How to list the last three new files. it should be sorted on the basis of the lat created status?.
You can use following
$ ls -rt /tmp/ | tail -3
ls will list all the files
-r reverse order while sorting
-t sort by modification time, newest first
tail will show you 10 lines by default from last
-n, --lines=K
output the last K lines, instead of the last 10; or use -n +K to output starting with the Kth
Related
How do I keep the latest 8 backup files and delete the older one
backup-Y-M-D.zip
backup-Y-M-D.zip
backup-Y-M-D.zip
backup-Y-M-D.zip
.
.
backup-Y-M-D.zip
There are about 80 files having .zip extension all I wanted to do is to keep latest 8 files according to the date on which created. I also tried logrotate but failed to rotate logs as it is not doing anything. Below down is the config file of logrotate.
/root/test/*.zip {
daily
missingok
extension .zip
rotate 4
nocompress
}
If the naming convention is guaranteed you could just rely on the alphabetical ordering of the files when expanding a glob pattern to get the oldest or newest files. According to Filename Expansion:
After word splitting, unless the -f option has been set (see The Set Builtin), Bash scans each word for the characters ‘*’, ‘?’, and ‘[’. If one of these characters appears, and is not quoted, then the word is regarded as a pattern, and replaced with an alphabetically sorted list of filenames matching the pattern (see Pattern Matching).
Demo:
[user#hostname]$ touch backup-2022-06-14.zip backup-2022-06-13.zip backup-2021-07-04.zip
[user#hostname]$ echo *
backup-2021-07-04.zip backup-2022-06-13.zip backup-2022-06-14.zip
You can leverage this to get a list of files other than the last N elements:
[user#hostname]$ all_files=(*)
[user#hostname]$ old_files=( "${all_files[#]:0:${#all_files[#]}-1}" ) #change -1 to -8 if you want to keep the 8 newest
[user#hostname]$ echo "${old_files[#]}"
backup-2021-07-04.zip backup-2022-06-13.zip
And then do whatever you want with that list, such as remove it with rm "${old_files[#]}".
One way to do this is with the following one-liner, ran from the directory where the logs are located:
ls -t | head -n -8 | xargs --no-run-if-empty rm
Explanation:
ls -t - lists all the files in order from youngest to oldest
head -n -8 - gets all the files except the first 8
xargs --no-run-if-empty rm - deletes the selected files if there are any, preventing errors if you ever have fewer than 8 logs
If you want to set this up to run automatically every day, giving you peace of mind in case your server is offline on the 7th day of a cycle and misses the one week mark, run crontab -e and add the following to your jobs:
0 0 * * * cd yourDirNameHere && ls -t | head -n -8 | xargs --no-run-if-empty rm
Then the log cleaner will be ran every night at midnight.
On Linux, how do I derive a long list of items sorted by size (largest to smallest)?
Is this correct:
$ ls -laShr /var/~
To list all files and sort them by size, use the -S option. By default, it displays output in descending order (biggest to smallest in size).
$ ls -laS /var/~
And to sort in reverse order, add the -r flag as follows.
$ ls -laShr /var/~
U can use
$ ls -lShr /var/
l- for long listing
S- for sorting by size(default descending)
h- for human-readable format
r- for a reverse sort
how can I use program as a key to sort in Unix shell? In other words to sort output of 'ls' (or any other program) by return value of a program applied on each line.
I'll give two example solutions:
A one-line command that is simpler and therefore something I'd try use first.
A bash script that allows sorting a list by output from an arbitrary bash function that reads each line of the list as input.
Example 1 (without executing command on each line)
If the question is how to, in general, sort outputs of programs like ls, below is an example specific to ls that sorts by inode. However, every program may have its own idiosyncrasies when generating its output so this example may have to be adapted:
ls -ail /home/user/ | tail -n+2 | tr -s ' ' | sort -t' ' -k1,1 -g
Here are the different parts of this command broken down:
ls -ail /home/user/
Lists all (-a) files in directory /home/user/ in list (-l) format with inode (-i).
tail -n+1
Cuts off first line from ls output.
tr -s ' '
Combines (-s) multiple spaces (' ') for sort.
sort -t ' ' -k 1 -g
Sorts list by first (1) field of integers (-g) separated by one space (' ').
