Is there a simple way, in a pretty standard UNIX environment with bash, to run a command to delete all but the most recent X files from a directory?
To give a bit more of a concrete example, imagine some cron job writing out a file (say, a log file or a tar-ed up backup) to a directory every hour. I'd like a way to have another cron job running which would remove the oldest files in that directory until there are less than, say, 5.
And just to be clear, there's only one file present, it should never be deleted.
The problems with the existing answers:
inability to handle filenames with embedded spaces or newlines.
in the case of solutions that invoke rm directly on an unquoted command substitution (rm `...`), there's an added risk of unintended globbing.
inability to distinguish between files and directories (i.e., if directories happened to be among the 5 most recently modified filesystem items, you'd effectively retain fewer than 5 files, and applying rm to directories will fail).
wnoise's answer addresses these issues, but the solution is GNU-specific (and quite complex).
Here's a pragmatic, POSIX-compliant solution that comes with only one caveat: it cannot handle filenames with embedded newlines - but I don't consider that a real-world concern for most people.
For the record, here's the explanation for why it's generally not a good idea to parse ls output: http://mywiki.wooledge.org/ParsingLs
ls -tp | grep -v '/$' | tail -n +6 | xargs -I {} rm -- {}
Note: This command operates in the current directory; to target a directory explicitly, use a subshell ((...)) with cd:
(cd /path/to && ls -tp | grep -v '/$' | tail -n +6 | xargs -I {} rm -- {})
The same applies analogously to the commands below.
The above is inefficient, because xargs has to invoke rm separately for each filename.
However, your platform's specific xargs implementation may allow you to solve this problem:
A solution that works with GNU xargs is to use -d '\n', which makes xargs consider each input line a separate argument, yet passes as many arguments as will fit on a command line at once:
ls -tp | grep -v '/$' | tail -n +6 | xargs -d '\n' -r rm --
Note: Option -r (--no-run-if-empty) ensures that rm is not invoked if there's no input.
A solution that works with both GNU xargs and BSD xargs (including on macOS) - though technically still not POSIX-compliant - is to use -0 to handle NUL-separated input, after first translating newlines to NUL (0x0) chars., which also passes (typically) all filenames at once:
ls -tp | grep -v '/$' | tail -n +6 | tr '\n' '\0' | xargs -0 rm --
Explanation:
ls -tp prints the names of filesystem items sorted by how recently they were modified , in descending order (most recently modified items first) (-t), with directories printed with a trailing / to mark them as such (-p).
Note: It is the fact that ls -tp always outputs file / directory names only, not full paths, that necessitates the subshell approach mentioned above for targeting a directory other than the current one ((cd /path/to && ls -tp ...)).
grep -v '/$' then weeds out directories from the resulting listing, by omitting (-v) lines that have a trailing / (/$).
Caveat: Since a symlink that points to a directory is technically not itself a directory, such symlinks will not be excluded.
tail -n +6 skips the first 5 entries in the listing, in effect returning all but the 5 most recently modified files, if any.
Note that in order to exclude N files, N+1 must be passed to tail -n +.
xargs -I {} rm -- {} (and its variations) then invokes on rm on all these files; if there are no matches at all, xargs won't do anything.
xargs -I {} rm -- {} defines placeholder {} that represents each input line as a whole, so rm is then invoked once for each input line, but with filenames with embedded spaces handled correctly.
-- in all cases ensures that any filenames that happen to start with - aren't mistaken for options by rm.
A variation on the original problem, in case the matching files need to be processed individually or collected in a shell array:
# One by one, in a shell loop (POSIX-compliant):
ls -tp | grep -v '/$' | tail -n +6 | while IFS= read -r f; do echo "$f"; done
# One by one, but using a Bash process substitution (<(...),
# so that the variables inside the `while` loop remain in scope:
while IFS= read -r f; do echo "$f"; done < <(ls -tp | grep -v '/$' | tail -n +6)
# Collecting the matches in a Bash *array*:
IFS=$'\n' read -d '' -ra files < <(ls -tp | grep -v '/$' | tail -n +6)
printf '%s\n' "${files[#]}" # print array elements
Remove all but 5 (or whatever number) of the most recent files in a directory.
