无法通过 ImportError [Odoo 9.0]

Can't get past ImportError [Odoo 9.0]

我刚刚 运行 浏览了 v9 的 "Building a Module" 文档,并且正在跟进。

我的代码看起来与此处的示例相同:https://www.odoo.com/documentation/9.0/howtos/backend.html

当我启动服务器时,这是我的错误:

Traceback (most recent call last):

File "/usr/lib/python2.7/dist-packages/werkzeug/serving.py", line 177, in run_wsgi

execute(self.server.app)

File "/usr/lib/python2.7/dist-packages/werkzeug/serving.py", line 165, in execute

application_iter = app(environ, start_response)

File "/usr/lib/python2.7/dist-packages/openerp/service/server.py", line 245, in app

return self.app(e, s)

File "/usr/lib/python2.7/dist-packages/openerp/service/wsgi_server.py", line 184, in application

return application_unproxied(environ, start_response)

File "/usr/lib/python2.7/dist-packages/openerp/service/wsgi_server.py", line 170, in application_unproxied

result = handler(environ, start_response)

File "/usr/lib/python2.7/dist-packages/openerp/http.py", line 1488, in __call__

return self.dispatch(environ, start_response)

File "/usr/lib/python2.7/dist-packages/openerp/http.py", line 1462, in __call__

return self.app(environ, start_wrapped)

File "/usr/lib/python2.7/dist-packages/werkzeug/wsgi.py", line 579, in __call__

return self.app(environ, start_response)

File "/usr/lib/python2.7/dist-packages/openerp/http.py", line 1637, in dispatch

ir_http = request.registry['ir.http']

File "/usr/lib/python2.7/dist-packages/openerp/http.py", line 360, in registry

return openerp.modules.registry.RegistryManager.get(self.db) if self.db else None

File "/usr/lib/python2.7/dist-packages/openerp/modules/registry.py", line 354, in get

update_module)

File "/usr/lib/python2.7/dist-packages/openerp/modules/registry.py", line 385, in new

openerp.modules.load_modules(registry._db, force_demo, status, update_module)

File "/usr/lib/python2.7/dist-packages/openerp/modules/loading.py", line 334, in load_modules

force, status, report, loaded_modules, update_module)

File "/usr/lib/python2.7/dist-packages/openerp/modules/loading.py", line 237, in load_marked_modules

loaded, processed = load_module_graph(cr, graph, progressdict, report=report, skip_modules=loaded_modules, perform_checks=perform_checks)

File "/usr/lib/python2.7/dist-packages/openerp/modules/loading.py", line 123, in load_module_graph

load_openerp_module(package.name)

File "/usr/lib/python2.7/dist-packages/openerp/modules/module.py", line 324, in load_openerp_module

__import__('openerp.addons.' + module_name)

File "/usr/lib/python2.7/dist-packages/openerp/modules/module.py", line 61, in load_module

mod = imp.load_module('openerp.addons.' + module_part, f, path, descr)

File "/home/lslaz/odoo-dev/addons/openacademy/__init__.py", line 2, in <module>

from . import models

ImportError: cannot import name models

^C2016-02-03 19:54:08,626 3329 INFO ? openerp.service.server: Initiating shutdown

2016-02-03 19:54:08,626 3329 INFO ? openerp.service.server: Hit CTRL-C again or send a second signal to force the shutdown.

如果我从 init.py 中删除 "from . import models",我会得到一个不同的错误:

ParseError: "Invalid model name in the action definition.
None" while parsing /home/lslaz/odoo-dev/addons/openacademy/views/views.xml:8, near

<record model="ir.actions.act_window" id="course_list_action">
    <field name="name">Courses</field>
    <field name="res_model">openacademy.course</field>
    <field name="view_type">form</field>
    <field name="view_mode">tree,form</field>
    <field name="help" type="html">
        <p class="oe_view_nocontent_create">
            Create the first course
        </p>
    </field>
</record>

请帮忙!我相信这两件事都是应该的,但在我弄清楚问题出在哪里之前我无法前进。

尝试从行 from . import models 中删除 .,使其显示为 from import models

我不知道为什么,但在我的例子中,Odoo 在新应用程序中生成类似 .py 文件的模型,但不像文件夹。

如果你想为你的模型使用文件夹,你需要:

1) 像这样在 __init__.py 中导入模型:

# -*- coding: utf-8 -*-

import models

2) 在文件夹 models 中创建 __init__.py 并像这样导入所有模型:

# -*- coding: utf-8 -*-

import my_model_one
import one_more_model
# etc.

我知道这有点晚了,但我也在学习教程,这是我解决这个问题的方法:

创建 wizard.py 文件时,教程未显示必要的导入。但是你必须导入以下内容:

# -*- coding: utf-8 -*-
from openerp import fields, models, api

class Wizard(models.TransientModel): ...

希望这对某人有所帮助。