In other words, how to combine tail and find/grep command in bash.
I want to find all the files(including the files in subdirectories) in my repo have a specific word in the last line, say FIX in the last line. I tried grep -Rl "FIX" to display all the files containing "FIX", but I don't know how to combine the tail command in it. Anyone can help??
Run tail on all the files at once and then grep the output for FIX. Since tail prepends each line with the corresponding file name when given multiple file names, that's all you have to do.
find -type f -exec tail -n1 {} + | grep FIX
Or use ** to find all files and subdirectories, then run tail on each of them one at a time:
shopt -s globstar
for file in **; do
[[ -f $file ]] && tail -n1 "$file" | grep -q FIX && echo "$file"
done
Or use find to find all matches and pipe it to a while read loop:
find -type f -print0 | while IFS= read -rd '' file; do
tail -n1 "$file" | grep -q FIX && echo "$file"
done
Or do the same thing but with -exec + and an explicit sub-shell:
find -type f -exec sh -c 'for file; do tail -n1 "$file" | grep -q FIX && echo "$file"; done' sh {} +
If you want to know if the last line matches a pattern, use sed and restrict the match to the last line with $. sed doesn't easily give a return value or do pretty printing of the filename like grep, but it gets the job done.
find . -exec sh -c "sed -n '$ { /FIX/p; }' {} | grep -q . " \; -print
Here, we use -n to suppress printing, and then print (with /p) only when the last line matches the pattern /FIX/. The output is piped to grep to get a return value that find uses to decide whether or not to -print the name.
Or, you can avoid using grep for the return by doing something like:
find . -exec awk 'END{ exit ! match($0, "FIX")}' {} \; -print
have a folder structure as shown below ./all_files
-rwxrwxrwx reference_file.txt
drwxrwxrwx file1.txt
drwxrwxrwx file2.txt
drwxrwxrwx file3.txt
reference_file.txt has filenames as shown below
$cat reference_file.txt
file1.txt
file2.txt
data in file1.txt and file2.txt are as shown below:
$cat file1.txt
step_1
step_2
step_3
Now, I have to take particular step say step2 from each file
Note1: file name must present in reference_file.txt
Note2: step2 is not line no:2 always.
Note3: search should perform recursively.
I have used below script:
#!/bin/sh
for i in cat reference_file.txt;
do
find . -type f -name $i | grep -v 'FS*' | xargs grep -F 'step_2'
done<reference_file.txt
after using above code i got no output.
# bash -x script.sh
+ for i in cat reference_file.txt
+ find . -type f -name **cat**
+ xargs grep -F 'step_2'
+ for i in cat **reference_file.txt**
+ find . -type f -name reference_file.txt
+ xargs grep -F 'step_2'
Added New requirement:
target=step_XX_2 where XX can be anything and should be skipped for search.. so that desire ouput will be.. step_ab_2 step_cd_2 step_ef_2
I think this is what you are trying to achieve. Please let me know:
EDIT: my previous version did not search recursively.
Further edits: Note that using process substitution for find means that this script MUST be run under bash and not sh.
Further edit for change in specification: note the change to target and the -E option to grep instead of -F.
#!/bin/bash
target='step_.*?_?2'
while read -r name
do
# EDIT: exclude certain directories
if [[ $name == "old1" || $name == "old2" ]]
then
# do the next iteration of the loop
continue
fi
while read -r fname
do
if [[ $fname != FS* ]]
then
# Display the filename (grep -H is not in POSIX)
if out=$(grep -E "$target" "$fname")
then
echo "$fname: $out"
fi
fi
done < <(find . -type f -name "$name")
done < reference_file.txt
Note that your trace (bash -x) uses bash but your #! line uses sh. They are different - you should be consistent with the shell you are using.
So, I have dropped the xargs, that reads strings standard input and executes a program using the strings as argument. Since we already have the argument strings for grep we don't need it.
Your grep -v 'FS*' probably doesn't do what you expect. The regular expression FS* means "F followed by zero or more S's". Not the same as a shell pattern matching (globbing). In my solution I have used FS* because I am using the shell, not grep.
