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Writing Controllers in openERP modules.

By
P Mittal
on 5/27/13, 11:37 PM 18,479 views

I have 3 questions that might require a detailed answer. (Links to tutorials or good illustrations would be really helpful)

  1. How to map URL's to Controller functions. (like URLDispatcher in Django)?
  2. How to write Controllers in OpenERP? (an additional comparison with Django Views would be great).
  3. I see something like this in a controller.

class MyController(oeweb.Controller):

  _cp_path = '/some-path'

Does _cp_path denote the corresponding URL, assigned to this Controller?

hi all  ! i want to upload file in my page, but request cannot get file upload. please tell me why? detail :  */ /*line code*/ cgi.FieldStorage(fp=fp, environ=environ) /*cannot get file upload Tks ! my code: /temp = """<html><body>/ /<form method="post" action="/sa/save">/ /<p>File: <input type="file" name="myfile"></p>/ /<p><input type="submit" value="Upload"></p>/ /</form>/ /</body></html>"""/ /class simplecontroller(http.Controller):/ /_cp_path = '/sa'/ /@http./ httprequest [1] /def index(self, req, **kwargs):/ /return temp/ /@http./ httprequest [2] /def save(self, req, **kwargs):/ /environ = req./ httprequest [3] /.environ/ /fp=environ['wsgi.input']/ */fs = cgi.FieldStorage(fp=fp, environ=environ)# page load forever when run to this line/* /f = fs.getfirst('myfile')/ /filepath =req.params.get('myfile')/ /fout = open(filepath,'w') / /fout.write(f.file.read()) / /fout.close() / /return "upload successfully"/ [1] http://terp.vn/forum/tags/httprequest/ [2] http://terp.vn/forum/tags/httprequest/ [3] http://terp.vn/forum/tags/httprequest/

Nguyễn Xuân Huy
on 6/12/14, 1:12 AM
5

Gopakumar N G

--Gopakumar N G--
1198
| 5 3 7
Cochin, India
--Gopakumar N G--

Senior Software Engineer at BroadTech IT Solutions Pvt. Ltd.

Gopakumar N G
On 11/13/13, 8:14 AM

Web controllers are classes in OpenERP able to catch the http requests sent by any browser. They allow to generate html pages to be served like any web server, implement new methods to be used by the Javascript client, etc.

Controllers File By convention the controllers should be placed in the controllers directory of the module. Example: web_example

├── controllers
│   ├── __init__.py
│   └── my_controllers.py
├── __init__.py
└── __openerp__.py

In __init__.py you must add:

import controllers

And here is the content of controllers/__init__.py:

import my_controllers

Now you can put the following content in controllers/my_controllers.py:

import openerp.http as http
from openerp.http import request

Controller Declaration

In your controllers file, you can now declare a controller this way:

class MyController(http.Controller):

    @http.route('/my_url/some_html', type="http")
    def some_html(self):
        return "<h1>This is a test</h1>"

    @http.route('/my_url/some_json', type="json")
    def some_json(self):
        return {"sample_dictionary": "This is a sample JSON dictionary"}

A controller must inherit from http.Controller. Each time you define a method with @http.route() it defines a url to match. As example, the some_html() method will be called a client query the /my_url/some_html url.

Refer Web Controllers

7
Michael Watchorn
On 2/25/14, 2:13 PM

The link "Web Controllers" posted by Gopakumar N G is ONLY applicable to people developing with TRUNK and NOT for people developing controllers meant for V7.
I am not fully skilled at the art of web controllers (yet!), so I can't answer Question 1.
Question 3 : Yes, _cp_path denotes the path. The thing to remember is that the function name becomes the last part of the URL (unless you're using a function named index, which is the default page served).

Question 2: Here's my example code:


# -*- coding: utf-8 -*-
import openerp.addons.web.http as http

class SimpleController(http.Controller):
    _cp_path = '/sc'

    @http.httprequest
    def index(self, req, s_action=None, **kw):

        SIMPLE_TEMPLATE = """
<html>
    <head>
        <meta http-equiv="X-UA-Compatible" content="IE=edge,chrome=1"/>
        <meta http-equiv="content-type" content="text/html; charset=utf-8" />
        <title>OpenERP V7 Simple Web Controller Example using HttpRequest</title>
    </head>
    <body>
        <form action="/sc/Page2" method="POST">
            <button name="GoToPage2" value="*">Go To Page 2</button>
        </form>
    </body>
</html>
"""
        return SIMPLE_TEMPLATE

