[linux] Recursively find all files newer than a given time

Given a time_t:

? date -ur 1312603983
Sat  6 Aug 2011 04:13:03 UTC

I'm looking for a bash one-liner that lists all files newer. The comparison should take the timezone into account.

Something like

find . --newer 1312603983

But with a time_t instead of a file.

Maybe someone can use it. Find all files which were modified within a certain time frame recursively, just run:

find . -type f -newermt "2013-06-01" \! -newermt "2013-06-20"

Assuming a modern release, find -newermt is powerful:

find -newermt '10 minutes ago' ## other units work too, see `Date input formats`

or, if you want to specify a time_t (seconds since epoch):

find -newermt @1568670245

For reference, -newermt is not directly listed in the man page for find. Instead, it is shown as -newerXY, where XY are placeholders for mt. Other replacements are legal, but not applicable for this solution.

From man find -newerXY:

Time specifications are interpreted as for the argument to the -d option of GNU date.

So the following are equivalent to the initial example:

find -newermt "$(date '+%Y-%m-%d %H:%M:%S' -d '10 minutes ago')" ## long form using 'date'
find -newermt "@$(date +%s -d '10 minutes ago')" ## short form using 'date' -- notice '@'

The date -d (and find -newermt) arguments are quite flexible, but the documentation is obscure. Here's one source that seems to be on point: Date input formats

You can also do this without a marker file.

The %s format to date is seconds since the epoch. find's -mmin flag takes an argument in minutes, so divide the difference in seconds by 60. And the "-" in front of age means find files whose last modification is less than age.

time=1312603983
now=$(date +'%s')
((age = (now - time) / 60))
find . -type f -mmin -$age

With newer versions of gnu find you can use -newermt, which makes it trivial.

Given a unix timestamp (seconds since epoch) of 1494500000, do:

find . -type f -newermt "$(date '+%Y-%m-%d %H:%M:%S' -d @1494500000)"

To grep those files for "foo":

find . -type f -newermt "$(date '+%Y-%m-%d %H:%M:%S' -d @1494500000)" -exec grep -H 'foo' '{}' \;

You can find every file what is created/modified in the last day, use this example:

find /directory -newermt $(date +%Y-%m-%d -d '1 day ago') -type f -print

for finding everything in the last week, use '1 week ago' or '7 day ago' anything you want

So there's another way (and it is portable to some extent_

(python <<EOF
import fnmatch
import os
import os.path as path
import time

matches = []
def find(dirname=None, newerThan=3*24*3600, olderThan=None):
    for root, dirnames, filenames in os.walk(dirname or '.'):
        for filename in fnmatch.filter(filenames, '*'):
            filepath = os.path.join(root, filename)
            matches.append(path)
            ts_now = time.time()
            newer = ts_now - path.getmtime(filepath) < newerThan
            older = ts_now - path.getmtime(filepath) > newerThan
            if newerThan and newer or olderThan and older: print filepath
    for dirname in dirnames:
        if dirname not in ['.', '..']:
            print 'dir:', dirname
            find(dirname)
find('.')
EOF
) | xargs -I '{}' echo found file modified within 3 days '{}'