rm `ls -t | awk 'NR>5'`
(ls -t|head -n 5;ls)|sort|uniq -u|xargs rm
This version supports names with spaces:
(ls -t|head -n 5;ls)|sort|uniq -u|sed -e 's,.*,"&",g'|xargs rm
Simpler variant of thelsdj's answer:
ls -tr | head -n -5 | xargs --no-run-if-empty rm
ls -tr displays all the files, oldest first (-t newest first, -r reverse).
head -n -5 displays all but the 5 last lines (ie the 5 newest files).
xargs rm calls rm for each selected file.
find . -maxdepth 1 -type f -printf '%T# %p\0' | sort -r -z -n | awk 'BEGIN { RS="\0"; ORS="\0"; FS="" } NR > 5 { sub("^[0-9]*(.[0-9]*)? ", ""); print }' | xargs -0 rm -f
Requires GNU find for -printf, and GNU sort for -z, and GNU awk for "\0", and GNU xargs for -0, but handles files with embedded newlines or spaces.
All these answers fail when there are directories in the current directory. Here's something that works:
find . -maxdepth 1 -type f | xargs -x ls -t | awk 'NR>5' | xargs -L1 rm
This:
works when there are directories in the current directory
tries to remove each file even if the previous one couldn't be removed (due to permissions, etc.)
fails safe when the number of files in the current directory is excessive and xargs would normally screw you over (the -x)
doesn't cater for spaces in filenames (perhaps you're using the wrong OS?)
ls -tQ | tail -n+4 | xargs rm
List filenames by modification time, quoting each filename. Exclude first 3 (3 most recent). Remove remaining.
EDIT after helpful comment from mklement0 (thanks!): corrected -n+3 argument, and note this will not work as expected if filenames contain newlines and/or the directory contains subdirectories.
Ignoring newlines is ignoring security and good coding. wnoise had the only good answer. Here is a variation on his that puts the filenames in an array $x
while IFS= read -rd ''; do
x+=("${REPLY#* }");
done < <(find . -maxdepth 1 -printf '%T# %p\0' | sort -r -z -n )
For Linux (GNU tools), an efficient & robust way to keep the n newest files in the current directory while removing the rest:
n=5
find . -maxdepth 1 -type f -printf '%T# %p\0' |
sort -z -nrt ' ' -k1,1 |
sed -z -e "1,${n}d" -e 's/[^ ]* //' |
xargs -0r rm -f
For BSD, find doesn't have the -printf predicate, stat can't output NULL bytes, and sed + awk can't handle NULL-delimited records.
Here's a solution that doesn't support newlines in paths but that safeguards against them by filtering them out:
#!/bin/bash
n=5
find . -maxdepth 1 -type f ! -path $'*\n*' -exec stat -f '%.9Fm %N' {} + |
sort -nrt ' ' -k1,1 |
awk -v n="$n" -F'^[^ ]* ' 'NR > n {printf "%s%c", $2, 0}' |
xargs -0 rm -f
note: I'm using bash because of the $'\n' notation. For sh you can define a variable containing a literal newline and use it instead.
Solution for UNIX & Linux (inspired from AIX/HP-UX/SunOS/BSD/Linux ls -b):
Some platforms don't provide find -printf, nor stat, nor support NUL-delimited records with stat/sort/awk/sed/xargs. That's why using perl is probably the most portable way to tackle the problem, because it is available by default in almost every OS.
I could have written the whole thing in perl but I didn't. I only use it for substituting stat and for encoding-decoding-escaping the filenames. The core logic is the same as the previous solutions and is implemented with POSIX tools.
note: perl's default stat has a resolution of a second, but starting from perl-5.8.9 you can get sub-second resolution with the stat function of the module Time::HiRes (when both the OS and the filesystem support it). That's what I'm using here; if your perl doesn't provide it then you can remove the ‑MTime::HiRes=stat from the command line.
