Rm and Egrep -v combo - linux

I want to remove all the logs except the current log and the log before that.
These log files are created after 20 minutes.So the files names are like
abc_23_19_10_3341.log
abc_23_19_30_3342.log
abc_23_19_50_3241.log
abc_23_20_10_3421.log
where 23 is today's date(might include yesterday's date also)
19 is the hour(7 o clock),10,30,50,10 are the minutes.
In this case i want i want to keep abc_23_20_10_3421.log which is the current log(which is currently being writen) and abc_23_19_50_3241.log(the previous one)
and remove the rest.
I got it to work by creating a folder,putting the first files in that folder and removing the files and then deleting it.But that's too long...
I also tried this
files_nodelete=`ls -t | head -n 2 | tr '\n' '|'`
rm *.txt | egrep -v "$files_nodelete"
but it didnt work.But if i put ls instead of rm it works.
I am an amateur in linux.So please suggest a simple idea..or a logic..xargs rm i tried but it didnt work.
Also read about mtime,but seems abit complicated since I am new to linux
Working on a solaris system

Try the logadm tool in Solaris, it might be the simplest way to rotate logs. If you just want to get things done, it will do it.
http://docs.oracle.com/cd/E23823_01/html/816-5166/logadm-1m.html

If you want a solution similar (but working) to your try this:
ls abc*.log | sort | head -n-2 | xargs rm
ls abc*.log: list all files, matching the pattern abc*.log
sort: sorts this list lexicographical (by name) from oldes to to newest logfile
head -n-2: return all but the last two entry in the list (you can give -n a negativ count too)
xargs rm: compose the rm command with the entries from stdin
If there are two or less files in the directory, this command will return an error like
rm: missing operand
and will not delete any files.

It is usually not a good idea to use ls to point to files. Some files may cause havoc (files which have a [Newline] or a weird character in their name are the usual exemples ....).
Using shell globs : Here is an interresting way : we count the files newer than the one we are about to remove!
pattern='abc*.log'
for i in $pattern ; do
[ -f "$i" ] || break ;
#determine if this is the most recent file, in the current directory
# [I add -maxdepth 1 to limit the find to only that directory, no subdirs]
if [ $(find . -maxdepth 1 -name "$pattern" -type f -newer "$i" -print0 | tr -cd '\000' | tr '\000' '+' | wc -c) -gt 1 ];
then
#there are 2 files more recent than $i that match the pattern
#we can delete $i
echo rm "$i" # remove the echo only when you are 100% sure that you want to delete all those files !
else
echo "$i is one of the 2 most recent files matching '${pattern}', I keep it"
fi
done
I only use the globbing mechanism to feed filenames to "find", and just use the terminating "0" of the -printf0 to count the outputed filenames (thus I have no problems with any special characters in those filenames, I just need to know how many files were outputted)
tr -cd "\000" will keep only the \000, ie the terminating NUL character outputed by print0. Then I translate each \000 to a single + character, and I count them with the wc -c. If I see 0, "$i" was the most recent file. If I see 1, "$i" was the one just a bit older (so the find sees only the most recent one). And if I see more than 1, it means the 2 files (mathching the pattern) that we want to keep are newer than "$i", so we can delete "$i"
I'm sure someone will step in with a better one, but the idea could be reused, I guess...

Thanks guyz for all the answers.
I found my answer
files=`ls -t *.txt | head -n 2 | tr '\n' '|' | rev |cut -c 2- |rev`
rm `ls -t | egrep -v "$files"`
Thank you for the help

