CODEWITHSUNDEEP
pythondjango
User100696
User100696

Reputation: 707

Django - Passing variables from a function to another

I need some of your help please,

I'm working with pysftp this is working great but now I'm trying to make it work to my project in Django worked great in console but I want to get the data from a form so I won't need to use the console to do that.

here's my view:

def sftp_form(request):
    if request.method == 'POST':
        form = sftpForm(request.POST or None)
        if form.is_valid():
            data = form.cleaned_data
            host = data['host']
            usuario = data['usuario']
            clave = data['clave']
            print host
            print usuario
            print clave
    else:
        form=sftpForm()
    return render(request, 'sftp.html', {'form':form})


def SFTP_subir():
    host = raw_input('ingrese el host: ') # I want form's host here.
    usuario = raw_input('ingrese el usuario: ')# I want form's usuario here.
    clave = raw_input('ingrese la clave: ')# I want form's clave here.
    try:
        transferencia = sftp.Connection(host=host, username=usuario, password=clave)

        remotepath= 'remotepath'
        localpath="mylocalpath"

        transferencia.put(localpath,remotepath)

        print ('\n' + 'Sucess.')

    except Exception, e:
        print str(e)

as you can see in my code sftp_subir() it's asking me for host,usuario and clave from console, but I want to make it work with sftp_form() host,usuario and clave.

Upvotes: 1

Views: 2953

Answers (3)

Juan D. Gómez
Juan D. Gómez

Reputation: 499

In your view:

def sftp_form(request):
    if request.method == 'POST':
        form = sftpForm(request.POST or None)
        if form.is_valid():
            data = form.cleaned_data
            host = data['host']
            usuario = data['usuario']
            clave = data['clave']
            print host
            print usuario
            print clave
            SFTP_subir(host, usuario, clave) # here you invoke the function, passing variables as arguments 
    else:
        form=sftpForm()
    return render(request, 'sftp.html', {'form':form})

Then refactor your function to receive those params:

def SFTP_subir(host, usuario, clave):
    try:
        transferencia = sftp.Connection(host=host, username=usuario, password=clave)

        remotepath= 'remotepath'
        localpath="mylocalpath"

        transferencia.put(localpath,remotepath)

        print ('\n' + 'Sucess.')

    except Exception, e:
        print str(e)

Upvotes: 1

sureshvv
sureshvv

Reputation: 4422

you can do the

sftp.connect(...)
...
<4 lines following>

inside the request.method == "POST" block instead of your print statements.

Upvotes: 0

e4c5
e4c5

Reputation: 53734

There seem to be a slight mixup here, you can't use raw_input in a django web app. If you using Django as a CLI you can't use an HTTP request. As @sayse suggested in the comments, if you are using a view in a web app all you need to do is to define your second function to be one that accepts paramers

def sftp_form(request):
    if request.method == 'POST':
        form = sftpForm(request.POST or None)
        if form.is_valid():
            data = form.cleaned_data
            host = data['host']
            usuario = data['usuario']
            clave = data['clave']
            SFTP_subir(hosts, usuario,clave)

    else:
        form=sftpForm()
    return render(request, 'sftp.html', {'form':form})


def SFTP_subir(hosts, usuario,clave):
    try:
        transferencia = sftp.Connection(host=host, username=usuario, password=clave)

        remotepath= 'remotepath'
        localpath="mylocalpath"

        transferencia.put(localpath,remotepath)

        print ('\n' + 'Sucess.')

    except Exception, e:
        print str(e)

Once you make this code you still have a long way to go because your SFTP method doesn't return any usefull response.

Upvotes: 2

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