n=5
find . '(' -name '.' -o -prune ')' -type f -exec \
perl -MTime::HiRes=stat -le '
foreach (#ARGV) {
#st = stat($_);
if ( #st > 0 ) {
s/([\\\n])/sprintf( "\\%03o", ord($1) )/ge;
print sprintf( "%.9f %s", $st[9], $_ );
}
else { print STDERR "stat: $_: $!"; }
}
' {} + |
sort -nrt ' ' -k1,1 |
sed -e "1,${n}d" -e 's/[^ ]* //' |
perl -l -ne '
s/\\([0-7]{3})/chr(oct($1))/ge;
s/(["\n])/"\\$1"/g;
print "\"$_\"";
' |
xargs -E '' sh -c '[ "$#" -gt 0 ] && rm -f "$#"' sh
Explanations:
For each file found, the first perl gets the modification time and outputs it along the encoded filename (each newline and backslash characters are replaced with the literals \012 and \134 respectively).
Now each time filename is guaranteed to be single-line, so POSIX sort and sed can safely work with this stream.
The second perl decodes the filenames and escapes them for POSIX xargs.
Lastly, xargs calls rm for deleting the files. The sh command is a trick that prevents xargs from running rm when there's no files to delete.
I realize this is an old thread, but maybe someone will benefit from this. This command will find files in the current directory :
for F in $(find . -maxdepth 1 -type f -name "*_srv_logs_*.tar.gz" -printf '%T# %p\n' | sort -r -z -n | tail -n+5 | awk '{ print $2; }'); do rm $F; done
This is a little more robust than some of the previous answers as it allows to limit your search domain to files matching expressions. First, find files matching whatever conditions you want. Print those files with the timestamps next to them.
find . -maxdepth 1 -type f -name "*_srv_logs_*.tar.gz" -printf '%T# %p\n'
Next, sort them by the timestamps:
sort -r -z -n
Then, knock off the 4 most recent files from the list:
tail -n+5
Grab the 2nd column (the filename, not the timestamp):
awk '{ print $2; }'
And then wrap that whole thing up into a for statement:
for F in $(); do rm $F; done
This may be a more verbose command, but I had much better luck being able to target conditional files and execute more complex commands against them.
If the filenames don't have spaces, this will work:
ls -C1 -t| awk 'NR>5'|xargs rm
If the filenames do have spaces, something like
ls -C1 -t | awk 'NR>5' | sed -e "s/^/rm '/" -e "s/$/'/" | sh
Basic logic:
get a listing of the files in time order, one column
get all but the first 5 (n=5 for this example)
first version: send those to rm
second version: gen a script that will remove them properly
With zsh
Assuming you don't care about present directories and you will not have more than 999 files (choose a bigger number if you want, or create a while loop).
[ 6 -le `ls *(.)|wc -l` ] && rm *(.om[6,999])
In *(.om[6,999]), the . means files, the o means sort order up, the m means by date of modification (put a for access time or c for inode change), the [6,999] chooses a range of file, so doesn't rm the 5 first.
Adaptation of #mklement0's excellent answer with some parameters and without needing to navigate to the folder containing the files to be deleted...
TARGET_FOLDER="/my/folder/path"
FILES_KEEP=5
ls -tp "$TARGET_FOLDER"**/* | grep -v '/$' | tail -n +$((FILES_KEEP+1)) | xargs -d '\n' -r rm --
[Ref(s).: https://stackoverflow.com/a/3572628/3223785 ]
Thanks! 😉
found interesting cmd in Sed-Onliners - Delete last 3 lines - fnd it perfect for another way to skin the cat (okay not) but idea:
#!/bin/bash
# sed cmd chng #2 to value file wish to retain
cd /opt/depot
ls -1 MyMintFiles*.zip > BigList
sed -n -e :a -e '1,2!{P;N;D;};N;ba' BigList > DeList
for i in `cat DeList`
do
echo "Deleted $i"
rm -f $i
#echo "File(s) gonzo "
#read junk
done
exit 0
Removes all but the 10 latest (most recents) files
ls -t1 | head -n $(echo $(ls -1 | wc -l) - 10 | bc) | xargs rm
If less than 10 files no file is removed and you will have :
error head: illegal line count -- 0
To count files with bash
I needed an elegant solution for the busybox (router), all xargs or array solutions were useless to me - no such command available there. find and mtime is not the proper answer as we are talking about 10 items and not necessarily 10 days. Espo's answer was the shortest and cleanest and likely the most unversal one.