Related

List file using ls to find meet the condition

I am writing a batch program to delete all file in a directory with condition in filename.
In the directory there's a large number of text file (~ hundreds of thousand of files) with filename fixed as "abc" + date
abc_20180820.txt
abc_20180821.txt
abc_20180822.txt
abc_20180823.txt
abc_20180824.txt
The program try to grep all the file, compare the date to a fixed-date, delete it if filename's date < fixed date.
But the problem is it took so long to handle that large amount of file (~1 hour to delete 300k files).
My question: Is there a way to compare the date when running ls command? Not get all file in a list then compare to delete, but list only file already meet the condition then delete. I think that will have better performance.
My code is
TARGET_DATE = "5-12"
DEL_DATE = "20180823"
ls -t | grep "[0-9]\{8\}".txt\$ > ${LIST}
for EACH_FILE in `cat ${LIST}` ;
do
DATE=`echo ${EACH_FILE} | cut -c${TARGET_DATE }`
COMPARE=`expr "${DATE}" \< "${DEL_DATE}"`
if [ $COMPARE -eq 1 ] ;
then
rm -f ${EACH_FILE}
fi
done
Found some similar problem but I dont know how to get it done
List file using ls with a condition and process/grep files that only whitespaces
Here is a refactoring which gets rid of the pesky ls. Looping over a large directory is still going to be somewhat slow.
# Use lowercase for private variables
# to avoid clobbering a reserved system variable
# You can't have spaces around the equals sign
del_date="20180823"
# No need for ls here
# No need for a temporary file
for filename in *[0-9][0-9][0-9][0-9][0-9][0-9][0-9][0-9].txt
do
# Avoid external process; use the shell's parameter substitution
date=${filename%.txt}
# This could fail if the file name contains literal shell metacharacters!
date=${date#${date%?????????}}
# Avoid expr
if [ "$date" -lt "$del_date" ]; then
# Just print the file name, null-terminated for xargs
printf '%s\0' "$filename"
fi
done |
# For efficiency, do batch delete
xargs -r0 rm
The wildcard expansion will still take a fair amount of time because the shell will sort the list of filenames. A better solution is probably to refactor this into a find command which avoids the sorting.
find . -maxdepth 1 -type f \( \
-name '*1[89][0-9][0-9][0-9][0-9][0-9][0-9].txt' \
-o -name '*201[0-7][0-9][0-9][0-9][0-9].txt' \
-o -name '*20180[1-7][0-9][0-9].txt ' \
-o -name '*201808[01][0-9].txt' \
-o -name '*2018082[0-2].txt' \
\) -delete
You could do something like:
rm 201[0-7]*.txt # remove all files from 2010-2017
rm 20180[1-4]*.txt # remove all files from Jan-Apr 2018
# And so on
...
to remove a large number of files. Then your code would run faster.
Yes it takes a lot of time if you have so many files in one folder.
It is bad idea to keep so many files in one folder. Even simple ls or find will be killing storage. And if you have some scripts which iterate over your files, you are for sure killing storage.
So after you wait for one hour to clean it. Take time and make better folders structure. It is good idea to sort files according to years/month/days ... possibly hours
e.g.
somefolder/2018/08/24/...files here
Then you can easily delete, move compress ... whole month or year.
I found a solution in this thread.
https://unix.stackexchange.com/questions/199554/get-files-with-a-name-containing-a-date-value-less-than-or-equal-to-a-given-inpu
The awk command is so powerful, only take me ~1 minute to deal with hundreds of thousand of files (1/10 compare to the loop).
ls | awk -v date="$DEL_DATE" '$0 <= date' | xargs rm -vrf
I can even count, copy, move with that command with the fastest answer I've ever seen.
COUNT="$(ls | awk -v date="${DEL_DATE}" '$0 <= target' | xargs rm -vrf | wc -l)"

Operating on multiple results from find command in bash

Hi I'm a novice linux user. I'm trying to use the find command in bash to search through a given directory, each containing multiple files of the same name but with varying content, to find a maximum value within the files.
Initially I wasn't taking the directory as input and knew the file wouldn't be less than 2 directories deep so I was using nested loops as follows:
prev_value=0
for i in <directory_name> ; do
if [ -d "$i" ]; then
cd $i
for j in "$i"/* ; do
if [ -d "$j" ]; then
cd $j
curr_value=`grep "<keyword>" <filename>.txt | cut -c32-33` #gets value I'm comparing
if [ $curr_value -lt $prev_value ]; then
curr_value=$prev_value
else
prev_value=$curr_value
fi
fi
done
fi
done
echo $prev_value
Obviously that's not going to cut it now. I've looked into the -exec option of find but since find is producing a vast amount of results I'm just not sure how to handle the variable assignment and comparisons. Any help would be appreciated, thanks.
find "${DIRECTORY}" -name "${FILENAME}.txt" -print0 | xargs -0 -L 1 grep "${KEYWORD}" | cut -c32-33 | sort -nr | head -n1
We find the filenames that are named FILENAME.txt (FILENAME is a bash variable) that exist under DIRECTORY.
We print them all out, separated by nulls (this avoids any problems with certain characters in directory or file names).
Then we read them all in again using xargs, and pass the null-separated (-0) values as arguments to grep, launching one grep for each filename (-L 1 - let's be POSIX-compliant here). (I do that to avoid grep printing the filenames, which would screw up cut).
Then we sort all the results, numerically (-n), in descending order (-r).
Finally, we take the first line (head -n1) of the sorted numbers - which will be the maximum.
P.S. If you have 4 CPU cores you can try adding the -P 4 option to xargs to try to make the grep part of it run faster.

