Move all photos from Google takeout into one directory

  mkdir photos
  mv -v **/*.jpg **/*.mp4 photos

  shopt -s globstar
  mkdir photos
  mv -v **/*.jpg **/*.mp4 photos

Zsh:

  for i in **/*.jpg; do
    date=$(exif $i --machine-readable --tag=0x9003 | head -c 10 | tr : -)
    if [[ -n $date ]]; then
      mkdir -p $date
      mv -v $i $date
    else
      mkdir -p Unknown
      mv -v $i Unknown
      # (Check other EXIF values, e.g. DateTimeOriginal.)
    fi
  done

I think you could achieve the same result with just the find command:

`find ./takeout -iname “.jpg” -exec mv {} newdir`

If you're not using Windows, you should probably learn to use some basic shell commands.

Also, line 19 has a bug.

Also, readdir('foo', {withFileTypes: true}) will return a list of directory entries so you can check the type without doing a stat. Also, sync versions are easier.

    fs.readdirSync(DOCS_DIR, {withFileTypes: true}).forEach((file) => {
        if(file.name.startsWith('.') || !file.isDirectory()){ return; }
        ...
    });

    import fs from 'fs';
    import path from 'path';

    copyMedia('Takeout');

    function copyMedia(dir) {
      fs.readdirSync(dir, {withFileTypes: true}).forEach((file) => {
        if (file.isDirectory()) {
          copyMedia(path.join(dir, file.name));
        } else {
          if (file.name.toLowerCase().endsWith('.jpg') || file.name.toLowerCase().endsWith('.mp4')) {
            fs.cpSync(path.join(dir, file.name), path.join('photos', file.name));
          }
        }
      });
    }

    find takeout -name ".jpg" -exec mv {} dest_dir

I think most cameras get to IMG_9999 and cycle back to IMG_0000