Error with spaces and when no files are to be deleted are both simply solved the standard way:
rm "$(ls -td *.tar | awk 'NR>7')" 2>&-
Bit more educational version: We can do it all if we use awk differently. Normally, I use this method to pass (return) variables from the awk to the sh. As we read all the time that can not be done, I beg to differ: here is the method.
Example for .tar files with no problem regarding the spaces in the filename. To test, replace "rm" with the "ls".
eval $(ls -td *.tar | awk 'NR>7 { print "rm \"" $0 "\""}')
Explanation:
ls -td *.tar lists all .tar files sorted by the time. To apply to all the files in the current folder, remove the "d *.tar" part
awk 'NR>7... skips the first 7 lines
print "rm \"" $0 "\"" constructs a line: rm "file name"
eval executes it
Since we are using rm, I would not use the above command in a script! Wiser usage is:
(cd /FolderToDeleteWithin && eval $(ls -td *.tar | awk 'NR>7 { print "rm \"" $0 "\""}'))
In the case of using ls -t command will not do any harm on such silly examples as: touch 'foo " bar' and touch 'hello * world'. Not that we ever create files with such names in real life!
Sidenote. If we wanted to pass a variable to the sh this way, we would simply modify the print (simple form, no spaces tolerated):
print "VarName="$1
to set the variable VarName to the value of $1. Multiple variables can be created in one go. This VarName becomes a normal sh variable and can be normally used in a script or shell afterwards. So, to create variables with awk and give them back to the shell:
eval $(ls -td *.tar | awk 'NR>7 { print "VarName=\""$1"\"" }'); echo "$VarName"
leaveCount=5
fileCount=$(ls -1 *.log | wc -l)
tailCount=$((fileCount - leaveCount))
# avoid negative tail argument
[[ $tailCount < 0 ]] && tailCount=0
ls -t *.log | tail -$tailCount | xargs rm -f
I made this into a bash shell script. Usage: keep NUM DIR where NUM is the number of files to keep and DIR is the directory to scrub.
#!/bin/bash
# Keep last N files by date.
# Usage: keep NUMBER DIRECTORY
echo ""
if [ $# -lt 2 ]; then
echo "Usage: $0 NUMFILES DIR"
echo "Keep last N newest files."
exit 1
fi
if [ ! -e $2 ]; then
echo "ERROR: directory '$1' does not exist"
exit 1
fi
if [ ! -d $2 ]; then
echo "ERROR: '$1' is not a directory"
exit 1
fi
pushd $2 > /dev/null
ls -tp | grep -v '/' | tail -n +"$1" | xargs -I {} rm -- {}
popd > /dev/null
echo "Done. Kept $1 most recent files in $2."
ls $2|wc -l
Modified version of the answer of #Fabien if you want to specify a path. Useful if you're running the script elsewhere.
ls -tr /path/foo/ | head -n -5 | xargs -I% --no-run-if-empty rm /path/foo/%
I'm passing two positional args to a script to run, both args are a path, and while in the scenario analyzing the paths, the problem is sometimes there is some path like: m i sc . . . . .. . . it has dots and spaces, and sometimes even we have a backslash in dir names.
It is so tried to get arguments via two procedures, directly and via at sign.
SOURCE_ARG=$1
DESTINATION_ARG=$2
and
ARG_COUNT=0
for POSITIONAL_ARGUMENTS in "${#}"
do
((ARG_COUNT++))
ARGUMENT_ARRAY[$ARG_COUNT]=$POSITIONAL_ARGUMENTS
done
In the loop, I iterate through the result of commands that have forwarded to them.
while IFS= read -r dir
do
echo "${ARGUMENT_ARRAY[1]}"
echo "${dir}"
while IFS= read -r item
do
# do some stuff
done < <(ls -A "$dir"/)
done < <(du -hP "$SOURCE_ARG" | awk '{$1=""; print $0}' | grep -v "^.$" | sed "s/^ //g")
when i use echo "${ARGUMENT_ARRAY[1]}" i get the same path as i need to check but when using loop iteration varible as dir in here ->echo "${dir}" i got all the spaces escaped, since other commands for that path could not do their jobs.
