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I need an XML-RPC snippet in python that confirms and/or starts an MO (v 6.1).

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Abbandona
Autore Risposta migliore

1 Setup

import xmlrpclib

HOST = 'localhost'
PORT = 8069
DB = 'demo'
USER = 'admin'
PASS = 'admin'

url = 'http://%s:%d/xmlrpc/'%(HOST,PORT)
object_proxy = xmlrpclib.ServerProxy(url+'object') 
common_proxy = xmlrpclib.ServerProxy(url+'common')

2 Login

uid = common_proxy.login(DB,USER,PASS)
print "Logged in as %s (uid:%d)"%(USER,uid)

def execute(*args):
    return object_proxy.execute(DB, uid, PASS, *args)

3 Search

mo_ids = execute('mrp.production', 'search', [('state','=','draft')])
print mo_ids

4 Confirm & Produce

for mo in mo_ids:

    # confirm  MOs
    object_proxy.exec_workflow(DB,uid, PASS, 'mrp.production', 'button_confirm', mo)
    # You may want to add a step for Check availability or Force availability 

    # produce MOs
    object_proxy.exec_workflow(DB,uid, PASS, 'mrp.production', 'button_produce', mo)
Avatar
Abbandona

Hey Ray,

Thanks for you post.

What would the last step be?

object_proxy.exec_workflow(DB,uid, PASS, 'mrp.production', 'button_produce_done', mo

would be to easy..:)

I need to find a way to call "action_produce". Then i would be able to change status and finish the production

cheers!

Risposta migliore

Hey Ray,

Thanks for you post.

What would the last step be?

object_proxy.exec_workflow(DB,uid, PASS, 'mrp.production', 'button_produce_done', mo

would be to easy..:)

I need to find a way to call "action_produce". Then i would be able to change status and finish the production

cheers!

Avatar
Abbandona
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