Suppose I've got a directory that looks like:
-rw-r--r-- 1 some-user wheel 0 file1
-rw-r--r-- 1 some-user wheel 257 file2
-rw-r--r-- 1 some-user wheel 0 file3
-rwxr-xr-x 1 some-user wheel 212 file4
-rw-r--r-- 1 some-user wheel 2012 file5
.... more files here.
If it's relevant, assume that the names of the files are more random than just file#.
How do I remove only the files that are empty (meaning that the file has 0 bytes in it) in a directory, using rm and grep or sed in some form?
The easiest way is to run find with -empty test and -delete action, e.g.:
find -type f -empty -delete
The command finds all files (-type f) in the current directory and its subdirectories, tests if the matched files are empty, and applies -delete action, if -empty returns true.
If you want to restrict the operation to specific levels of depth, use -mindepth and -maxdepth global options.
The command is:
cd DirectoryWithTheFiles
rm -f $(find . -size 0)
I have a chache directory setup on Amazon EBS. I'm using it as a cache for an S3FS mounted file system that holds virtual backup tapes. The tapes are being used with Bacula.
The backup to the S3 mounted directory would be way to slow to be usable without some form of local cache. The storage on S3 is of course virtually limitless. So I need to clear out the /cache directory every so often.
I want to be able to delete tape files in that directory that are older than 15 minutes.
So I tried this command:
[root#ops:~] #find /cache/jf-backup/ -type f -daystart -mmin +15
/cache/jf-backup/jf-backup-tape-0073
/cache/jf-backup/jf-backup-tape-0074
And it does find the files. Howewver they are not older than 15 minutes:
[root#ops:~] #ls -l /cache/jf-backup/
total 6199968
-rw-------. 1 root root 5368688607 Feb 25 14:39 jf-backup-tape-0073
-rw-------. 1 root root 980074496 Feb 25 14:42 jf-backup-tape-0074
[root#ops:~] #date
Thu Feb 25 14:46:59 EST 2016
How can I get the find command to only find files that are older than 15 minutes? Once I do I want to delete those files with a command like this:
find /cache/jf-backup/ -type f -daystart -mmin +15 -exec rm -rf {} \;
From #Jan at https://unix.stackexchange.com/questions/155184/how-to-find-and-delete-files-older-than-specific-days-in-unix, try
find /PATH/TO/FILES -type f -mmin +15 -exec rm -f {} +
I'm trying to find files modified in last x minutes, for example in the last hour. Many forums and tutorials on the net suggest to use the find command with the -mmin option, like this:
find . -mmin -60 |xargs ls -l
However, this command did not work for me as expected. As you can see from the following listing, it also shows files modified earlier than 1 hour ago:
-rw------- 1 user user 9065 Oct 28 23:13 1446070435.V902I67a5567M283852.harvester
-rw------- 1 user user 1331 Oct 29 01:10 1446077402.V902I67a5b34M538793.harvester
-rw------- 1 user user 1615 Oct 29 01:36 1446078983.V902I67a5b35M267251.harvester
-rw------- 1 user user 72365 Oct 29 02:27 1446082022.V902I67a5b36M873811.harvester
-rw------- 1 user user 69102 Oct 29 02:27 1446082024.V902I67a5b37M142247.harvester
-rw------- 1 user user 2611 Oct 29 02:34 1446082482.V902I67a5b38M258101.harvester
-rw------- 1 user user 2612 Oct 29 02:34 1446082485.V902I67a5b39M607107.harvester
-rw------- 1 user user 2600 Oct 29 02:34 1446082488.V902I67a5b3aM465574.harvester
-rw------- 1 user user 10779 Oct 29 03:27 1446085622.V902I67a5b3bM110329.harvester
-rw------- 1 user user 5836 Oct 29 03:27 1446085623.V902I67a5b3cM254104.harvester
-rw------- 1 user user 8970 Oct 29 04:27 1446089232.V902I67a5b3dM936339.harvester
-rw------- 1 user user 165393 Oct 29 06:10 1446095400.V902I67a5b3eM290158.harvester
-rw------- 1 user user 105054 Oct 29 06:10 1446095430.V902I67a5b3fM265065.harvester
-rw------- 1 user user 1615 Oct 29 06:24 1446096244.V902I67a5b40M55701.harvester
-rw------- 1 user user 1620 Oct 29 06:24 1446096292.V902I67a5b41M337769.harvester
-rw------- 1 user user 10436 Oct 29 06:36 1446096973.V902I67a5b42M707215.harvester
-rw------- 1 user user 7150 Oct 29 06:36 1446097019.V902I67a5b43M415731.harvester
-rw------- 1 user user 4357 Oct 29 06:39 1446097194.V902I67a5b56M446687.harvester
