自定义pyramid_formalchemy表单

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.

los*_*rje 5

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......

  1. 例如,如果您有一个模型User,那么您将需要一个名为UserFieldSet的FieldSet.

  2. UserFieldSet必须是实例而不是类.

  3. 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.