Clearing archive files with linux bash script

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-

Script for renaming files with logical

Someone has very kindly help get me started on a mass rename script for renaming PDF files.
As you can see I need to add a bit of logical to stop the below happening - so something like add a unique number to a duplicate file name?
rename 's/^(.{5}).*(\..*)$/$1$2/' *
rename -n 's/^(.{5}).*(\..*)$/$1$2/' *
Annexes 123114345234525.pdf renamed as Annex.pdf
Annexes 123114432452352.pdf renamed as Annex.pdf
Hope this makes sense?
Thanks
for i in *
do
x='' # counter
j="${i:0:2}" # new name
e="${i##*.}" # ext
while [ -e "$j$x" ] # try to find other name
do
((x++)) # inc counter
done
mv "$i" "$j$x" # rename
done
before
$ ls
he.pdf hejjj.pdf hello.pdf wo.pdf workd.pdf world.pdf
after
$ ls
he.pdf he1.pdf he2.pdf wo.pdf wo1.pdf wo2.pdf
This should check whether there will be any duplicates:
rename -n [...] | grep -o ' renamed as .*' | sort | uniq -d
If you get any output of the form renamed as [...], then you have a collision.
Of course, this won't work in a couple corner cases - If your files contain newlines or the literal string renamed as, for example.
As noted in my answer on your previous question:
for f in *.pdf; do
tmp=`echo $f | sed -r 's/^(.{5}).*(\..*)$/$1$2/'`
mv -b ./"$f" ./"$tmp"
done
That will make backups of deleted or overwritten files. A better alternative would be this script:
#!/bin/bash
for f in $*; do
tar -rvf /tmp/backup.tar $f
tmp=`echo $f | sed -r 's/^(.{5}).*(\..*)$/$1$2/'`
i=1
while [ -e tmp ]; do
tmp=`echo $tmp | sed "s/\./-$i/"`
i+=1
done
mv -b ./"$f" ./"$tmp"
done
Run the script like this:
find . -exec thescript '{}' \;
The find command gives you lots of options for specifing which files to run on, works recursively, and passes all the filenames in to the script. The script backs all file up with tar (uncompressed) and then renames them.
This isn't the best script, since it isn't smart enough to avoid the manual loop and check for identical file names.

