How do I get the one by one files from SFTP server and move them do another folder in Ubuntu bash script?
#!bin/sh
FOLDER=/home/SFTP/Folder1/
sftp SFTP#ip_address
cd /home/FSTP/Folder1/
for file in "$FOLDER"*
<<EOF
cd /home/local/Folder1
get $file
EOF
mv $file /home/SFTP/Done
done
I know it's not right, but i've tried my best and if anyone can help me, i will appreciate it. Thanks in advance.
OpenSSH sftp is not very powerful client for such tasks. You would have to run it twice. First to collect list of files, use the list to generate list of commands, and execute those in a second run.
Something like this:
# Collect list of files
files=`sftp -b - user#example.com <<EOF
cd /source/folder
ls
EOF`
files=`echo $files|sed "s/.*sftp> ls//"`
# Use the list to generate list of commands for the second run
(
echo cd /source/folder
for file in $files; do
echo get $file
echo rename $file /backup/folder/$file
done
) | sftp -b - user#example.com
Before you run the script on production files, I suggest, you first output the generated command list to a file to check, if the results are as expected.
Just replace the last line with:
) > commands.txt
Maybe use SFTP internal command.
sftp get -r $remote_path $local_path
OR with the -f option to flush files to disk
sftp get -rf $remote_path $local_path
Related
I am trying to execute the scp command in such a way that it can copy .csv files from source to sink, except a few specific CSV file.
For example in the source folder I am having four files:
file1.csv, file2.csv, file3.csv, file4.csv
Out of those four files, I want to copy all files, except file4.csv, to the sink location.
When I was using the below scp command:
scp /tmp/source/*.csv /tmp/sink/
It would copy all the four CSV files to the sink location.
How can I achieve the same by using the scp command or through writing a shell script?
You can use rsync with the --exclude switch, e.g.
rsync /tmp/source/*.csv /tmp/sink/ --exclude file4.csv
Bash has an extended globbing feature which allows for this. On many installations, you have to separately enable this feature with
shopt -e extglob
With that in place, you can
scp tmp/source/(!fnord*).csv /tmp/sink/
to copy all *.csv files except fnord.csv.
This is a shell feature; the shell will expand the glob to a list of matching files - scp will have no idea how that argument list was generated.
As mentioned in your comment, rsync is not an option for you. The solution presented by tripleee works only if the source is on the client side. Here I present a solution using ssh and tar. tar does have the --exclude flag, which allows us to exclude patterns:
from server to client:
$ ssh user#server 'tar -cf - --exclude "file4.csv" /path/to/dir/*csv' \
| tar -xf - --transform='s#.*/##' -C /path/to/destination
This essentially creates a tar-ball which is send over /dev/stdout which we pipe into a tar extract. To mimick scp we need to remove the full path using --transform (See U&L). Optionally you can add the destination directory.
from client to server:
We do essentially the same, but reverse the roles:
$ tar -cf - --exclude "file4.csv" /path/to/dir/*csv \
| ssh user#server 'tar -xf - --transform="s#.*/##" -C /path/to/destination'
You could use a bash array to collect your larger set, then remove the items you don't want. For example:
files=( /tmp/src/*.csv )
for i in "${!files[#]}"; do
[[ ${files[$i]} = *file4.csv ]] && unset files[$i]
done
scp "${files[#]}" host:/tmp/sink/
Note that our for loop steps through array indices rather than values, so that we'll have the right input for the unset command if we need it.
I need to download only defined files with wget and ftp.
For example:
1.I retrieve all files using:
echo ls -R | ftp ftp://user:password#host > ./list.txt
2.Then I will parse the result and get a list with absolute paths for each file:
/path/to-the/file-1
/path/to-the/file-2
etc.
3.And now I need to download all files from the result list using wget and ftp.
And I don't want to create a separate FTP session for each file download process.
Please give your advice. Thank you.
Update:
For recursive download I'm using it: wget -r ftp://user:password#host:/ -nH -P /download/path. It works great, but I need to pass a file with a list of remote files for downloading via FTP with one FTP session.
Sorry, I missed the "single session" part when I commented. I think you need to have your script generate a second script to run a single FTP session.
So, your script will not do any FTP itself, it will just write another script that does the transfers. So, it will write a script that does this
ftp -n <SOMEADDRESS> <<EOS
quote USER <USERNAME>
quote PASS <PASSWORD>
bin
get file1 localname1
get file2 localname2
...
get fileN localnameN
quit
EOS
Then it will execute that script, by doing:
bash < thatScript
So your script will look like this:
#!/bin/bash
ScriptName=funkyFTPer
cat - <<END > $ScriptName
ftp -n 192.168.0.1 <<EOS
quote USER freddy
quote PASS frog
END
# Your selection code goes here ***PHNQZ***
echo get file1 localname1 >> $ScriptName
echo get file2 localname2 >> $ScriptName
echo get fileN localnameN >> $ScriptName
echo quit >> $ScriptName
echo EOS >> $ScriptName
echo "Now run bash < $ScriptName"
Then delete the script as it contains your password. Or you can put the password in your .netrc file.
