I try to create a script who delete all the olds files except the three more recent files on my backup directory with lftp.
I have try to do this with ls -1tr who return all the files in ascending date order, and after I do a head -$NB_BACKUP_TO_RM ($NB_BACKUP_TO_RM is the numbers of files that I want to delete in my lists), this two commands return the correct files.
After this I want to remove all of them, so I do a xargs rm --, but Bash returns that the files don't exist... I think this command is not running into the remote directory, but in the local directory, and I don't know what I can do for delete this files (of my return lists).
Here is the full code:
MAX_BACKUP=3
NB_BACKUP=$(lftp -e "ls -1tr $REMOTE_DIR/full_backup_ftp* | wc -l ; quit" -u $USER,$PASSWORD $HOST)
if (( $NB_BACKUP > $MAX_BACKUP ))
then
NB_BACKUP_TO_RM=$(($NB_BACKUP-$MAX_BACKUP))
REMOVE=$(lftp -e "ls -1tr $REMOTE_DIR/full_backup_ftp* | head -$NB_BACKUP_TO_RM | xargs rm -- ; quit" -u $USER,$PASSWORD $HOST)
echo $REMOVE
fi
Have you an idea of the problem? How can I delete the files of my lists (after ls -1tr $REMOTE_DIR/full_backup_ftp* and head -$NB_BACKUP_TO_RM)
Thanks for your help
Starting SFTP connection can be time consuming. Slightly modified solution to avoid multiple lftp sessions below. It will perform much better the the alternative solution, especially if large number of files have to be purged.
Basically, leveraging lftp flexibility to mix lftp command with external commands. It creates a command file with a series of 'rm' (leveraging head ,xargs, ...), and executing those commands INSIDE the same lftp session.
Also note that lftp 'ls' does not allow wildcard, use 'cls' instead
Make sure you test this carefully, because of potential removal of important files
lftp -e $USER,$PASSWORD $HOST <<__CMD__
cls -1tr $REMOTE_DIR/full_backup_ftp* | head -$NB_BACKUP_TO_RM | xarg -I{} echo rm {} > rm_list.txt
source rm_list.txt
__CMD__
Or with one liner, using 'lftp' ability to execute dynamically generated command (source -e). It eliminate the temporary file.
lftp -e $USER,$PASSWORD $HOST <<__CMD__
source -e 'cls -1tr $REMOTE_DIR/full_backup_ftp* | head -$NB_BACKUP_TO_RM | xarg -I{} echo rm {}'
__CMD__
Looks xargs is unknown cmd for lftp after man lftp. And xargs rm is deleting local files not remote files.
so please use xargs as below, it works for me.
lftp -e "ls -1tr $REMOTE_DIR/full_backup_ftp*; quit" -u $USER,$PASSWORD $HOST | head -$NB_BACKUP_TO_RM | xargs -I {} lftp -e 'rm '{}'; quit' -u $USER,$PASSWORD $HOST
Related
I have multiple file in SFTP server from which I need to copy only latest file. I have written sample code but in that I am passing filename. What logic I need to add that it identify the latest file from sftp and copy it into my local?
In SFTP server -
my_data_20220428.csv
my_data_20220504.csv
my_data_20220501.csv
my_data_20220429.csv
The code which I am running-
datadir="/script/data"
cd ${datadir}
rm -f ${datadir}/my_data*.csv
rm -f ${logfile}
lftp<<END_SCRIPT
open sftp://${sftphost}
user ${sftpuser} ${sftppassword}
cd ${sftpfolder}
lcd $datadir
mget my_data_20220504.csv
bye
END_SCRIPT
what changes I need to do it automatically pick the latest file from server without hardcoding the filename?
You can try this script mainly copied from your sample, so it is expected that the variables have already been created.
#!/usr/bin/env bash
datadir="/script/data"
rm -f "$datadir"/my_data*.csv
rm -f "$logfile"
new=$(echo "ls -halt $sftpfolder" | lftp -u "${sftpuser}","${sftppassword}" sftp://"${sftphost}" | sed -n '/my_data/s/.* \(.*\)/\1/p' | head -1)
lftp -u "${sftpuser}","${sftppassword}" sftp://"${sftphost}" << --EOF--
cd "$sftpfolder"
lcd "$datadir"
get "$new"
bye
--EOF--
You could try:
latest=$(lftp "sftp://$sftpuser:$sftppassword#$sftphost" \
-e "cd $sftpfolder; glob rels -1t *.csv; bye" |
head -1)
lftp "sftp://$sftpuser:$sftppassword#myhost" \
-e "cd $sftpfolder; mget $latest; bye"
am trying to execute the following command:
$ ssh root#10.10.10.50 "tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )"
ls: cannot access /var/log/alert_ARCDB.log: No such file or directory
tail: cannot follow `-' by name
notice the error returned, when i login to ssh separately and then execute
tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )"
see the below:
# ls -t /var/log/alert_ARCDB.log | head -n1
/var/log/alert_ARCDB.log
why is that happening and how to fix it. am trying to do this in one line as i don't want to create a script file.
