los*_*rje 2 python formalchemy pyramid
我在pyramid_formalchemy中创建自定义表单时遇到问题.我怀疑包中有一个错误,并想确认我没有遗漏任何东西.我的设置如下:
def includeme(config):
config.include('pyramid_formalchemy')
# Adding the jquery libraries
config.include('fa.jquery')
# Adding the package specific routes
config.include('myapp.web.formalchemy.faroutes')
config.formalchemy_admin('admin',
models=[User],
forms=faforms,
session_factory=session,
view='fa.jquery.pyramid.ModelView',
factory='myapp.model.RootFactory')
config.formalchemy_model('user',
model='myapp.model.user.User',
session_factory=session,
view='fa.jquery.pyramid.ModelView',
factory='myapp.model.RootFactory')
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faforms是一个包含我的自定义表单的模块:
from myapp.model.user import User
from formalchemy import FieldSet
from formalchemy import Grid
class UserFieldSet(FieldSet):
def __init__(self):
FieldSet.__init__(self, User)
self.configure()
class UserGrid(Grid):
def __init__(self):
Grid.__init__(self, User)
self.configure()
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如果我评论上面的两个类,formalchemy工作正常.我可以查看用户,我可以编辑它们.
当我把这两个课程放进去遇到问题时.问题是pyramid_formalchemy从模块的命名空间中获取UserGrid和UserFieldSet,然后尝试使用它们,就好像它们是实例化的类一样.这打破了事情.另一方面,如果pyramid_formalchemy没有找到将动态创建类的类并实例化它们.我相信有问题的代码在pyramid_formalchemy/views.py中,第236行从get_grid()函数开始:
def get_grid(self):
"""return a Grid object"""
request = self.request
model_name = request.model_name
form_name = '%sGrid' % model_name
if hasattr(request.forms, form_name):
g = getattr(request.forms, form_name) <-- g is a class type not an
g.engine = g.engine or self.engine <-- instance!
g.readonly = True <-- why is it not instantiated?
g._request = self.request
self.update_grid(g)
return g
model = self.context.get_model() <-- UserGrid not found in faforms
grid = self.grid_class(model) <-- module.
grid.engine = self.engine <-- so a Grid is instantiated
if not isinstance(request.forms, list):
# add default grid to form module eg: caching
setattr(request.forms, form_name, grid)
grid = grid.copy()
grid._request = self.request
self.update_grid(grid)
return grid
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在这里,您可以看到是否找不到匹配的网格(或字段集),它将被实例化,但如果找到它,将直接使用类类型,但实际上不会实例化.
这有什么想法?我设置错了吗?
基本上我使用CSRF令牌,所以我需要自定义我的表单以从会话中获取令牌.
谢谢.
pyramid_formalchemy对您如何设置表单文件做出了一些(大多数是未记录的)假设.以下是我遇到并解决的陷阱......
例如,如果您有一个模型User,那么您将需要一个名为UserFieldSet的FieldSet.
UserFieldSet必须是实例而不是类.
小心你的进口,或者你将打破pyramid_formalchemy所做的假设.如果您有一个模型类User,则导入User的包,但不导入User类本身.然后通过在引用前加上包名称来引用该类.
以下是使用上述3点的工作示例.
from myapp.model import user
from formalchemy import Field
from formalchemy import FieldSet
from formalchemy import Grid
UserFieldSet = FieldSet(user.User)
UserFieldSet.configure()
UserGrid = Grid(user.User)
UserGrid.configure()
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您还可以向formalchemy小组发布更多信息.