linux-shell: renaming files to creation time

Question

In the linux shell:

for f in *.jpg
do
    mv -n "$f" "$(date -r "$f" +"%Y%m%d_%H%M%S").jpg"
done

Explanation:

for f in *.jpg do

This starts the loop over all jpeg files. A feature of this is that it will work with all file names, even ones with spaces, tabs or other difficult characters in the names.
mv -n "$f" "$(date -r "$f" +"%Y%m%d_%H%M%S").jpg"

This renames the file. It uses the -r option which tells date to display the date of the file rather than the current date. The specification +"%Y%m%d_%H%M%S" tells date to format it as you specified.

The file name, $f, is placed in double quotes where ever it is used. This assures that odd file names will not cause errors.

The -n option to mv tells move never to overwrite an existing file.
done

This completes the loop.

For interactive use, you may prefer that the command is all on one line. In that case, use:

for f in *.jpg; do mv -n "$f" "$(date -r "$f" +"%Y%m%d_%H%M%S").jpg"; done

To name the file based on the EXIF Create Date (instead of the file system date), we need exiftool or equivalent:

for f in *.jpg
do
    mv -n "$f" "$(exiftool -d "%Y%m%d_%H%M%S" -CreateDate "$f" | awk '{print $4".jpg"}')"
done

Explanation:

The above is quite similar to the commands for the file date but with the use of exiftool and awk to extract the EXIF image Create Date.

The exiftool command provides the date in a format like:
```
$ exiftool -d "%Y%m%d_%H%M%S"  -CreateDate sample.jpg
Create Date                     : 20121027_181338
```
The actual date that we want is the fourth field in the output.
We pass the exiftool output to awk so that it can extract the field that we want:
```
awk '{print $4".jpg"}'
```
This selects the date field and also adds on the .jpg extension.