如何在xaml中绑定到当前riacontext用户

发布于 2024-09-03 00:34:55 字数 847 浏览 1 评论 0原文

我有一个具有“getuserbyguid”方法的数据上下文，我想传入当前登录的user.userid作为参数，但我不知道如何通过xaml绑定到当前登录的用户。我尝试过 {Binding Path=User.UserId} 但没有任何运气。我正在使用内置的 riaservices 身份验证方法，因此用户信息应该在 riacontext 中公开，还是我错了？

例如，我有这个，

<riaControls:DomainDataSource x:Name="FollowingGridData" AutoLoad="True" QueryName="GetUsersFollowedByIDQuery" LoadSize="20">
        <riaControls:DomainDataSource.DomainContext>
            <my:NotesDomainContext />
        </riaControls:DomainDataSource.DomainContext>
        <riaControls:DomainDataSource.QueryParameters>
            <riaControls:Parameter ParameterName="userguid" Value="{Binding Path=User.UserId}" />
        </riaControls:DomainDataSource.QueryParameters>
    </riaControls:DomainDataSource>

但它给了我一个错误，说它不是一个 guid，这意味着它不能正确绑定

原文

I Have a datacontext that has a "getuserbyguid" method, i want to pass in the current logged in user.userid as a parameter, but I don't know how to bind to the current logged in user through xaml. I've tried {Binding Path=User.UserId} but without any luck. I'm using the built in riaservices authentication methods, so the userinfo should be exposed in the riacontext, or am I wrong about this?

I have this for instance

<riaControls:DomainDataSource x:Name="FollowingGridData" AutoLoad="True" QueryName="GetUsersFollowedByIDQuery" LoadSize="20">
        <riaControls:DomainDataSource.DomainContext>
            <my:NotesDomainContext />
        </riaControls:DomainDataSource.DomainContext>
        <riaControls:DomainDataSource.QueryParameters>
            <riaControls:Parameter ParameterName="userguid" Value="{Binding Path=User.UserId}" />
        </riaControls:DomainDataSource.QueryParameters>
    </riaControls:DomainDataSource>

But it gives me an error saying that it's not a guid, meaning that it must not be binding correctly

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