How to create directory if doesn’t exists in sftp

sftpshell

I want to create a directory if it doesn't exists after login to sftp server.

test.sh

sftp name@example.com << EOF
mkdir test
put test.xml
bye
EOF

Now i call test.sh and upload different files each time to test folder. When running this

mkdir test

First time it works and second time it throws Couldn't create directory: Failure error?

How to create a directory if doesn't exists and if exists don't create directory in sftp.

Best Solution

man 1 sftp (from openssh-client package):

-b batchfile

    Batch mode reads a series of commands from an input
    batchfile instead of stdin. Since it lacks user
    interaction it should be used in conjunction with
    non-interactive authentication. A batchfile of ‘-’
    may be used to indicate standard input. sftp will
    abort if any of the following commands fail: get,
    put, reget, reput, rename, ln, rm, mkdir, chdir, ls,
    lchdir, chmod, chown, chgrp, lpwd, df, symlink, and
    lmkdir. Termination on error can be suppressed on a
    command by command basis by prefixing the command
    with a ‘-’ character (for example, -rm /tmp/blah*).

So:

{
  echo -mkdir dir1
  echo -mkdir dir1/dir2
  echo -mkdir dir1/dir2/dir3
} | sftp -b - $user@$host

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

To check if a directory exists in a shell script, you can use the following:

if [ -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY exists.
fi

Or to check if a directory doesn't exist:

if [ ! -d "$DIRECTORY" ]; then
  # Control will enter here if $DIRECTORY doesn't exist.
fi

However, as Jon Ericson points out, subsequent commands may not work as intended if you do not take into account that a symbolic link to a directory will also pass this check. E.g. running this:

ln -s "$ACTUAL_DIR" "$SYMLINK"
if [ -d "$SYMLINK" ]; then 
  rmdir "$SYMLINK" 
fi

Will produce the error message:

rmdir: failed to remove `symlink': Not a directory

So symbolic links may have to be treated differently, if subsequent commands expect directories:

if [ -d "$LINK_OR_DIR" ]; then 
  if [ -L "$LINK_OR_DIR" ]; then
    # It is a symlink!
    # Symbolic link specific commands go here.
    rm "$LINK_OR_DIR"
  else
    # It's a directory!
    # Directory command goes here.
    rmdir "$LINK_OR_DIR"
  fi
fi

Take particular note of the double-quotes used to wrap the variables. The reason for this is explained by 8jean in another answer.

If the variables contain spaces or other unusual characters it will probably cause the script to fail.

Python – How to execute a program or call a system command

Use the subprocess module in the standard library:

import subprocess
subprocess.run(["ls", "-l"])

The advantage of subprocess.run over os.system is that it is more flexible (you can get the stdout, stderr, the "real" status code, better error handling, etc...).

Even the documentation for os.system recommends using subprocess instead:

The subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using this function. See the Replacing Older Functions with the subprocess Module section in the subprocess documentation for some helpful recipes.

On Python 3.4 and earlier, use subprocess.call instead of .run:

subprocess.call(["ls", "-l"])

Best Solution

Related Solutions

Bash – How to check if a directory exists in a Bash shell script

Python – How to execute a program or call a system command

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