I found the slimier post from STO but those does not filter files with extension. So writing again.
I an writing a shell script to keep last (most latest) 3 .txt files in directory and wants to remove all other .txt files.
For Example... In Directory "Home" I have following files.
test.txt
my.txt
image.jpg
test.avi
sample.txt
country.txt
study.txt
When I run linux script, output should be like as below....
Keep File (keep only last 3 .txt files only)
test.txt
my.txt
image.jpg
test.avi
sample.txt
Delete File
country.txt
study.txt
Thanks
List entries by ctime (newest first), skip the first three items, delete the rest:
ls -c *.txt | tail -n +4 | xargs rm
Related
I have two directories, one called clients and another called test, inside the directory called clients I have some folders, I need a shell script that reads the name of the folders inside clients and creates .txt files with the same name inside the folder test, I am very new to shell and I have no idea how to do this, could you guys help me please?
Try using xargs with ls. ls -F displays all files in the directory client, but then displays the folders with an extra / at the end. the grep uses the extra / in the output of ls -F to only pass folders to the next command. Then, sed 's/\///g removes the extra / from grep, and passes the names to xargs. xargs will then pass the folders to the % symbol, and then make text files with the names.
ls -F client | grep / | sed 's/\///g' | xargs -I % touch tests/%.txt
Goal: I am trying to find the last two modified log files in a directory and would like to compare between them and write the difference to a file. Also,in the diff file I don't want to see any common lines between them.
If your directory only contains log files (Specifically, no other directories)
diff $(ls -tA | head -2 | tail -1) $(ls -tA | head -1) > outfile
What's going on here:
ls -tA lists all files in the directory organized by date. You can modify this if you have a more specific type of file that you're looking for, such as all files that end in .log would be: ls -tA *.log . Note that this includes hidden files because of the -A flag.
head -2 | tail -1 grabs the second line, which is the second newest file, and head -1 grabs the first line, which is the newest file. There's probably a better way to do this step, but this does the job.
diff then receives two file names, with the first being the second newest file in the directory, and the second being the newest file in the directory.
> outfile prints the output from diff to a file named outfile so you can change the word outfile to whatever you want (Except to the name of one of your two log files, because it will be axed by the output).
I have a list of directories within directories and this is what I am trying to attempt:
find a specific file format which is .xml
within all these .xml files, read the contents in the files and remove line 3
For line 3, its string is as follows: dxflib <Name of whatever folder it is in>.dxb
I tried using find -name "*.xml" | xargs grep -v "dxflib" in the terminal (I am using linux) and I found out that while my code works and it displays the results, it did not overwrite the changes to the file.
And as I googled online, it is mentioned that I will need to add in >> output.txt etc
And hence, are there anyways in which I can make it to save / overwrite its own file?
Removes third line in file:
sed -i '3d' file
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-
On my computer running Ubuntu, I have a folder full of hundreds files all named "index.html.n" where n starts at one and continues upwards. Some of those files are actual html files, some are image files (png and jpg), and some of them are zip files.
My goal is to permanently remove every single file except the zip archives. I assume it's some combination of rm and file, but I'm not sure of the exact syntax.
If it fits into your argument list and no filenames contain colon a simple pipe with xargs should do:
file * | grep -vi zip | cut -d: -f1 | tr '\n' '\0' | xargs -0 rm
First find to find matching file, then file to get file types. sed eliminates other file types and also removes everything but the filenames from the output of file. lastly, rm for deleting:
find -name 'index.html.[0-9]*' | \
xargs file | \
sed -n 's/\([^:]*\): Zip archive.*/\1/p' |
xargs rm
I would run:
for f in in index.html.*
do
file "$f" | grep -qi zip
[ $? -ne 0 ] && rm -i "$f"
done
and remove -i option if you feel confident enough
Here's the approach I'd use; it's not entirely automated, but it's less error-prone than some other approaches.
file * > cleanup.sh
or
file index.html.* > cleanup.sh
This generates a list of all files (excluding dot files), or of all index.html.* files, in your current directory and writes the list to cleanup.sh.
Using your favorite text editor (mine happens to be vim), edit cleanup.sh:
Add #!/bin/sh as the first line
Delete all lines containing the string "Zip archive"
On each line, delete everything from the : to the end of the line (in vim, :%s/:.*$//)
Replace the beginning of each line with "rm" followed by a space
Exit your editor, updating the file.
chmod +x cleanup.sh
You should now have a shell script that will delete everything except zip files.
Carefully inspect the script before running it. Look out for typos, and for files whose names contain shell metacharacters. You might need to add quotation marks to the file names.
(Note that if you do this as a one-line shell command, you don't have the opportunity to inspect the list of files you're going to delete before you actually delete them.)
Once you're satisfied that your script is correct, run
./cleanup.sh
from your shell prompt.
for i in index.html.*
do
$type = file $i;
if [[ ! $file =~ "Zip" ]]
then
rm $file
fi
done
Change the rm to a ls for testing purposes.