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 .
Related
When we list files in unix using ls -l command the output is a table with space as a separator, for example the following
(jupyter-lab) ➜ mylab ls -l
total 2
drwxr-sr-x. 2 hs0424 ragr 0 Feb 1 12:17 A bad directory
drwxr-sr-x. 2 hs0424 ragr 0 Feb 1 12:18 A very bad directory
I want to convert to a tab separated file (.tsv), just changing spaces to \t, such as ls -l | sed -E 's/ +/\t/g' would not work since filenames contain spaces. Do we have better solution ?
Hard to show expected output with tabs but if we use \t as a replacement of tab, I want something as follows,
(jupyter-lab) ➜ mylab ls -l
total 2
drwxr-sr-x.\t2\ths0424\tragr\t0\tFeb 1\t12:17\tA bad directory
drwxr-sr-x.\t2\ths0424\tragr\t0\tFeb 1\t12:18\tA very bad directory
(Edit 1)
We can assume access to GNU tools
Use GNU find -printf or stat, either of which let you provide an arbitrary format string, instead of ls.
find . -mindepth 1 -maxdepth 1 -printf '%M\t%y\t%g\t%G\t%u\t%U\t%f\t%l\n'
or
# for normal cases
stat --printf='%A\t%G\t%g\t%U\t%u\t%n\n' *
# for directories where filenames could exceed command line length limit
printf '%s\0' * | xargs -0 stat --printf='%A\t%G\t%g\t%U\t%u\t%n\n'
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 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'
I am using
ls -ltr /homedir/mydirectory/work/ |tail -n 1|cut -d ' ' -f 10
But this is a very crude way of getting the desired result.And also its unreliable.
The output I get on simply executing
ls -ltr /homedir/mydirectory/work/ |tail -n 1
is
-rw-r--r-- 1 user pusers 1764 Apr 1 12:06 firstfile.xml
So here I get the file name.
But if the output on doing the above command is like
-rw-r--r-- 100 user pusers 1764 Apr 1 12:06 firstfile.xml
the first command fails ! And understandably as I am cutting the result from the 10th character which does not hold valid now.
So how to refine it.
Why do you use the -l flag for ls if you don't need it? Make ls simply output the filenames if you don't need more information instead of trying to "parse" its non-unified output (raping poor text processing utilities...).
LAST_MODIFIED_FILE=`ls -tr | tail -n 1`
If you really want to achieve this using your method, then, use awk instead of cut
ls -ltr /var/log/ |tail -n 1| awk '{print $9}'
Extended user user529758 answer which can give result as per file name
use below commnad as per the file name
ls -tr Filename* | tail -n 1
I was looking for a linux command line one-liner to rename a bunch of files in one shot.
pattern1.a pattern1.b pattern1.c ...
Once the command is executed I should get
pattern2.a pattern2.b pattern2.c ...
for i in pattern1.*; do mv -- "$i" "${i/pattern1/pattern2}"; done
Before you run it, stick an echo in front of the mv to see what it would do.
If you happen to be using Linux, you may also have a perl script at /usr/bin/rename (sometimes installed as "prename") which can rename files based on more complex patterns than shell globbing permits.
The /usr/bin/rename on one of my systems is documented here. It could be used like this:
rename "s/pattern1/pattern2/" pattern1.*
A number of other Linux environments seem to have a different rename that might be used like this:
rename pattern1 pattern2 pattern1.*
Check man rename on your system for details.
Plenty of ways to skin this cat. If you'd prefer your pattern to be a regex rather than a fileglob, and you'd like to do the change recursively you could use something like this:
find . -print | sed -ne '/^\.\/pattern1\(\..*\)/s//mv "&" "pattern2\1"/p'
As Kerrek suggested with his answer, this one first shows you what it would do. Pipe the output through a shell (i.e. add | sh to the end) once you're comfortable with the commands.
This works for me:
[ghoti#pc ~]$ ls -l foo.*
-rw-r--r-- 1 ghoti wheel 0 Mar 26 13:59 foo.php
-rw-r--r-- 1 ghoti wheel 0 Mar 26 13:59 foo.txt
[ghoti#pc ~]$ find . -print | sed -ne '/^\.\/foo\(\..*\)/s//mv "&" "bar\1"/p'
mv "./foo.txt" "bar.txt"
mv "./foo.php" "bar.php"
[ghoti#pc ~]$ find . -print | sed -ne '/^\.\/foo\(\..*\)/s//mv "&" "bar\1"/p' | sh
[ghoti#pc ~]$ ls -l foo.* bar.*
ls: foo.*: No such file or directory
-rw-r--r-- 1 ghoti wheel 0 Mar 26 13:59 bar.php
-rw-r--r-- 1 ghoti wheel 0 Mar 26 13:59 bar.txt
[ghoti#pc ~]$