-rw------- 1 user user 4283 Oct 29 06:39 1446097195.V902I67a5b57M957052.harvester
-rw------- 1 user user 4393 Oct 29 06:39 1446097197.V902I67a5b58M774506.harvester
-rw------- 1 user user 4264 Oct 29 06:39 1446097198.V902I67a5b59M532213.harvester
-rw------- 1 user user 4272 Oct 29 06:40 1446097201.V902I67a5b5aM534679.harvester
-rw------- 1 user user 4274 Oct 29 06:40 1446097228.V902I67a5b5dM363553.harvester
-rw------- 1 user user 20905 Oct 29 06:44 1446097455.V902I67a5b5eM918314.harvester
Actually, it just listed all files in the current directory. We can take one of these files as an example and check if its modification time is really as displayed by the ls command:
stat 1446070435.V902I67a5567M283852.harvester
File: ‘1446070435.V902I67a5567M283852.harvester’
Size: 9065 Blocks: 24 IO Block: 4096 regular file
Device: 902h/2306d Inode: 108680551 Links: 1
Access: (0600/-rw-------) Uid: ( 1001/ user) Gid: ( 1027/ user)
Access: 2015-10-28 23:13:55.281515368 +0100
Modify: 2015-10-28 23:13:55.281515368 +0100
Change: 2015-10-28 23:13:55.313515539 +0100
As we can see, this file was definitely last modified earlier than 1 hour ago! I also tried find -mmin 60 or find -mmin +60, but it did not work either.
Why is this happening and how to use the find command correctly?
I can reproduce your problem if there are no files in the directory that were modified in the last hour. In that case, find . -mmin -60 returns nothing. The command find . -mmin -60 |xargs ls -l, however, returns every file in the directory which is consistent with what happens when ls -l is run without an argument.
To make sure that ls -l is only run when a file is found, try:
find . -mmin -60 -type f -exec ls -l {} +
The problem is that
find . -mmin -60
outputs:
.
./file1
./file2
Note the line with one dot?
That makes ls list the whole directory exactly the same as when ls -l . is executed.
One solution is to list only files (not directories):
find . -mmin -60 -type f | xargs ls -l
But it is better to use directly the option -exec of find:
find . -mmin -60 -type f -exec ls -l {} \;
Or just:
find . -mmin -60 -type f -ls
Which, by the way is safe even including directories:
find . -mmin -60 -ls
To search for files in /target_directory and all its sub-directories, that have been modified in the last 60 minutes:
$ find /target_directory -type f -mmin -60
To find the most recently modified files, sorted in the reverse order of update time (i.e., the most recently updated files first):
$ find /etc -type f -printf '%TY-%Tm-%Td %TT %p\n' | sort -r
Manual of find:
Numeric arguments can be specified as
+n for greater than n,
-n for less than n,
n for exactly n.
-amin n
File was last accessed n minutes ago.
-anewer file
File was last accessed more recently than file was modified. If file is a symbolic link and the -H option or the -L option is in effect, the access time of the file it points to is always
used.
-atime n
File was last accessed n*24 hours ago. When find figures out how many 24-hour periods ago the file was last accessed, any fractional part is ignored, so to match -atime +1, a file has to
have been accessed at least two days ago.
-cmin n
File's status was last changed n minutes ago.
-cnewer file
File's status was last changed more recently than file was modified. If file is a symbolic link and the -H option or the -L option is in effect, the status-change time of the file it points
to is always used.
-ctime n
File's status was last changed n*24 hours ago. See the comments for -atime to understand how rounding affects the interpretation of file status change times.
Example:
find /dir -cmin -60 # creation time
find /dir -mmin -60 # modification time
find /dir -amin -60 # access time
I am working through the same need and I believe your timeframe is incorrect.
Try these:
15min change: find . -mtime -.01
1hr change: find . -mtime -.04
12 hr change: find . -mtime -.5
You should be using 24 hours as your base. The number after -mtime should be relative to 24 hours. Thus -.5 is the equivalent of 12 hours, because 12 hours is half of 24 hours.
