将值解析为Shell Bash脚本

Question

I am trying to make a simple bash script for a command so i can make it reusable for me the command has been taken from another post to convert videos with ffmpeg then upload.

Here is what i have so far.

#!/bin/bash
MOVIES=$1
EXT=$2
BUCKET=$3
find "$MOVIES" -name "*.$EXT" -exec sh -c 'ffmpeg -i "$0" -c:v libx264 -crf 22 -c:a libfaac -movflags faststart "${0%%.mov}.mp4" && s3cmd put "${0%%.mov}.mp4" "$BUCKET""$(basename "${0%%.mov}.mp4")"' {} \;
exit;

I am having issues with the third var if i run it like this.

sh batch.sh ~/Documents/screengrabs/ mov s3://bucket/

I am getting a error it encodes properly.

ERROR: Parameter problem: Destination must be S3Uri. Got: file://49A22352-9F41-48B9-BF97-610CBF699025-630-0000055828D0D55F.mp4

This means it is not parsing the $3 parameter $BUCKET properly.

Any help this is my first bash script attempt.

Thanks

Update still not working

#!/bin/bash
MOVIES="$1"
EXT="$2"
BUCKET="$3"

find "$MOVIES" -name "*.${EXT}" -exec sh -c 'ffmpeg -i "$0" -c:v libx264 -crf 22 -c:a libfaac -movflags faststart "${0%%.mov}.mp4" && s3cmd put "${0%%.mov}.mp4" "${BUCKET}/sam.mp4"' {} \;

exit;

WORKING

#!/bin/bash
MOVIES=$1
EXT=$2
export BUCKET=$3
find "$MOVIES" -name "*.$EXT" -exec sh -c 'ffmpeg -i "$0" -c:v libx264 -crf 22 -c:a libfaac -movflags faststart "${0%%.mov}.mp4" && s3cmd put "${0%%.mov}.mp4" "$BUCKET""$(basename "${0%%.mov}.mp4")"' {} \;
exit;

Answer 1

The problem is that $0 is the name of the script that is executed (in this case a temporary file created by sh from the argument to -c ), not the first argument to the resulting script. Note that the error message does not mention a file with sam in the name, so it is being generated by the ${0%%.mov}.mp4 argument to s3cmd , not the one that uses $BUCKET . Replace it with $1 . Also, you need to export BUCKET so that the shell started by find has access to it. (The literal string $BUCKET , not its value, is passed to sh via its -c option because the string is rightly single-quoted.)

(UPDATE: as Etan pointed out, $0 is correct.)

#!/bin/bash
MOVIES=$1
EXT=$2
export BUCKET=$3
find "$MOVIES" -name "*.$EXT" -exec sh -c 'ffmpeg -i "$0" -c:v libx264 -crf 22 -c:a libfaac -movflags faststart "${1%%.mov}.mp4" && s3cmd put "${0%%.mov}.mp4" "$BUCKET""$(basename "${0%%.mov}.mp4")"' {} \;
exit;

In bash 4, however, I would just write

#!/bin/bash
MOVIES=$1
EXT=$2
BUCKET=$3

shopt -s globstar
for movie in "$MOVIES"/**/*."$EXT"; do
    mp4movie=${movie%%.mov}.mp4
    ffmpeg -i "$movie" -c:v libx264 -crf 22 -c:a libfaac -movflags faststart "$mp4movie" &&
      s3cmd put "$mp4movie" "$BUCKET/$mp4movie"
done

This is much easier to get right than trying to cram a small script into a single string for sh -c .

---

Manual recursive search with bash 3.2:

process_file () {
    movie=$1
    mp4movie=${movie%%.mov}.mp4
    ffmpeg -i "$movie" -c:v libx264 -crf 22 -c:a libfaac -movflags faststart "$mp4movie" &&
      s3cmd put "$mp4movie" "$BUCKET/$mp4movie"
}

process_directory () {
    dir=$1
    for name in "$dir"/*; do
        if [[ -d "$name" ]]; then
            process_directory "$name"
        elif [[ -f $name && $name = *.$EXT ]]; then 
            process_file "$name"
        fi
    done
}

MOVIES=$1
EXT=$2
BUCKET=$3
process_directory "$MOVIES"

Answer 2

@chepner's answer about using a "native" shell script is a good one. That being said, and as I just posted in a comment on the OP, I believe the issue here is simply a variable scope issue.

If you edit the script to use "$1" instead of "$BUCKET" in the script passed to sh -c and add "$BUCKET" between the {} and \\; I believe the script will work (though I haven't fully evaluated the script in other respects).