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问题:
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?
回答1:
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 {} +
回答2:
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
回答3:
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
回答4:
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.
回答5:
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
回答6:
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
}
回答7:
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
回答8:
this command may be help you sir
find -type f -mtime -60