How can I find files/folders that were modified w/in the last 2 days and order them by size?
I have: find . -mtime -2
thx
This works for regular files:
find . -mtime -2 -type f| xargs ls -al | sort -n -k 5
This seems to work for folders too:
find . -mtime -2 | xargs ls -al -d | sort -n -k 5
Related
I need to find the most recently modified file in a directory that matches 3.0 MB.
First Attempt
ls -t /home/weather/some.cool*.file | head -n +1 | grep "3.0M"
Second Attempt
find /home/weather/ -maxdepth 1 -type f -name "some.cool*.file" -size 3M -exec ls -t "{}" +; | head -n +1
Am I close?
I hope this is of some use -
ls -ltr --block-size=MB | grep 3MB
The latest modified files will be displayed at the bottom of the output.
The -r flag shows the output in reverse order and the --block-size=MB will show the size of files in MB.
This should work:
ls -lh --sort=time /path/to/directory/*.file | grep "3.0M" | head -n =1
My application logs will be created in below folders in linux system.
Folder 1: 100001_1001
folder 2 : 200001_1002
folder 3 :300061_1003
folder 4: 300001_1004
folder 5 :400011_1008
want to delete all files except the latest file in above folders and want to add this to cron job.
i tried below one not working need help
30 1 * * * ls -lt /abc/cde/etc/100* | awk '{if(NR!=1) print $9}' | xargs -i rm -rf {} \;
30 1 * * * ls -lt /abc/cde/etc/200* | awk '{if(NR!=1) print $9}' | xargs -i rm -rf {} \;
30 1 * * * ls -lt /abc/cde/etc/300* | awk '{if(NR!=1) print $9}' | xargs -i rm -rf {} \;
30 1 * * * ls -lt /abc/cde/etc/400* | awk '{if(NR!=1) print $9}' | xargs -i rm -rf {} \;
You can use this pipeline consisting all gnu utilities (so that we can also handle file paths with special characters, whitespaces and glob characters)
find /parent/log/dir -type f -name '*.zip' -printf '%T#\t%p\0' |
sort -zk1,1rn |
cut -zf2 |
tail -z -n +2 |
xargs -0 rm -f
Using a slightly modified approach to your own:
find /abc/cde/etc/100* -printf "%A+\t%p\n" | sort -k1,1r| awk 'NR!=1{print $2}' | xargs -i rm "{}"
The find version doesn't suffer the lack of paths, so this MIGHT work (I don't know anything about the directory structure, and whether 100* points at a directory, a file or a group of files ...
You should use find, instead. It has a -delete action that deletes he files it found that match you specification. Warning: it is very easy to go wrong with -delete. Test your command first. Example, to find all files named *.zip under a/b/c (and only files):
find a/b/c -depth -name '*.zip' -type f -print
This is the test, it will print all files that the final command will delete (do not forget the -depth, it is important). And once you are sure, the command that does the deletion is:
find a/b/c -depth -name '*.zip' -type f -delete
find also has options to select files by last modification date, by size... You could, for instance, find all files that were modified at least 24 hours ago:
find a/b/c -depth -type f -mtime +0 -print
and, after careful check, delete them:
find a/b/c -depth -type f -mtime +0 -delete
How can I find, let's say, the 100 oldest accessed files? I've tried the following, but it just prints random accessed files.
find /home/you -iname "*.pdf" -atime -100000 -type f | tail -n100
find /home/you -iname '*.pdf' -printf '%A# %p\n' | sort -n | head -n 100
You could use the stat command
stat -c '%X %n' *.pdf | sort -n | head -n100
How can I display all files greater than 10k bytes in my current directory and it's subdirectories.