As regards creating directories locally, you can do that in the first script using mkdir -p. The -p has the advantage that it creates all directories in between in one go and doesn't get upset if they already exist.
So, just looking at the area of code where it says ***PHNQZ*** above, let's say your code decides you need file freddy/frog/c.txt, you could do:
remotename="freddy/frog/c.txt"
localdir=${remotename%/*} # Get just directory part using "bash Parameter Substitution"
mkdir -p "$localdir" # make directory and all parts in between
I am trying to use scp to copy large log files from a remote server. However I want only the lines in remote log file that has a string "Fail".
This is how I am doing it currently
scp user#ip:remote_folder/logfile* /localfolder
This copies all the files starting with logfile in remote server to my local folder. The files are pretty large and I need to copy only the lines in those log file, containing the string "Fail" from remote server. Can any body tell me how to do this? Can I use cat or grep command?
Use grep on the remote machine and filter the output into file name and content:
#!/usr/bin/env bash
BASEDIR=~/temp/log
IFS=$'\n'
for match in `ssh user#ip grep -r Fail "remote_folder/logfile*"`
do
IFS=: read file line <<< $match
mkdir -p `dirname $BASEDIR/$file`
echo $line >> $BASEDIR/$file
done
You might want to look at an explanation to IFS in combination with read.
ssh user#ip grep Fail remote_folder/logfile*
The goal is I want to monitor one directory from different server. For example the remote server is user#host.
I have list.txt that contents list of file that will be moved. And list.txt located in a remote server.
Currently I have this code.
ssh user#host cat /full-path/list.txt |
{
while read line;
do mv user#host:/full-path/$line user#host:/full-path/done/;
done;
}
When I run the code above, error exists. There's no such file or directory.
But when I log in to user#host and cat one file randomly from list.txt, the file exists.
The while loop runs on the local server. You need to put the script in quotes so it's an argument to the ssh command.
... Or a here document, like this:
ssh user#host <<':'
while read line; do
mv /full-path/"$line" /full-path/done/
done </full-path/list.txt
:
... or more succinctly
ssh user#host 'cd /full-path && xargs -a list.txt mv -t done'
Notice also the absence of a useless cat and the local file name resolution (mv would have no idea about the SSH remote path syntax you were trying to use).
I have a bash script that basically should work like below:
get build number from user and put it in buildNum var
prep the build on local machine by calling a local script with buildNum as it's argument
sftp the prepped zip file to remote server1
Do this:
ssh -v $server1 <<EOF
rm -rvf path1-on-remote-server1/*
cd path2-on-remote-server1
unzip ../prepped-zip-file-$buildNum.zip
exit
EOF
sftp the prepped zip file to remote server2
The problem i am having is that on the forth step of number 4, $buildNum is not known to the remote server and it fails.
I tried the following two solutions and both failed:
use double quotes "unzip ../prepped-zip-file-$buildNum.zip" which resulted in "unzip ../prepped-zip-file-11.6.zip: Command not found.
tried to get the build number again from the user during the SSH session which failed again by not even waiting for my input and looking for a zip file without the build number at the end of the name, as the var was empty,
i did :
ssh -v $server1 <<EOF
rm -rvf path1-on-remote-server1/*
cd path2-on-remote-server1
echo "enter build num once more: "
read bNum
unzip ../prepped-zip-file-$bNum.zip
exit
EOF
Any suggestions on how to achieve what i am after.
Thanks in advance
Are you sure this doesn't work?
ssh -v $server1 <<EOF
rm -rvf path1-on-remote-server1/*
cd path2-on-remote-server1
unzip ../prepped-zip-file-$buildNum.zip
exit
EOF
When I try it on my machine, with cat instead of ssh -v $server1 for testing, the variable does get substituted into the here-document, just as if the entire document had been on a command line. The remove shell never needs to know there was a variable in the first place.
Actually, though, you may want to give the remote command on the ssh command line rather than redirecting the standard input. This would be more robust in case some of the parts of it unexpectedly decide to try reading from stdin:
ssh -v $server1 "rm -rvf path1-on-remote-server1/*
cd path2-on-remote-server1
unzip ../prepped-zip-file-$buildNum.zip"
(Note that multi-line double-quoted strings are okay with bash).
#!/bin/sh
printf "Enter Build Number: "
read BUILD_NUM
cat << EOF | ssh $server1
hostname
echo "${BUILD_NUM}"
uptime
EOF
That works for me.