Thanks a lot
Shell parameter expansion happens before command execution.
Here's a simple example. If I type...
ls "$HOME"
...the shell replaces $HOME with the path to my home directory first, then runs something like ls /home/larsks. The ls command has no idea that the command line originally had $HOME.
If we look at your command...
$ ssh root#10.10.10.50 "tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )"
...we see that you're in exactly the same situation. The $(ls -t ...) expression is expanded before ssh is executed. In other words, that command is running your local system.
You can inhibit the shell expansion on your local system by using single quotes. For example, running:
echo '$HOME'
Will produce:
$HOME
So you can run:
ssh root#10.10.10.50 'tail -F -n 1 $(ls -t /var/log/alert_ARCDB.log | head -n1 )'
But there's another problem here. If /var/log/alert_ARCDB.log is a file, your command makes no sense: calling ls -t on a single file gets you nothing.
If alert-ARCDB.log is a directory, you have a different problem. The result of ls /some/directory is a list of filenames without any directory prefix. If I run something like:
ls -t /tmp
I will get output like
file1
file2
If I do this:
tail $(ls -t /tmp | head -1)
I end up with a command that looks like:
tail file1
And that will fail, because there is no file1 in my current directory.
One approach would be to pipe the commands you want to perform to ssh. One simple way to achieve that is to first create a function that will echo the commands you want executed :
remote_commands()
{
echo 'cd /var/log/alert_ARCDB.log'
echo 'tail -F -n 1 "$(ls -t | head -n1 )"'
}
The cd will allow you to use the relative path listed by ls. The single quotes make sure that everything will be sent as-is to the remote shell, with no local expansion occurring.
Then you can do
ssh root#10.10.10.50 bash < <(remote_commands)
This assumes alert_ARCDB.log is a directory (or else I am not sure why you would want to add head -n1 after that).
I'm building a little bash script to run another bash script that's found in multiple directories. Here's the code:
cd /home/mainuser/CaseStudies/
grep -R -o --include="Auto.sh" [\w] | wc -l
When I execute just that part, it finds the same file 5 times in each folder. So instead of getting 49 results, I get 245. I've written a recursive bash script before and I used it as a template for this problem:
grep -R -o --include=*.class [\w] | wc -l
This code has always worked perfectly, without any duplication. I've tried running the first code with and without the " ", I've tried -r as well. I've read through the bash documentation and I can't seem to find a way to prevent, or even why I'm getting, this duplication. Any thoughts on how to get around this?
As a separate, but related question, if I could launch Auto.sh inside of each directory so that the output of Auto.sh was dumped into that directory; without having to place Auto.sh in each folder. That would probably be much more efficient that what I'm currently doing and it would also probably fix my current duplication problem.
This is the code for Auto.sh:
#!/bin/bash
index=1
cd /home/mainuser/CaseStudies/
grep -R -o --include=*.class [\w] | wc -l
grep -R -o --include=*.class [\w] |awk '{print $3}' > out.txt
while read LINE; do
echo 'Path '$LINE > 'Outputs/ClassOut'$index'.txt'
javap -c $LINE >> 'Outputs/ClassOut'$index'.txt'
index=$((index+1))
done <out.txt
Preferably I would like to make it dump only the javap outputs for the application its currently looking at. Since those .class files could be in any number of sub-directories, I'm not sure how to make them all dump in the top folder, without executing a modified Auto.sh in the top directory of each application.
Ok, so to fix the multiple find:
grep -R -o --include="Auto.sh" [\w] | wc -l
Should be:
grep -R -l --include=Auto.sh '\w' | wc -l
The reason this was happening, was that it was looking for instances of the letter w in Auto.sh. Which occurred 5 times in the file.
However, the overall fix that doesn't require having to place Auto.sh in every directory, is something like this:
MAIN_DIR=/home/mainuser/CaseStudies/
cd $MAIN_DIR
ls -d */ > DirectoryList.txt
while read LINE; do
cd $LINE
mkdir ProjectOutputs
bash /home/mainuser/Auto.sh
cd $MAIN_DIR
done <DirectoryList.txt
That calls this Auto.sh code:
index=1
grep -R -o --include=*.class '\w' | wc -l
grep -R -o --include=*.class '\w' | awk '{print $3}' > ProjectOutputs.txt
while read LINE; do
echo 'Path '$LINE > 'ProjectOutputs/ClassOut'$index'.txt'
javap -c $LINE >> 'ProjectOutputs/ClassOut'$index'.txt'
index=$((index+1))
done <ProjectOutputs.txt
Thanks again for everyone's help!
