Metadata-Version: 1.1
Name: ozpy
Version: 1.0
Summary: UNKNOWN
Home-page: https://github.com/iomarmochtar/ozpy
Author: Imam Omar Mochtar
Author-email: iomarmochtar@gmail.com
License: GPL
Description: OZPY
        ====
        
        Python (2.7+ including 3.X) library for accessing Zimbra SOAP (https://wiki.zimbra.com/wiki/SOAP_API_Reference_Material_Beginning_with_ZCS_8) by using builtin Python library (no dependency required).
        Currently this library split into 2 parts: Zmprov and Mailbox.
        
        not all zmprov command(s) has been implemented, because i add them only based on customer/project needs
        
        but you can add your own by extending **OZSoap** which is base of **Zmprov** and **Mailbox**
        for example creating new COS (Class Of Service)
        
        .. code-block:: python
        
        	from ozpy.base import OZSoap
        
        	class NewClass(OZSoap):
        
        		def create_cos(self, name):
        			body = {"name": [{
        			  "_content": name
        			}]}
        			return self.send("CreateCos", body)
        
        or directly call the soap method (by omitting Request suffix)
        
        .. code-block:: python
        
        	# zmsoap_obj is an instance from class OZSoap
        
        	zmsoap_obj.CreateCos(
        		name=[{"_content": "barudong"}]
        	)
        
        you can use **zmsoap** to get the parameters in soap body by using **--verbose** and **--json**
        
        .. code-block:: bash
        
        	zmsoap -z CreateCosRequest/name=new_cos  --json --verbose
        
        Examples
        --------
        
        fetch all account
        
        .. code-block:: python
        
        	from ozpy.zmprov import Zmprov
        
        	zmprov = Zmprov(
        		username="admin@mail.com",
        		password="superpassword",
        		soapurl="https://192.168.113.75:7071/service/admin/soap"
        	)
        	print zmprov.gaa()
        
        
        Sending email
        
        .. code-block:: python
        
        	from ozpy.mailbox import Mailbox
        
        	mbx = Mailbox(
        		username="user1@mail.com",
        		password="superpassword",
        		soapurl="https://192.168.113.75/service/soap"
        	)
        	mbx.sendMail('admin@mail.com', 'This is subject', 'Email content')
        
Keywords: Zimbra Python library
Platform: UNKNOWN
Classifier: Programming Language :: Python :: 2.7
Classifier: Programming Language :: Python :: 3.7
Classifier: Natural Language :: English