Example 2 (executing command with each line as input)
Here is a more adaptable example in a bash script I worked up to show how the list of files generated from ls -a1 can be fed into bash function getinode which uses stat to output the inode for each file. A while loop repeats this process for each file, saving in comma-delimited format the data by repeatedly appending a variable named OUTPUT which at the end is sorted by sort using the first field.
The important part is that the function getinode can be anything, so long as it outputs a string. I set up getinode to receive a file path as input (first argument $1) and to then output the inode to stdout via echo $INODE. The script calls getinode via $(getinode "$FILEPATH").
#!/bin/bash
# Usage: lsinodesort.sh [file]
# Refs/attrib:
# [1]: How to sort a csv file by sorting on a single field. https://stackoverflow.com/a/44744800
# [2]: How to read a while loop variable. https://stackoverflow.com/a/16854326
WORKDIR="$1" # read directory from first argument
getinode() {
# Usage: getinode [path]
INODE="$(stat "$1" --format=%i)"
echo $INODE
}
if [ -d "$WORKDIR" ]; then
LINES="$(ls -a1 "$WORKDIR")" # save `ls` output to variable LINES
else
exit 1; # not a valid directory
fi
while read line; do
path="$WORKDIR"/"$line" # Determine path.
if [ -f "$path" ]; then # Check if path is a file.
FILEPATH="$path"
FILENAME="$(basename "$path")" # Determine filename from path.
FILEINODE=$(getinode "$FILEPATH") # Get inode.
OUTPUT="$FILEINODE"",""$FILENAME""\n""$OUTPUT" ; # Append inode and file name to OUTPUT
fi
done <<< "$LINES" # See [2].
OUTPUT=$(printf "${OUTPUT}" | sort -t, -k1,1) # sort OUTPUT. See [1]
OUTPUT="inode","filename""\n""$OUTPUT"
printf "${OUTPUT}\n" # print final OUTPUT.
When I run it on my own home folder I get output like this:
inode,filename
3932162,.bashrc
3932165,.bash_logout
3932382,.zshrc
3932454,.gitconfig
3933234,.bash_aliases
3933512,.profile
3933612,.viminfo
I'm not sure to understand your question, so I'll try to rephrase it first.
If I'm not mistaken, you want to sort the output of a program (it may be ls or any other command in a Unix shell).
I'll suggest using the pipeline feature available on Unix shell.
For instance, you can sort the output of the ls command using :
ls /home | sort
This feature is available but not limited to the ls command.
By the way, there are optional flags you can use for sorting ls command results if that's your specific use case :
ls -S # for sorting by file size
ls -t # for sorting by modification time
You can also append the --reverse or -r flag for displaying the result in reverse order.
As for the sort function, there are also flags allowing to customize your result as per your needs :
sort -n # for sorting numerically instead of alphabetically
sort -k5 # for sorting based on the 5th column
sort -t "," # for using the comma as a field separator
You can combine all of them like that for sorting the output of ‘ls -l‘ command on the basis of field 2,5 (Numeric) and 9 (Non-Numeric/alphabetically).
ls -l /home/$USER | sort -t "," -nk2,5 -k9
sort function examples
Here is my problem,
I have a folder where is stored multiple files with a specific format:
Name_of_file.TypeMM-DD-YYYY-HH:MM
where MM-DD-YYYY-HH:MM is the time of its creation. There could be multiple files with the same name but not the same time of course.
What i want is a script that can keep the 3 newest version of each file.
So, I found one example there:
Deleting oldest files with shell
But I don't want to delete a number of files but to keep a certain number of newer files. Is there a way to get that find command, parse in the Name_of_file and keep the 3 newest???
Here is the code I've tried yet, but it's not exactly what I need.
find /the/folder -type f -name 'Name_of_file.Type*' -mtime +3 -delete
Thanks for help!
So i decided to add my final solution in case anyone liked to get it. It's a combination of the 2 solutions given.
ls -r | grep -P "(.+)\d{4}-\d{2}-\d{2}-\d{2}:\d{2}" | awk 'NR > 3' | xargs rm
One line, super efficiant. If anything changes on the pattern of date or name just change the grep -P pattern to match it. This way you are sure that only the files fitting this pattern will get deleted.
Can you be extra, extra sure that the timestamp on the file is the exact same timestamp on the file name? If they're off a bit, do you care?
The ls command can sort files by timestamp order. You could do something like this:
$ ls -t | awk 'NR > 3' | xargs rm
THe ls -t lists the files by modification time where the newest are first.