Actually, there's more than one issue here. The main one is that xargs by default executes the command you specified, even when no arguments have been passed. To change that you might use a GNU extension to xargs:
--no-run-if-empty
-r
If the standard input does not contain any nonblanks, do not run the command. Normally, the command is run once even if there is no input. This option is a GNU extension.
Simple example:
find . -mmin -60 | xargs -r ls -l
But this might match to all subdirectories, including . (the current directory), and ls will list each of them individually. So the output will be a mess. Solution: pass -d to ls, which prohibits listing the directory contents:
find . -mmin -60 | xargs -r ls -ld
Now you don't like . (the current directory) in your list? Solution: exclude the first directory level (0) from find output:
find . -mindepth 1 -mmin -60 | xargs -r ls -ld
Now you'd need only the files in your list? Solution: exclude the directories:
find . -type f -mmin -60 | xargs -r ls -l
Now you have some files with names containing white space, quote marks, or backslashes? Solution: use null-terminated output (find) and input (xargs) (these are also GNU extensions, afaik):
find . -type f -mmin -60 -print0 | xargs -r0 ls -l
This may work for you. I used it for cleaning folders during deployments for deleting old deployment files.
clean_anyfolder() {
local temp2="$1/**"; //PATH
temp3=( $(ls -d $temp2 -t | grep "`date | awk '{print $2" "$3}'`") )
j=0;
while [ $j -lt ${#temp3[#]} ]
do
echo "to be removed ${temp3[$j]}"
delete_file_or_folder ${temp3[$j]} 0 //DELETE HERE
fi
j=`expr $j + 1`
done
}
this command may be help you sir
find -type f -mtime -60
I need to combine these to commands in order to have a sorted list by date created with the specified "filename".
I know that sorting files by date can be achieved with:
ls -lrt
and finding a file by name with
find . -name "filename*"
I don't know how to combine these two. I tried with a pipeline but I don't get the right result.
[EDIT]
Not sorted
find . -name "filename" -printf '%TY:%Tm:%Td %TH:%Tm %h/%f\n' | sort
Forget xargs. "Find" and "sort" are all the tools you need.
My best guess would be to use xargs:
find . -name 'filename*' -print0 | xargs -0 /bin/ls -ltr
There's an upper limit on the number of arguments, but it shouldn't be a problem unless they occupy more than 32kB (read more here), in which case you will get blocks of sorted files :)
find . -name "filename" -exec ls --full-time \{\} \; | cut -d' ' -f7- | sort
You might have to adjust the cut command depending on what your version of ls outputs.
Check the below-shared command:
1) List Files directory with Last Modified Date/Time
To list files and shows the last modified files at top, we will use -lt options with ls command.
$ ls -lt /run
output
total 24
-rw-rw-r--. 1 root utmp 2304 Sep 8 14:58 utmp
-rw-r--r--. 1 root root 4 Sep 8 12:41 dhclient-eth0.pid
drwxr-xr-x. 4 root root 100 Sep 8 03:31 lock
drwxr-xr-x. 3 root root 60 Sep 7 23:11 user
drwxr-xr-x. 7 root root 160 Aug 26 14:59 udev
drwxr-xr-x. 2 root root 60 Aug 21 13:18 tuned
https://linoxide.com/linux-how-to/how-sort-files-date-using-ls-command-linux/
I'd like to list out all files with modification dates in the last n days (or even simply after Y-m-d) in a directory. It must work recursively through all subdirectories as well.
How can I do this?
Ideal output:
file.txt Mar 26 15:15
file2.txt Mar 27 01:15
Acceptable output:
file.txt
file2.txt
Answered! (Thanks for all the help)
$ find . -type f -mtime -1 -exec ls -lah {} \;
-rw-r--rw- 1 apache apache 18K Mar 26 08:22 ./file1.txt
-rw-r--rw- 1 apache apache 12K Mar 26 09:23 ./dir1/file2.txt
-rw-r--rw- 1 apache apache 16K Mar 26 10:24 ./dir1/dir2/file3.txt
find . -type f -mtime -1 -exec ls -l {} \;
will list all files within last 24 hours, with a long listing just to confirm modification date
use :
find . -mtime +1
For more informations, see
man find
find dir -mtime +1 -print
That will find all files in dir and subdirectories that were modified 1 day ago or before that.