Looping through 'find' files and 'grep' to file character set result (standard input)

Question

I'm using find to analyze a list of files. I want it to identify the files that are not ASCII files.

So this is what I've tried on the command line:

for me in `find 2022*`; do file -i $me | grep -L 'ascii'  ; done;

Output:

(standard input)
(standard input)
(standard input)
(standard input)
(standard input)
(standard input)
(standard input)

The number of standard input, which is 7, correctly indicates that it actually finds the right files out of the 200 submitted, but I need the name of the file itself.

How do I do this?

Answer 1

file is already printing the filename, you have to cut it after a simple grep. If your files are into the same directory:

file -i * | grep -v 'charset.*ascii' | cut -d: -f1

And if you match files into more subdirectories, into a bash shell:

shopt -s globstar
file -i 2022** | grep -v 'charset.*ascii' | cut -d: -f1

Or using find:

find 2022* -type f -exec sh -c 'file -i "$@" | grep -v 'ascii' | cut -d: -f1' sh {} +

---

There are some edge cases, the most probable could be a colon into a filename (: is used by file and grep to separate the filename from the rest) or filenames with newlines or a filename matching charset.*ascii while the file is not ascii. Here is another version to handle these cases also (assuming null separation supported):

file -00i * | awk -v RS='\0' -v ORS='\n' 'NF%2{f=$0;next} !/ascii/{print f}'

file with -00 puts a null byte at the end of the filename and at the end of the line. So we test only even lines, and if it doesn't match, we print the previous line (the filename).

Answer 2

You're piping data into grep's stdin, so grep does not have a filename to print.

I'm assuming 2022* are the filenames, not directories.

find . -type f -name '2022*' -exec sh -c '
    for file; do
        file -bi "$file" | grep -q ascii || echo "$file"
    done
' sh '{}' +