There is an input like:
folder1
folder2
folder3
...
foldern
I would like to iterate over taking multiple lines at once and processes each line, remove the first / (and more but for now this is enough) and echo the. Iterating over in bash with a single thread can be slow sometimes. The alternative way of doing this would be splitting up the input file to N pieces and run the same script with different input and output N times, at the end you can merge the results.
I was wondering if this is possible with xargs.
Update 1:
Input:
/a/b/c
/d/f/e
/h/i/j
Output:
mkdir a/b/c
mkdir d/f/e
mkdir h/i/j
Script:
for i in $(<test); do
echo mkdir $(echo $i | sed 's/\///') ;
done
Doing it with xargs does not work as I would expect:
xargs -a test -I line --max-procs=2 echo mkdir $(echo $line | sed 's/\///')
Obviously I need a way to execute the sed on the input for each line, but using $() does not work.
You probably want:
--max-procs=max-procs, -P max-procs
Run up to max-procs processes at a time; the default is 1. If
max-procs is 0, xargs will run as many processes as possible at
a time. Use the -n option with -P; otherwise chances are that
only one exec will be done.
http://unixhelp.ed.ac.uk/CGI/man-cgi?xargs
With GNU Parallel you can do:
cat file | perl -pe s:/:: | parallel mkdir -p
or:
cat file | parallel mkdir -p {= s:/:: =}
As being quite a newbie in linux, I have the follwing question.
I have list of files (this time resulting from svn status) and i want to create a script to loop them all and replace tabs with 4 spaces.
So I want from
....
D HTML/templates/t_bla.tpl
M HTML/templates/t_list_markt.tpl
M HTML/templates/t_vip.tpl
M HTML/templates/upsell.tpl
M HTML/templates/t_warranty.tpl
M HTML/templates/top.tpl
A + HTML/templates/t_r1.tpl
....
to something like
for i in <files>; expand -t4;do cp $i /tmp/x;expand -t4 /tmp/x > $i;done;
but I dont know how to do that...
You can use this command:
svn st | cut -c8- | xargs ls
This will cut the first 8 characters leaving only a list of file names, without Subversion flags. You can also add grep before cut to filter only some type of changes, like /^M/. xargs will pass the list of files as arguments to a given command (ls in this case).
I would use sed, like so:
for i in files
do
sed -i 's/\t/ /' "$i"
done
That big block in there is four spaces. ;-)
I haven't tested that, but it should work. And I'd back up your files just in case. The -i flag means that it will do the replacements on the files in-place, but if it messes up, you'll want to be able to restore them.
This assumes that $files contains the filenames. However, you can also use Adam's approach at grabbing the filenames, just use the sed command above without the "$i".
Not asking for any votes, but for the record I'll post the combined answer from #Adam Byrtek and #Dan Fego:
svn st | cut -c8- | xargs sed -i 's/\t/ /'
I could not test it with real subversion output, but this should do the job:
svn st | cut -c8- | while read file; do expand -t4 $file > "$file-temp"; mv "$file-temp" "$file"; done
svn st | cut -c8- will generate a list of files without subversion flags. read will then save each entry in the variable $file and expand is used to replace the tabs with four spaces in each file.
Not quite what you're asking, but perhaps you should be looking into commit hooks in subversion?
You could create a hook to block check-ins of any code that contains tabs at the start of a line, or contains tabs at all.
In the repo directory on your subversion server there'll be a directory called hooks. Put something in there which is executable called 'pre-commit' and it'll be run before anything is allowed to be committed. It can return a status to block the commit if you wish.
Here's what I have to stop php files with syntax errors being checked in:
#!/bin/bash
REPOS="$1"
TXN="$2"
PHP="/usr/bin/php"
SVNLOOK=/usr/bin/svnlook
$SVNLOOK log -t "$TXN" "$REPOS" | grep "[a-zA-Z0-9]" > /dev/null
if [ $? -ne 0 ]
then
echo 1>&2
echo "You must enter a comment" 1>&2
exit 1
fi
CHANGED=`$SVNLOOK changed -t "$TXN" "$REPOS" | awk '{print $2}'`
for LINE in $CHANGED
do
FILE=`echo $LINE | egrep \\.php$`
if [ $? == 0 ]
then
MESSAGE=`$SVNLOOK cat -t "$TXN" "$REPOS" "${FILE}" | $PHP -l`
if [ $? -ne 0 ]
then
echo 1>&2
echo "***********************************" 1>&2
echo "PHP error in: ${FILE}:" 1>&2
echo "$MESSAGE" | sed "s| -| $FILE|g" 1>&2
echo "***********************************" 1>&2
exit 1
fi
fi
done