recursively "normalize" filenames

i mean getting rid of special chars in filenames, etc.
i have made a script, that can recursively rename files [http://pastebin.com/raw.php?i=kXeHbDQw]:
e.g.: before:
THIS i.s my file (1).txt
after running the script:
This-i-s-my-file-1.txt
Ok. here it is:
But: when i wanted to test it "fully", with filenames like this:
¤¥¦§¨©ª«¬®¯°±²³´µ¶·¸¹º»¼½¾¿ÀÂÃÄÅÆÇÈÊËÌÎÏÐÑÒÔÕ×ØÙUÛUÝÞßàâãäåæçèêëìîïðñòôõ÷øùûýþÿ.txt
áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&'()*+,:;<=>?#[\]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£.txt
it fails [http://pastebin.com/raw.php?i=iu8Pwrnr]:
$ sh renamer.sh directorythathasthefiles
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£': No such file or directory
mv: cannot stat `./áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†....and so on
$
so "mv" can't handle special chars.. :\
i worked on it for many hours..
does anyone has a working one? [that can handle chars [filenames] in that 2 lines too?]
mv handles special characters just fine. Your script doesn't.
In no particular order:
You are using find to find all directories, and ls each directory separately.
Why use for DEPTH in... if you can do exactly the same with one command?
find -maxdepth 100 -type d
Which makes the arbitrary depth limit unnecessary
find -type d
Don't ever parse the output of ls, especially if you can let find handle that, too
find -not -type d
Make sure it works in the worst possible case:
find -not -type d -print0 | while read -r -d '' FILENAME; do
This stops read from eating certain escapes and choking on filenames with new-line characters.
You are repeating the entire ls | replace cycle for every single character. Don't - it kills performance. Loop over each directory all files once, and just use multiple sed's, or multiple replacements in one sed command.
sed 's/á/a/g; s/í/i/g; ...'
(I was going to suggest sed 'y/áí/ai/', but unfortunately that doesn't seem to work with Unicode. Perhaps perl -CS -Mutf8 -pe 'y/áí/ai/' would.)
You're still thinking in ASCII: "other special chars - ASCII Codes 33.. ..255". Don't.
These days, most systems use Unicode in UTF-8 encoding, which has a much wider range of "special" characters - so big that listing them out one by one becomes pointless. (It is even multibyte - "e" is one byte, "ė" is three bytes.)
True ASCII has 128 characters. What you currently have in mind are the ISO 8859 character sets (sometimes called "ANSI") - in particular, ISO 8859-1. But they go all the way up to 8859-16, and only the "ASCII" part stays the same.
echo -n $(command) is rather useless.
There are much easier ways to find the directory and basename given a path. For example, you can do
directory=$(dirname "$path")
oldnname=$(basename "$path")
# filter $oldname
mv "$path" "$directory/$newname"
Do not use egrep to check for errors. Check the program's return code. (Like you already do with cd.)
And instead of filtering out other errors, do...
if [[ -e $directory/$newname ]]; then
echo "target already exists, skipping: $oldname -> $newname"
continue
else
mv "$path" "$directory/$newname"
fi
The ton of sed 's/------------/-/g' calls can be changed to a single regexp:
sed -r 's/-{2,}/-/g'
The [ ]s in tr [foo] [bar] are unnecessary. They just cause tr to replace [ to [, and ] to ].
Seriously?
echo "$FOLDERNAME" | sed "s/$/\//g"
How about this instead?
echo "$FOLDERNAME/"
And finally, use detox.
Try something like:
find . -print0 -type f | awk 'BEGIN {RS="\x00"} { printf "%s\x00", $0; gsub("[^[:alnum:]]", "-"); printf "%s\0", $0 }' | xargs -0 -L 2 mv
Use of xargs(1) will ensure that each filename passed exactly as one parameter. awk(1) is used to add new filename right after old one.
One more trick: sed -e 's/-+/-/g' will replace groups of more than one "-" with exactly one.
Assuming the rest of your script is right, your problem is that you are using read but you should use read -r. Notice how the backslash disappeared:
áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&'()*+,:;<=>?#[\]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£.txt
áíüűúöőóéÁÍÜŰÚÖŐÓÉ!"#$%&\'()*+,:;<=>?#[]^_`{|}~€‚ƒ„…†‡ˆ‰Š‹ŒŽ‘’“”•–—˜™š›œžŸ¡¢£
Ugh...
Some tips to clean up your script:
** Use sed to do translation on multiple characters at once, that'll clean things up and make it easier to manage:
dev:~$ echo 'áàaieeé!.txt' | sed -e 's/[áàã]/a/g; s/[éè]/e/g'
aaaieee!.txt
** rather than renaming the file for each change, run all your filters then do one move
$ NEWNAME='áàaieeé!.txt'
$ NEWNAME="$(echo "$NEWNAME" | sed -e 's/[áàã]/a/g; s/[éè]/e/g')"
$ NEWNAME="$(echo "$NEWNAME" | sed -e 's/aa*/a/g')"
$ echo $NEWNAME
aieee!.txt
** rather than doing a ls | read ... loop, use:
for OLDNAME in $DIR/*; do
blah
blah
blah
done
** separate out your path traversal and renaming logic into two scripts. One script finds the files which need to be renamed, one script handles the normalization of a single file. Once you learn the 'find' command, you'll realize you can toss the first script :)

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