How do I delete file by filename that are set as dates?

Question

I want to delete file not by date access or created, but by filename. The filenames will be dates and I want to have a cronjob run once a week that will purge filename dates older than 7 days. I could do a

find /my/directory -type f -name '*file-name.yyyy-mm-dd.qz' -delete

But I would have to change the script on a weekly basis to run. I would like to avoid having to modify the job every week.

Answer 1

Here is a more robust form that correctly handles spaces (or even newlines) in filenames and directory names.

find . -type f -name '*.[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9].qz' -exec sh -c 'fdate="${1%.qz}"; fdate="${fdate##*.}"; [ "$fdate" "<" "$(date +%F -d "7 days ago")" ] && rm "$1"' find-sh {} \;

---

This involves a lot of shell trickery that might look alien to some people, so let's break it down:

Starting in the current directory, recursively find all regular files...

find . -type f

...whose names end in the exact pattern ".YYYY-MM-DD.qz"...

-name '*.[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9].qz'

...then, run a shell command on each matching file (note the single quotes)...

-exec sh -c '

...which first strips off the trailing ".qz"...

fdate="${1%.qz}";

...then strips off the leading extra part, leaving only "YYYY-MM-DD"...

fdate="${fdate##*.}";

...and compares that string to see if it sorts (lexically) earlier than "YYYY-MM-DD" of the date seven days ago...

[ "$fdate" "<" "$(date +%F -d "7 days ago")" ]

...and if so, removes the file...

&& rm "$1"'

...and we'll use "find-sh" as the "script name" (i.e. $0) to be used for error reporting...

find-sh

...and set the filename found by find to parameter one ($1) of the inline shell script.

{} \;

Answer 2

I can't post a comment, but here is the command to create empty test files:

for i in {0..50};do touch blah.`date "+%Y-%m-%d" -d "$i days ago"`.qz  ;done

I believe Dave's answer does the trick. (I needed to complete the for command with '; do')

A great one liner for cron.

Answer 3

You can use this:

find /my/directory -type f -name "*file-name.$(date -d '7 days ago' +%Y-%m-%d).qz" -delete 

This lets the shell tell us what date was 7 days a go: $(date -d '7 days ago' +%Y-%m-%d)

For example: if today is 2016-06-06 this will be: 2016-05-30

So, whenever your command is executed, that will take the date 7 days ago and so on.

I hope you understand me. If you need something, just comment.

Answer 4

I'll recommend a short script, basically gets all the filenames that have a date like format, then loops through them, checking if seven days ago is later than the date found in the filename, and if so, deletes it.

The date parsing basically makes an integer in the format YYYYMMDD and compares using that, which should work fine since the larger units have a higher factor of ten.

This is a rough script off the top of my head that I didn't get a chance to test.

for file in $(find /my/directory -type f -name "*file-name.[0-9][0-9][0-9][0-9]-[0-9][0-9]-[0-9][0-9].qz")
    if [ $(date -d '7 days ago' +"%Y%m%d") -ge $(date -d $(echo $(basename $file) | cut -d '.' -f 2) +"%Y%m%d") ]; then
        rm $file
    fi
 done