I know ls -t will list all files by modified time. But how can I limit these results to only the last n files?
Try using head or tail. If you want the 5 most-recently modified files:
ls -1t | head -5
The -1 (that's a one) says one file per line and the head says take the first 5 entries.
If you want the last 5 try
ls -1t | tail -5
The accepted answer lists only the filenames, but to get the top 5 files one can also use:
ls -lht | head -6
where:
-l outputs in a list format
-h makes output human readable (i.e. file sizes appear in kb, mb, etc.)
-t sorts output by placing most recently modified file first
head -6 will show 5 files because ls prints the block size in the first line of output.
I think this is a slightly more elegant and possibly more useful approach.
Example output:
total 26960312
-rw-r--r--# 1 user staff 1.2K 11 Jan 11:22 phone2.7.py
-rw-r--r--# 1 user staff 2.7M 10 Jan 15:26 03-cookies-1.pdf
-rw-r--r--# 1 user staff 9.2M 9 Jan 16:21 Wk1_sem.pdf
-rw-r--r--# 1 user staff 502K 8 Jan 10:20 lab-01.pdf
-rw-rw-rw-# 1 user staff 2.0M 5 Jan 22:06 0410-1.wmv
Use tail command:
ls -t | tail -n 5
By default ls -t sorts output from newest to oldest, so the combination of commands to use depends in which direction you want your output to be ordered.
For the newest 5 files ordered from newest to oldest, use head to take the first 5 lines of output:
ls -t | head -n 5
For the newest 5 files ordered from oldest to newest, use the -r switch to reverse ls's sort order, and use tail to take the last 5 lines of output:
ls -tr | tail -n 5
ls -t list files by creation time not last modified time. Use ls -ltc if you want to list files by last modified time from last to first(top to bottom). Thus to list the last n: ls -ltc | head ${n}
None of other answers worked for me. The results were both folders and files, which is not what I would expect.
The solution that worked for me was:
find . -type f -mmin -10 -ls
This lists in the current directory all the files modified in the last 10 minutes. It will not list last 5 files, but I think it might help nevertheless
if you want to watch as it process last five modified file and refresh in 2 secs and show total number of files at top use this:
watch 'ls -Art | wc -l ; ls -ltr | tail -n 5'
Related
I want to combine directories and sub-directories and sum-up the first column as follows:
original output:
8 ./.g/apps/panel/mon/lt/prefs
12 ./.g/apps/panel/mon/lt
40 ./.g/apps/panel/mon
44 ./.g/apps/panel
88 ./.g/apps
112 ./.g
4 ./.g
4 ./.pof
20 ./.local/share/applications
4 ./.local/share/m/packages
8 ./.local/share/m
4 ./.local/share/Trash/info
4 ./.local/share/Trash/files
12 ./.local/share/Trash
44 ./.local/share
new output:
308 ./.g
4 ./.pof
96 ./.local/share
the original command: du -k, and I'm trying with awk and cut commands but fails.
edit:- I got up to here:
du -k | awk '{print $1}' | cut -d "/" -f 1
Now, I'm struggling to merge similar lines and sum-up the first column.
p.s this is just an output example*
thank you.
Use du -d 1 to list accumulative content of 1 directory bellow current.
du -h -d 1
Provide a human readable count.
You can try with command:
du -sh *
Try
du -sk .g .pof .local/share
The -s switch is summary, that is, du will search all the files, all the way down the folders inside, and report just the grand total. (The -k switch print the size in kilobytes; thanks Romeo Ninov).
You have to manually specify each folder you want to know the grand total of.
If you type, for example
du -sk .
it will output just a single number, accounting for the current folder (and below) file sizes.
If you type
du -sk *
the result will depend on what your shell expands * to (usually all the files and folders not starting with a dot (.) in the current folder).
What is the best way of doing this? It's a 250GB Text file 1 word per line
Input:
123
123
123
456
456
874
875
875
8923
8932
8923
Output wanted:
123
456
874
875
8923
8932
I need to get 1 copy of each duplicated line I DON'T WANT if there are 2 of the SAME LINES, REMOVE BOTH, just remove 1, always keeping 1 unique line.
What I do now:
$ cat final.txt | sort | uniq > finalnoduplicates.txt
In a screen, this is working? I don't know, because when I check the size of output file, and it's 0:
123user#instance-1:~$ ls -l
total 243898460
-rw-rw-r-- 1 123user 249751990933 Sep 3 13:59 final.txt
-rw-rw-r-- 1 123user 0 Sep 3 14:26 finalnoduplicates.txt
123user#instance-1:~$
But when I check htop cpu value of the screen running this command is at 100%.
Am I doing something wrong?
You can do this using just sort.
$ sort -u final.txt > finalnoduplicates.txt
You can simplify this further and just have sort do all of it:
$ sort -u final.txt -o finalnoduplicates.txt
Finally, since your input file is purely just numerical data, you can tell sort via the -n switch this to further improve the overall performance of this task:
$ sort -nu final.txt -o finalnoduplicates.txt
sort's man page
-n, --numeric-sort
compare according to string numerical value
-u, --unique
with -c, check for strict ordering; without -c, output only the
first of an equal run
-o, --output=FILE
write result to FILE instead of standard output
I found out about this awesome tool called Duplicut. The entire point of the project was to combine the advantages of unique sorting and increasing the memory limit for wordlists.
It is pretty simple to install, this is the GitHub link
https://github.com/nil0x42/duplicut
I have a bunch of log files in a folder. When I cd into the folder and look at the files it looks something like this.