What I'm Asking for is that how can I get the output of $dir within the loop and as like as echo "${ARGUMENT_ARRAY[1]}" that i mentioned above(input with all spaces and backslashes)
Thanks to #Barmar in comments.
The only reason that filenames are without escapes (i.e. you see directories with no special character or special characters have been escaped) is because du is printing the filenames with escapes, so $dir variable would have escaped once and special characters are no longer available for the other loop iteration in my problem.
Now that we know the problem was raised by using du in my script:
while IFS= read -r dir
# do sth
done < <(du -hP "$SOURCE_ARG" | awk '{$1=""; print $0}' | grep -v "^.$" | sed "s/^ //g")
We can change the du to find and the problem is solved:
while IFS= read -r dir
# do sth
done < <(find "$SOURCE_ARG" -type d –)
PS 1:
Another problem raised as I wanted to print the lines to check them if they are ok or not (i.e. while debugging application) was with echo.
So be sure to try printf "%s\n" "$dir" instead of echo, as some versions of echo process escape sequences.
echo "${dir}"
printf "%s\n" "$dir"
PS 2:
Also If a filename has more than one space in a row, The way I used awk, was collapsing them into a single space.
awk '{$1=""; print $0}' | grep -v "^.$" | sed "s/^ //g"
I'm attempting to create a bash script that will grep a single file for two separate pieces of data, and print them to stdout.
So far this is what I have:
#!/bin/sh
cd /my/filePath/to/directory
APP=`grep -r --include "inputs.conf" "\[" | grep -oP '^[^\/]+'`
INPUT=`grep -r --include "inputs.conf" "\[" | grep -oP '\[[^\]]+'`
for i in $APP
do
{cd /opt/splunk/etc/deployment-apps
INPUT=`grep -r --include "inputs.conf" "\[" | grep -oP '\[[^\]]+'`
echo -n "$i | $INPUT"}
done
echo "";
exit
Which gives me an output printing the entire output of the first command (which is about 200 lines), then a |, then the other results from the second command. I was thinking I could create an array to do this, however I'm still learning bash.
This is an output example from the command without piping to grep:
TA-XA6x-Server/local/inputs.conf:[perfmon://Processor]
There are 200+ of these in a single execution, and I was looking to have the format be printed as something like this
app="TA-XA6x-Server/local/inputs.conf:" | input="[perfmon://Processor]"
There are essentially two pieces of information I'm attempting to stitch together:
the file path to the file
the contents of the file itself (the input)
Here is an example of the file path:
/opt/splunk/etc/deployment-apps/TA-XA6x-Server/local/inputs.conf
and this is an example of the inputs.conf file contents:
[perfmon://TCPv4]
The easy, mostly-working-ish approach is something like this:
#!/bin/bash
while IFS=: read -r name content; do
printf 'app="%s" | input="%s"\n' "$name" "$content"
done < <(grep -r --include "inputs.conf" "\[")
If you need to work reliably with all possible filenames (including names with colons or newlines) and have GNU grep available, consider the --null argument to grep and adjusting the read usage appropriately:
#!/bin/bash
while IFS= read -r -d '' name && IFS= read -r content; do
printf 'app="%s" | input="%s"\n' "$name" "$content"
done < <(grep -r --null --include "inputs.conf" "\[")
In bash I'd like to test a number of directories if they contain filenames outside of what I have whitelisted:
[a-z]
[A-Z]
[0-9]
[+-_ßäöüÄÖÜ.,]
[ ]
(ie. one space is ok, two or more in a row would not).
Trying
$ ls my/dir/ |grep --color=always -v [a][b][c][d][e][f][g][h][i][j][k][l][m][n][o][p][q][r][s][t][u][v][w][x][y][z][A][B][C][D][E][F][G][H][I][J][K][L][M][N][O][P][Q][R][S][T][U][V][W][X][Y][Z][0][1][2][3][4][5][6][7][8][9]
yields all files, even the one containing a # character for testing.
So does the "shorthand" of
$ ls my/dir/ |grep --color=always -v [a-zA-Z0-9]
(and yes, I know I'd have to include the [+-_ßäöüÄÖÜ., ] characters into this later).