Tried ls -size +10k but that didn't work.
find . -size +10k -exec ls -lh {} \+
the first part of this is identical to #sputnicks answer, and sucesffully finds all files in the directory over 10k (don't confuse k with K), my addition, the second part then executes ls -lh or ls that lists(-l) the files by human readable size(-h). negate the h if you prefer. of course the {} is the file itself, and the \+ is simply an alternative to \;
which in practice \; would repeat or:
ls -l found.file; ls -l found.file.2; ls -l found.file.3
where \+ display it as one statement or:
ls -l found.file found.file.2 found.file.3
more on \; vs + with find
Additionaly, you may want the listing ordered by size. Which is relatively easy to accomplish. I would at the -s option to ls, so ls -ls and then pipe it to sort -n to sort numerically
which would become:
find . -size +10k -exec ls -ls {} \+ | sort -n
or in reverse order add an -r :
find . -size +10k -exec ls -ls {} \+ | sort -nr
finally, your title says find biggest file in directory. You can do that by then piping the code to tail
find . -size +10k -exec ls -ls {} \+ | sort -n | tail -1
would find you the largest file in the directory and its sub directories.
note you could also sort files by size by using -S, and negate the need for sort. but to find the largest file you would need to use head so
find . -size +10k -exec ls -lS {} \+ | head -1
the benefit of doing it with -S and not sort is one, you don't have to type sort -n and two you can also use -h the human readable size option. which is one of my favorite to use, but is not available with older versisions of ls, for example we have an old centOs 4 server at work that doesn't have -h
Try doing this:
find . -size +10k -ls
And if you want to use the binary ls :
find . -size +10k -exec ls -l {} \;
I realize the assignment is likely long over. For anyone else:
You are overcomplicating.
find . -size +10k
I'll add to #matchew answer (not enough karma points to comment):
find . -size +10k -type f -maxdepth 1 -exec ls -lh {} \; > myLogFile.txt
-type f :specify regular file type
-maxdepth 1 :make sure it only find files in the current directory
You may use ls like that:
ls -lR | egrep -v '^d' | awk '$5>10240{print}'
Explanation:
ls -lR # list recursivly
egrep -v '^d' # only print lines which do not start with a 'd'. (files)
only print lines where the fifth column (size) is greater that 10240 bytes:
awk '$5>10240{print}'
How do I find all the files that were create only today and not in 24 hour period in unix/linux
On my Fedora 10 system, with findutils-4.4.0-1.fc10.i386:
find <path> -daystart -ctime 0 -print
The -daystart flag tells it to calculate from the start of today instead of from 24 hours ago.
Note however that this will actually list files created or modified in the last day. find has no options that look at the true creation date of the file.
find . -mtime -1 -type f -print
To find all files that are modified today only (since start of day only, i.e. 12 am), in current directory and its sub-directories:
touch -t `date +%m%d0000` /tmp/$$
find . -type f -newer /tmp/$$
rm /tmp/$$
Source
I use this with some frequency:
$ ls -altrh --time-style=+%D | grep $(date +%D)
After going through many posts I found the best one that really works
find $file_path -type f -name "*.txt" -mtime -1 -printf "%f\n"
This prints only the file name like
abc.txt not the /path/tofolder/abc.txt
Also also play around or customize with -mtime -1
This worked for me. Lists the files created on May 30 in the current directory.
ls -lt | grep 'May 30'
Use ls or find to have all the files that were created today.
Using ls : ls -ltr | grep "$(date '+%b %e')"
Using find : cd $YOUR_DIRECTORY; find . -ls 2>/dev/null| grep "$(date '+%b %e')"
find ./ -maxdepth 1 -type f -execdir basename '{}' ';' | grep `date +'%Y%m%d'`
You can use find and ls to accomplish with this:
find . -type f -exec ls -l {} \; | egrep "Aug 26";
It will find all files in this directory, display useful informations (-l) and filter the lines with some date you want... It may be a little bit slow, but still useful in some cases.
Just keep in mind there are 2 spaces between Aug and 26. Other wise your find command will not work.
find . -type f -exec ls -l {} \; | egrep "Aug 26";
If you're did something like accidentally rsync'd to the wrong directory, the above suggestions work to find new files, but for me, the easiest was connecting with an SFTP client like Transmit then ordering by date and deleting.
To get file before 24 hours execute below command:
find . -type f -mtime 1 -exec ls -l {} \;
To get files created today execute below command:
find . -type f -mtime -1 -exec ls -l {} \;
To Get files created before n days before, where +2 is before 2 days files in below command:
find . -type f -mtime +2 -exec ls -l {} \;