$ ls -lhat
-rw-r--r-- 1 root root 5.3K Sep 10 12:22 some_log_c48b72e8.log
-rw-r--r-- 1 root root 5.1M Sep 10 02:51 some_log_cebb6a28.log
-rw-r--r-- 1 root root 1.1K Aug 25 14:21 some_log_edc96130.log
-rw-r--r-- 1 root root 406K Aug 25 14:18 some_log_595c9c50.log
-rw-r--r-- 1 root root 65K Aug 24 16:00 some_log_36d179b3.log
-rw-r--r-- 1 root root 87K Aug 24 13:48 some_log_b29eb255.log
-rw-r--r-- 1 root root 13M Aug 22 11:55 some_log_eae54d84.log
-rw-r--r-- 1 root root 1.8M Aug 12 12:21 some_log_1aef4137.log
I want to look at the most recent messages in the most recent log file. I can now manually copy the name of the most recent log and then perform a tail on it and that will work.
$ tail -n 100 some_log_c48b72e8.log
This does involve manual labor so instead I would like to use bash-fu to do this.
I currently found this way to do it;
filename="$(ls -lat | sed -n 2p | tail -c 30)"; tail -n 100 $filename
It works, but I am bummed out that I need to save data into a variable to do it. Is it possible to do this in bash without saving intermediate results into a variable?
tail -n 100 "$(ls -at | head -n 1)"
You do not need ls to actually print timestamps, you just need to sort by them (ls -t). I added the -a option because it was in your original code, but note that this is not necessary unless your logfiles are "dot files", i.e. starting with a . (which they shouldn't).
Using ls this way saves you from parsing the output with sed and tail -c. (And you should not try to parse the output of ls.) Just pick the first file in the list (head -n 1), which is the newest. Putting it in quotation marks should save you from the more common "problems" like spaces in the filename. (If you have newlines or similar in your filenames, fix your filenames. :-D )
Instead of saving into a variable, you can use command substitution in-place.
A truly ls-free solution:
tail -n 100 < <(
for f in *; do
[[ $f -nt $newest ]] && newest=$f
done
cat "$newest"
)
There's no need to initialize newest, since any file will be newer than the null file named by the empty string.
It's a bit verbose, but it's guaranteed to work with any legal file name. Save it to a shell function for easier use:
tail_latest () {
dir=${1:-.}
size=${2:-100}
for f in "$dir"/*; do
[[ $f -nt $newest ]] && newest=$f
done
tail -f "$size" "$newest"
}
Some examples:
# Default of 100 lines from newest file in the current directory
tail_latest
# 200 lines from the newest file in another directory
tail_latest /some/log/dir 200
A plug for zsh: glob qualifiers let you sort the results of a glob directly, making it much easier to get the newest file.
tail -n 100 *(om[1,1])
om sorts the results by modification time (newest first). [1,1] limits the range of files matched to the first. (I think Y1 should do the same, but it kept giving me an "unknown file attribute" error.)
Without parsing ls, you'd use stat
tail -n 100 "$(stat -c "%Y %n" * | sort -nk1,1 | tail -1 | cut -d" " -f 2-)"
Will break if your filenames contain newlines.
version 2: newlines are OK
tail -n 100 "$(
stat --printf "%Y:%n\0" * |
sort -z -t: -k1,1nr |
{ IFS=: read -d '' time filename; echo "$filename"; }
)"
You can try this way also
ls -1t | head -n 1 | xargs tail -c 50
Explanation :
ls -1rht -- list the files based on modified time in reverse order.
tail -n 1 -- get the last one file
tail -c 50 -- show the last 50 character from the file.
I have the following folders in my base /var/www/.versions directory:
1435773881 Jul 1 21:04
1435774663 Jul 2 21:17
1435774856 Jul 3 21:20
1435775432 Jul 4 21:56
How could I remove all directories except most 10 recent with bash script?
This should do the trick, I believe?
rm -r $(ls -td /var/www/.versions/*/ | tac | head -n-10)
The idea: list (with ls) only directories ( that's the -d /var/www/.versions/*/) sorted by time with -t (oldest will be shown last).
Then, reverse the output using tac so the oldest directories are on top.
And then show them all except the last 10 lines with head and a negative argument to -n
Please, test with non-vital directories first ;-) You can change the rm -r by echo to see what would be removed.
You could use -rt option in ls for listing in reverse order of time.
rm -r $(ls -trd /var/www/.versions/*/ | head -n -10)
Also, be sure of you put / in the end of /var/www/.versions/*/ and that all directory names do not start with .
I am using ls -l -t to get a list of files in a directory ordered by time.
I would like to limit the search result to the top 2 files in the list.
Is this possible?
I've tried with grep and I struggled.
You can pipe it into head:
ls -l -t | head -3
Will give you top 3 lines (2 files and the total).
This will just give you the first 2 lines of files, skipping the size line:
ls -l -t | tail -n +2 | head -2
tail strips the first line, then head outputs the next 2 lines.
To avoid dealing with the top output line you can reverse the sort and get the last two lines
ls -ltr | tail -2
This is pretty safe, but depending what you'll do with those two file entries after you find them, you should read Parsing ls on the problems with using ls to get files and file information.
Or you could try just this
ls -1 -t | head -2
The -1 switch skips the title line.
You can use the head command to grab only the first two lines of output:
ls -l -t | head -2
You have to pipe through head.
ls -l -t | head -n 3
will output the two first results.
Try this:
ls -td -- * | head -n 2