    @http.httprequest
    def Page2(self, req, s_action=None, **kw):
        Page2_View = "<html><head></head><body>AND NOW YOU'RE ON PAGE 2</body></html>"
        return Page2_View

The folder definitions and __init__.py and __openerp__.py files are just as they are in the trunk documentation.
image description

It's just the contents of the 'my_controller.py' file that is different.
The following modules have controllers which go into more detail:

  • point_of_sale
  • web_diagram
  • web (see ./controllers/Testing.py)
    I still don't agree with the karma limitations. I have 4 applicable images I took, which I can't post because my karma is too low. Perhaps when my karma is high enough I will edit this post to include those pictures. In the mean time here's the point of them:
    EDIT: My karma is now high enough to post images:
    Default page served: (note the URL)

    image description

    Index page served, same as default: (note the URL)

    image description

    Click on the button and voila:

    image description

Note: I had to restart my openerp server, and update the module (in Settings | Installed Modules | Simple Web Controller Module | Upgrade ) every time I made a change. Or else I get "Not Found".

Good Job Michael.

Med Said BARA
on 2/25/14, 5:23 PM

How do you link to the custom controller action in a form view? Or more generally how do you link to these custom created controllers?

Ramy Wafa
on 4/14/14, 9:42 AM

Ramy - In my case it is a direct link from the users' Favorites in their browser. You could use a URL field, but I don't know how to do it from a button.

Michael Watchorn
on 4/23/14, 3:58 PM

Hi Michael, I having a html file when i click one button this html need to show how to do this...

sridhar
on 1/30/15, 6:19 AM
4

Davide Corio

--Davide Corio--
1822
| 4 3 7
Bricherasio, Italy
--Davide Corio--

Odoo Evangelist and Consultant

Davide Corio
On 5/28/13, 5:24 AM

_cp_path stands for "CherryPy Path", even though cherrypy isn't used anymore in openerp (replaced by werkzeug in v6.1).

anyway, _cp_path is used to map controllers, like in django.

For instance,

_cp_path = '/web/export/xls'

is mapped to the xls export function.

Unfortunately there's no doc about this topic. All you can find is here: https://doc.openerp.com/trunk/web/module/

You must decorate your controller "views" for them to be served. So within your class definition all view functions must be passed to the http object by preceding your view function definition with - @openerpweb.httprequest

mike
on 5/28/13, 4:35 PM
1
roman onosovski
On 6/10/14, 9:45 AM

Hello all! Thank you for this post as it helped me tremendously. My problem arises when I restart my openerp-server instance.. my module stays installed.. however the controller that worked prior to restart will not work until I update modules list.. then it starts responding again.. any ideas why? Thank you in advance

0

does this mean i need to use a web controller to make a get request to a local webservice? thanks

0
Cesar Lopez Aguillon
On 6/24/14, 9:37 AM

You must create the directory static/, so that the controller is saved.

addons/web/http.py line 566

    def load_addons(self):
        """ Load all addons from addons patch containg static files and
        controllers and configure them.  """


addons/web/controllers/main.py line 182:

def module_installed(req):
    # Candidates module the current heuristic is the /static dir
    loadable = openerpweb.addons_manifest.keys()
    modules = {}

 

Best regards,

CLA

 

0
Nguyễn Xuân Huy
On 6/12/14, 2:28 AM

Hi all,

Thank you for your post, i am have a issue when create page to upload file, i cannot get file upload from request.

Any one help me, thank very must !

my code:

import ...

temp = """<html><body>
<form method="post" action="/sa/save">
<p>File: <input type="file" name="myfile"></p>
<p><input type="submit" value="Upload"></p>
</form>
</body></html>"""

class simplecontroller(http.Controller):
_cp_path = '/sa'

@http.httprequest
def index(self, req, **kwargs):
return temp

@http.httprequest
def save(self, req, **kwargs):
environ = req.httprequest.environ
fp=environ['wsgi.input']
fs = cgi.FieldStorage(fp=fp, environ=environ)# page load forever when run to this line
f = fs.getfirst('myfile')
filepath =req.params.get('myfile')
fout = open(filepath,'w') 
fout.write(f.file.read()) 
fout.close() 

return "upload successfully"

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Asked: 5/27/13, 11:37 PM
Seen: 18479 times
Last updated: 3/28/16, 5:08 AM