CODEWITHSUNDEEP
bashshellsftp
SomeGuy
SomeGuy

Reputation: 163

Passing variables to sftp batch

I'm writing a script that needs to pass a variable to the sftp batch. I have been able to get some commands working based on other documentation I've searched out, but can't quite get to what I need.

The end-goal is to work similar to a file test operator on a remote server: ( if [ -f "$a" ] then :; else exit 0;)

Ultimately, I want the file to continue running the script if the file exists (:), or exit 0 if it does NOT exist (not exit 1). The remote machine is a Windows server, not Linux.

Here's what I have:

NOTE - the variable I'm trying to pass, $source_dir, changes based on the input parameter of the script that calls this function. This and the ls wildcard is the tricky part. I have been able to make it work when looking for a specific file, but not just "any" file.

source_dir=/this/directory/changes

RemoteCheck () {

    /bin/echo "cd $source_dir" > someBatch.txt
    /bin/echo "ls *" >> someBatch.txt

    /usr/bin/sftp -b someBatch.txt -oPort=${sftp_port} ${sftp_ip}

    exit_code=$?;
    if [ $exit_code -eq 0 ]; then
        :
    else
        exit 0
    fi
}

There may be a better way to do this, but I have searched multiple forums and have not yet found a way to manipulate this.

Upvotes: 0

Views: 2592

Answers (1)

Martin Prikryl
Martin Prikryl

Reputation: 202187

You cannot test for existence of any file using just exit code of the OpenSSH sftp.

You can redirect the sftp output to a file and parse it to see if there are any files.

You can use shell echo command to delimit the listing from the rest of the output like:

!echo listing-start
ls
!echo listing-end

Upvotes: 1

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