linux-shell: renaming files to creation time - linux

Question

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!

Answer 1

Naming based on file system date
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
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
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.

Answer 2

Thanks to #John1024 !
I needed to rename files with different extensions in the same time, according to last modification date :
for f in *; do
fn=$(basename "$f")
mv "$fn" "$(date -r "$f" +"%Y-%m-%d_%H-%M-%S")_$fn"
done
"DSC_0189.JPG" ➜ "2016-02-21_18-22-15_DSC_0189.JPG"
"MOV_0131.avi" ➜ "2016-01-01_20-30-31_MOV_0131.avi"
If you don't want to keep original filename :
mv "$fn" "$(date -r "$pathAndFileName" +"%Y-%m-%d_%H-%M-%S")"
Hope it helps noobs as me !

Answer 3

Try this
for file in `ls -1 *.jpg`; do name=`stat -c %y $file | awk -F"." '{ print $1 }' | sed -e "s/\-//g" -e "s/\://g" -e "s/[ ]/_/g"`.jpg; mv $file $name; done
Though there might be an easier way.

Answer 4

I created a shell script; I think it's mac only, linux might need other arguments.
#!/bin/bash
BASEDIR=$1;
for file in `ls -1 $BASEDIR`; do
TIMESTAMP=`stat -f "%B" $BASEDIR/$file`;
DATENAME=`date -r $TIMESTAMP +'%Y%m%d-%H%M%S'`-$file
mv -v $BASEDIR/$file $BASEDIR/$DATENAME;
done
when called with a directory path, moves all files in that directory to prepend the creation date of that file, like
../camera/P1210232.JPG -> ../camera/20220121-103456-P1210232.JPG

Answer 5

Change filename based on file creation time:
exiftool "-filename<FileCreateDate" -d %Y%m%d_%H%M%S%z%%-c.%%le input.jpg