x 1 user joe 0 Nov 11 19:07 test2*
-r--r----- 1 user joe 0 Nov 11 19:07 test3
-rwxr-x--- 1 user joe 0 Nov 11 19:07 test4*
-bash-4.1$ ./checkPerm test*
-rw-r--r--
-rw-r--r--
-r-xr-xr-x
-r--r-----
-rwxr-x---
-r--r-----
-rw-r--r--
-r-xr-xr-x
-r--r-----
-rwxr-x---
-bash-4.1$

What is going on in the loop?

Answer 1

It's your grep:

ls -l | grep 'test2*'

This will grep out anything starting with test since you're basically asking for anything starting with test that might end with 0 or more 2 s in it, as specified by the 2* .

To get your intended result, simply remove your loop and replace it with this:

ls -l "$@" | cut -d' ' -f1

Or keep your loop, but remove the grep:

ls -l $file | cut -d' ' -f1

Also, technically, none of those files are hidden. Hidden files in bash start with . , like .bashrc .

Answer 2

When you do the ls -l inside the loop and then grep the results, if there are files that contain test1 in the name, but not at the start, they are selected by the grep , giving you extra results. You could see that by doing:

ls -l | grep test

and seeing that there are many more entries than the 4 you get with ls -l test* .

Inside your loop, you should probably use just:

ls -ld "$file" | cut -d' ' -f1