list items correctly in bash with a space in their name

I have for example following Files:

test1.txt
test2.txt
test space.txt

I use the following command:

output=$(ls -l --time-style=long-iso | awk '{print $8, $6}')

And I get following output:

test1.txt 
test2.txt 
test YYYY-MM-DD

because of the space in its name, I get the wrong output.

Is it possible to get output like this?

"text1.txt" 
"text2.txt" 
"text space.txt"

This is why you never try to parse the output of ls. ls is designed to print to a terminal not to be used in scripts or to be parsed, there are other tools for that. You are also making it more complex yourself since you are using the -l --time-style=long-iso options but then your desired output is only the file name.

To get the list of files and directories in the current directory into a variable, the easiest way is to just use globs:

$ ls
 test1.txt   test2.txt  'test space.txt'
$ files=( * )
$ echo "Files: ${files[@]}"
Files: test1.txt test2.txt test space.txt
$ echo "The third file is '${files[2]}'"
The third file is 'test space.txt'

To have the file names quoted, do:

$ printf '"%s"\n' "${files[@]}"
"test1.txt"
"test2.txt"
"test space.txt"

As you can see, the $files array holds all of the files, with each of them stored as an element in the array. See here for a description of how bash arrays work, but briefly, you can list the entire array using ${array[@]} and individual entries using ${array[N]} where N is a number between 0 (the first element of the array) and X-1 where X is the number of elements in the array. In the example above, we have 3 elements in $files, so you have ${files[0]}, ${files[1]} and ${files[2]}.

Note that this approach even works with files whose name contains a newline (\n):

$ touch 'a file with a '$'\n''newline!'
$ ls
'a file with a '$'\n''newline!'   test1.txt   test2.txt  'test space.txt'
$ files=( * )
$ echo "File is: ${files[0]}"
File is: a file with a 
newline!

Now, if you also want to keep the file's modification date (which is not what your output shows, but is what I think your awk is attempting to do), you can do something like this:

$ files=( * )
$ for file in "${files[@]}"; do
   date=$(stat -c '%y' -- "$file" | cut -d ' ' -f1)
   printf 'Date: %s File: "%s"\n' "$date" "$file"
done
Date: 2023-08-15 File: "a file with a 
newline!"
Date: 2023-08-15 File: "test1.txt"
Date: 2023-08-15 File: "test2.txt"
Date: 2023-08-15 File: "test space.txt"

More reliably, find alone with its -printf action can give you the output you want from both ls and awk ... Like so:

find ! -name . -prune -printf '%P %TY-%Tm-%Td\n'

If you need to assign the output to a parameter, then an array would be the most appropriate while reading the output as NULL delimited (safer for unusual file/directory names with e.g. spaces/tabs/newlines) like so:

$ touch normalfile
$ touch 'filewith'$'\n''newline'
$ touch 'filewith'$'\t''tab'
$ touch 'filewith space'
$
$
$ readarray -d '' files < <(find ! -name . -prune -printf '%P %TY-%Tm-%Td\0')
$
$ printf '%s\n' "${files[@]}"
filewith space 2023-08-15
filewith    tab 2023-08-15
normalfile 2023-08-15
filewith
newline 2023-08-15

Notice that find will include hidden files and directories (Those starting with a dot like .filename) in the output, so if those are not desired then you might want to filter them out.

Please do note that the default separator for awk is a space, so you need to print all the variables, not only $8, but also $9 ... when you want the complete filename.

ls -l --time-style=long-iso test* | awk '{ t=index($0,$7); print substr($0,t+6), $6 }'

It will find the 7th column (the time) which is always 5 positions long, and will print the rest of the line substr($0,t+6), followed by the date $6.

You can make it handle even files whose name contains a newline character if you use the --zero option of ls and set awk's record separator to NULL EDIT2:

luuk@ZES:/mnt/d/TEMP$
luuk@ZES:/mnt/d/TEMP$ touch test$'\n'test.txt
luuk@ZES:/mnt/d/TEMP$ ls -l --time-style=long-iso test*
-rwxrwxrwx 1 luuk luuk 0 2023-08-15 21:04 'test'$'\n''test.txt'
luuk@ZES:/mnt/d/TEMP$ $ ls -l --zero --time-style=long-iso -- * | awk 'BEGIN{RS="\0"}{ t=index($0,$7); print substr($0,t+6), $6 }'
test
test.txt 2023-08-16

/mnt/d/TEMP is my TEMP directory on my Windows11 computer. (the special place for temporary files)

But/Conclusion: This might seem to work, but might also depend on other factors like the fact that this is testen under WSL. So, always TEST, before using anything you find on the internet...

I tried it with:

output=$(
    while IFS= read -r -d '' file; do
        filename=$(basename "$file")
        echo \"$filename\"
    done < <(find . -maxdepth 1 -type f -print0)
)

Now it works and i get the Output i want

"text1.txt" 
"text2.txt" 
"text space.txt"

Rather than using ls, use stat (man stat). You can output file info in any desired format.