I believe this question is duplicate of this
What you need is
#!/bin/sh
for i in `cat reference_file.txt`
do find . -type f -name $i | grep -v 'FS*' | xargs grep -F 'step_2'
done
See the backticks and Do Not read the file reference_file.txt twice.
I am trying to count lines in the files which matches to the given pattern. now the problem is it gives me only the number of lines. How can i get the file location or name along with number of matched lines?
command i am using now is
for i in $(find . -name 'foo.txt' | sed 's/\.\///g');
do
grep -l && -c '^>' $i;
done
so output i am expecting is like "file location/name number of lines matching"
grep can show you the file name if you specify it as a command-line argument. You can use xargs to invoke grep for each batch of filenames. It'll read the names from standard input and use them as command line arguments for grep.
find . | xargs grep -cH '^>'
Using your find command:
find . -name 'foo.txt' | sed 's/\.\///g' | xargs grep -cH '^>'
You capture the number of lines matching in a variable and test it:
n=$(grep -c '^>' "$i")
(( n > 0 )) && echo "$i:$n"
Actually, you don't even need the test: grep exits unsuccessfully if no matches found, so
n=$(grep -c '^>' "$i") && echo "$i:$n"
Actually, that's too much work. With GNU grep at least, use the -H option. A demo with a file I have lying around:
$ grep -c zero note.xml
16
$ grep -Hc zero note.xml
note.xml:16
I search for some text in some file list. I have the following command to print these lines:
ls -1 *.log | xargs tail --lines=10000 | grep text_for_search
The command output contains all of occurrences of text_for_search, but it hasn't information from which file the occurrences are. How to modify the command to provide this information too?
Actually log files are gigabytes in size, so it's essential to use tail --lines=10000 for each of them
You could just use a loop instead, which will keep track of the file name for you:
for file in *.log; do
if tail --lines=-10000 "$file" | grep -q text_for_search; then
echo "$file"
fi
done
The -q switch to grep suppresses the output, returning a 0 (success) exit code if the pattern is matched.
You can use find command:
find . -name "*.log" -exec grep text_for_search '{}' \;
grep will output filename and matched line. If you just need filenames - add -l switch to grep command.
'{}' - macro used for matched file name substitution in find's -exec command,
\; indicates end of arguments for command, called by exec
Replace your tail command with:
awk '{v[NR]=$0}END{for(i=NR-10000;i<=NR;i++)print FILENAME,v[i]}'
This above is just the replacement of the tail command except it adds a file name in the begining of each line.
You must avoid parsing ls output and use shell's for loop to iterate through all *.log files:
for f in *.log; do
awk -v c=$(wc -l < "$f") 'NR>c-10000 && /text_for_search/{print FILENAME ":" $0}' "$f"
done
EDIT:
You can use awk to search through all *.log files:
awk 'NR>=10000 && /text_for_search/ {print FILENAME ":" $0}' *.log
Is there a simple way, in a pretty standard UNIX environment with bash, to run a command to delete all but the most recent X files from a directory?
To give a bit more of a concrete example, imagine some cron job writing out a file (say, a log file or a tar-ed up backup) to a directory every hour. I'd like a way to have another cron job running which would remove the oldest files in that directory until there are less than, say, 5.
And just to be clear, there's only one file present, it should never be deleted.
The problems with the existing answers:
inability to handle filenames with embedded spaces or newlines.
in the case of solutions that invoke rm directly on an unquoted command substitution (rm `...`), there's an added risk of unintended globbing.
inability to distinguish between files and directories (i.e., if directories happened to be among the 5 most recently modified filesystem items, you'd effectively retain fewer than 5 files, and applying rm to directories will fail).
wnoise's answer addresses these issues, but the solution is GNU-specific (and quite complex).
Here's a pragmatic, POSIX-compliant solution that comes with only one caveat: it cannot handle filenames with embedded newlines - but I don't consider that a real-world concern for most people.
For the record, here's the explanation for why it's generally not a good idea to parse ls output: http://mywiki.wooledge.org/ParsingLs
ls -tp | grep -v '/$' | 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/%