I have tried using diff for this:
$ diff -y <(ls -1 my/dir/) <(ls -1 /my/dir/|tr -cd '[a-zA-Z0-9\n\r \-\.\,]')|grep --color=never '|'|cut -d "|" -f 1
which outputs the filenames which are outside of my tr list:
A File Containing Some Rainbo#ws.wav
But this seems rather clumsy... Any ideas for something, well, nicer?
$ find -name '*[^a-zA-Z0-9+\-_ßäöüÄÖÜ., ]*' -o -name '* *'
See man find. The -regex and -maxdepth options might also be interesting to you.
PS: Don't parse the output of ls.
Something to get you started:
#!/bin/bash
regex='^[a-zA-Z0-9.]+$'
while read line
do
if ! [[ $(basename "$line") =~ $regex ]]
then
echo $line
fi
done < <(find . -type f)
$ ls my/dir/ |grep --color=always -v [a][b][c][d][e][f][g][h][i][j][k][l][m][n][o][p][q][r][s][t][u][v][w][x][y][z][A][B][C][D][E][F][G][H][I][J][K][L][M][N][O][P][Q][R][S][T][U][V][W][X][Y][Z][0][1][2][3][4][5][6][7][8][9]
The above line will look for file names that don't contain the character sequence abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789, which is why it matches so many files.
try
$ ls my/dir/ |grep --color=always -v [^a-zA-Z0-9+-_ßäöüÄÖÜ. ]
$ ls my/dir/ |grep --color=always " "
The first line looks for files that contain other characters than the white-listed ones.
The second line looks for files that contain double space.
Inside a directory, how can I delete files that lack any of the words specified, so that only files that contain ALL the words are left? I tried to write a simple bash shell script using grep and rm commands, but I got lost. I am totally new to Linux, any help would be appreciated
How about:
grep -L foo *.txt | xargs rm
grep -L bar *.txt | xargs rm
If a file does not contain foo, then the first line will remove it.
If a file does not contain bar, then the second line will remove it.
Only files containing both foo and bar should be left
-L, --files-without-match
Suppress normal output; instead print the name of each input
file from which no output would normally have been printed. The
scanning will stop on the first match.
See also #Mykola Golubyev's post for placing in a loop.
list=`Word1 Word2 Word3 Word4 Word5`
for word in $list
grep -L $word *.txt | xargs rm
done
Addition to the answers above: Use the newline character as delimiter to handle file names with spaces!
grep -L $word $file | xargs -d '\n' rm
grep -L word | xargs rm
To do the same matching filenames (not the contents of files as most of the solutions above) you can use the following:
for file in `ls --color=never | grep -ve "\(foo\|bar\)"`
do
rm $file
done
As per comments:
for file in `ls`
shouldn't be used. The below does the same thing without using the ls
for file in *
do
if [ x`echo $file | grep -ve "\(test1\|test3\)"` == x ]; then
rm $file
fi
done
The -ve reverses the search for the regexp pattern for either foo or bar in the filename.
Any further words to be added to the list need to be separated by \|
e.g. one\|two\|three
First, remove the file-list:
rm flist
Then, for each of the words, add the file to the filelist if it contains that word:
grep -l WORD * >>flist
Then sort, uniqify and get a count:
sort flist | uniq -c >flist_with_count
All those files in flsit_with_count that don't have the number of words should be deleted. The format will be:
2 file1
7 file2
8 file3
8 file4
If there were 8 words, then file1 and file2 should be deleted. I'll leave the writing/testing of the script to you.
Okay, you convinced me, here's my script:
#!/bin/bash
rm -rf flist
for word in fopen fclose main ; do
grep -l ${word} *.c >>flist
done
rm $(sort flist | uniq -c | awk '$1 != 3 {print $2} {}')
This removes the files in the directory that didn't have all three words:
You could try something like this but it may break
if the patterns contain shell or grep meta characters:
(in this example one two three are the patterns)
for f in *; do
unset cmd
for p in one two three; do
cmd="fgrep \"$p\" \"$f\" && $cmd"
done
eval "$cmd" >/dev/null || rm "$f"
done
This will remove all files that doesn't contain words Ping or Sent
grep -L 'Ping\|Sent' * | xargs rm