（Linux）将变量格式化为列printf

Question

I'm trying to show a file name, size and type in 3 columns. The data just isn't showing correctly and I just can't see why.

Here's my code:

    fileName=$(ls -a ~/dirname)
    fileSize=$(find ~/dirname/* -printf '%s\n';)
    fileType=$(find ~/dirname/* | xargs file |  awk '{print $3}')

    printf " ----------------------------------------------------------\n"
    printf "%15s %15s %15s\n" "Name" "Size" "Type"
    printf " ----------------------------------------------------------\n"
    printf "%15s  %15s  %15s\n" "$fileName" "$fileSize" "$fileType"

The output is showing like: (Output as an image just to avoid stackoverflows formatting)

Answer 1

Your variables each hold the information for all the files, but you're printing them at one time, not separating them. Also, the order of files returned by find is not the same as the order of ls (it sorts them lexicographically).

You need to loop over the files and print each one on a different line:

printf "%15s %15s %15s\n" "Name" "Size" "Type"
printf " ----------------------------------------------------------\n"
ls -a ~/dirname | while read -r fileName
do
    fileSize=$(stat --format=%s "$fileName")
    fileType=$(file "$fileName" | awk '{print $3}')
    printf "%15s  %15s  %15s\n" "$fileName" "$fileSize" "$fileType"
done

Answer 2

I solved it with a little bit of help from Barmar identifying what I was doing wrong:

    for fileName in $(ls -a ~/dirname/*);
    do
            fileSize=$(stat --printf="%s" "$fileName")
            fileType=$(file "$fileName" | awk '{print $3}')
            printf "%15s  %15s  %15s\n" $(basename "$fileName") "$fileSize" "$fileType"
    done