Vfrdtyky

I'm writing a bash script that needs to delete old files.

It's currently implemented using :

find $LOCATION -name $REQUIRED_FILES -type f -mtime +1 -delete

This will delete of the files older than 1 day.

However, what if I need a finer resolution that 1 day, say like 6 hours old? Is there a nice clean way to do it, like there is using find and -mtime?

Does your find have the -mmin option? That can let you test the number of mins since last modification:

find $LOCATION -name $REQUIRED_FILES -type f -mmin +360 -delete

Or maybe look at using tmpwatch to do the same job. phjr also recommended tmpreaper in the comments.

You could to this trick: create a file 1 hour ago, and use the -newer file argument.

(Or use touch -t to create such a file).

Here is the approach that worked for me (and I don't see it being used above)

$ find /path/to/the/folder -name *.* -mmin +59 -delete > /dev/null

deleting all the files older than 59 minutes while leaving the folders intact.

For SunOS 5.10

 Example 6 Selecting a File Using 24-hour Mode





 The descriptions of -atime, -ctime, and -mtime use the  ter-

 minology n ``24-hour periods''. For example, a file accessed

 at 23:59 is selected by:





   example% find . -atime -1 -print









 at 00:01 the next day (less than 24 hours  later,  not  more

 than one day ago). The midnight boundary between days has no

 effect on the 24-hour calculation.

-mmin is for minutes.

Try looking at the man page.

man find

for more types.

If you do not have "-mmin" in your version of "find", then "-mtime -0.041667" gets pretty close to "within the last hour", so in your case, use:

-mtime +(X * 0.041667)

so, if X means 6 hours, then:

find . -mtime +0.25 -ls

works because 24 hours * 0.25 = 6 hours