linux-shell: renaming files to creation time


Good morning everybody,

for a website I’d like to rename files(pictures) in a folder from “1.jpg, 2.jpg, 3.jpg …” to “yyyymmdd_hhmmss.jpg” – so I’d like to read out the creation times an set this times as names for the pics. Does anybody have an idea how to do that for example with a linux-shell or with imagemagick?

Thank you!


Naming based on file system date

In the linux shell:

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


  • for f in *.jpg

    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

Naming based on EXIF Create Date

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


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.

Answered By – John1024

This Answer collected from stackoverflow, is licensed under cc by-sa 2.5 , cc by-sa 3.0 and cc by-sa 4.0

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