The `awk 'NR > 3' prints out the list of files except for the first three lines which are the three newest.
The xargs rm will remove the files that are older than the first three.
Now, this isn't the exact solution. There are possible problems with xargs because file names might contain weird characters or whitespace. If you can guarantee that's not the case, this should be okay.
Also, you probably want to group the files by name, and keep the last three. Hmm...
ls | sed 's/MM-DD-YYYY-HH:MM*$//' | sort -u | while read file
do
ls -t $file* | awk 'NR > 3' | xargs rm
done
The ls will list all of the files in the directory. The sed 's/\MM-DD-YYYY-HH:MM//' will remove the date time stamp from the files. Thesort -u` will make sure you only have the unique file names. Thus
file1.txt-01-12-1950
file2.txt-02-12-1978
file2.txt-03-12-1991
Will be reduced to just:
file1.txt
file2.txt
These are placed through the loop, and the ls $file* will list all of the files that start with the file name and suffix, but will pipe that to awk which will strip out the newest three, and pipe that to xargs rm that will delete all but the newest three.
Assuming we're using the date in the filename to date the archive file, and that is possible to change the date format to YYYY-MM-DD-HH:MM (as established in comments above), here's a quick and dirty shell script to keep the newest 3 versions of each file within the present working directory:
#!/bin/bash
KEEP=3 # number of versions to keep
while read FNAME; do
NODATE=${FNAME:0:-16} # get filename without the date (remove last 16 chars)
if [ "$NODATE" != "$LASTSEEN" ]; then # new file found
FOUND=1; LASTSEEN="$NODATE"
else # same file, different date
let FOUND="FOUND + 1"
if [ $FOUND -gt $KEEP ]; then
echo "- Deleting older file: $FNAME"
rm "$FNAME"
fi
fi
done < <(\ls -r | grep -P "(.+)\d{4}-\d{2}-\d{2}-\d{2}:\d{2}")
Example run:
[me#home]$ ls
another_file.txt2011-02-11-08:05
another_file.txt2012-12-09-23:13
delete_old.sh
not_an_archive.jpg
some_file.exe2011-12-12-12:11
some_file.exe2012-01-11-23:11
some_file.exe2012-12-10-00:11
some_file.exe2013-03-01-23:11
some_file.exe2013-03-01-23:12
[me#home]$ ./delete_old.sh
- Deleting older file: some_file.exe2012-01-11-23:11
- Deleting older file: some_file.exe2011-12-12-12:11
[me#home]$ ls
another_file.txt2011-02-11-08:05
another_file.txt2012-12-09-23:13
delete_old.sh
not_an_archive.jpg
some_file.exe2012-12-10-00:11
some_file.exe2013-03-01-23:11
some_file.exe2013-03-01-23:12
Essentially, but changing the file name to dates in the form to YYYY-MM-DD-HH:MM, a normal string sort (such as that done by ls) will automatically group similar files together sorted by date-time.
The ls -r on the last line simply lists all files within the current working directly print the results in reverse order so newer archive files appear first.
We pass the output through grep to extract only files that are in the correct format.
The output of that command combination is then looped through (see the while loop) and we can simply start deleting after 3 occurrences of the same filename (minus the date portion).
This pipeline will get you the 3 newest files (by modification time) in the current dir
stat -c $'%Y\t%n' file* | sort -n | tail -3 | cut -f 2-
To get all but the 3 newest:
stat -c $'%Y\t%n' file* | sort -rn | tail -n +4 | cut -f 2-
I have uploaded a file to a Linux computer. I do not know its name. So how to view files through their last created date attribute ?
ls -lat
will show a list of all files sorted by date. When listing with the -l flag using the -t flag sorts by date. If you only need the filename (for a script maybe) then try something like:
ls -lat | head -2 | tail -1 | awk '{print $9}'
This will list all files as before, get the first 2 rows (the first one will be something like 'total 260'), the get the last one (the one which shows the details of the file) and then get the 9th column which contains the filename.
find / -ctime -5
Will print the files created in the last five minutes. Increase the period one minute at a time to find your file.
Assuming you know the folder where you'll be searching it, the most easy solution is:
ls -t | head -1
# use -A in case the file can start with a dot
ls -tA | head -1
ls -t will sort by time, newest first (from ls --help itself)
head -1 will only keep 1 line at the top of anything
Use ls -lUt or ls -lUtr, as you wish. You can take a look at the ls command documentation typing man ls on a terminal.