与参数的构造函数记录？ [英] Record with parameterless constructor?

问题描述

我试图构建一个使用实体框架的Web应用程序（ASP.NET MVC3），我已经再次碰了壁。尝试运行在视图上集合foreach循环时，它抛出以下异常：

I'm trying to build a web app (ASP.NET MVC3) that uses Entity Framework, and I've once again hit a wall. It throws following exception when trying to run a foreach loop over the collection in the view:

System.InvalidOperationException：本
  类的GvG.Entities.News'没有
  参数的构造函数。

System.InvalidOperationException: The class 'GvG.Entities.News' has no parameterless constructor.

现在是我的问题，是否有可能在我的记录类型定义莫名其妙参数构造函数？

Now is my question, is it possible to somehow define a parameterless constructor on my record type?

在此刻我的记录类型如下：

My record type at the moment looks like:

type News = { 
    mutable ID:int; 
    mutable Author:string; 
    mutable Title:string; 
    mutable Content:string }

我所知道的，我可以创建烘烤等领域，而不是一类，但多数民众赞成什么，我试图避免的。

I'm aware of that I can create a class with baking-fields etc. instead, but thats what I'm trying to avoid.

推荐答案

我尝试了一些承包项目我做了F＃的团队在解决这个（espeically为实体框架），你可以找到的 F＃PowerPack的来源。它不完全测试＆放大器;你必须建立它自己。实验解决方案取代了与其他（可变）类型查询的所有F＃元组和F＃记录，然后转换结果返回给记录/元组。

I tried to solve this (espeically for Entity Framework) during some contracting project I did for the F# team and you can find an experimental solution in F# PowerPack sources. It is not fully tested & you'd have to build it yourself. The experimental solution replaces all F# tuples and F# records in the query with other (mutable) types and then transforms results back to records/tuples.

修改没看到你的问题定义类提，但我会离开这里的例子为别人谁可能来这里与同样的问题。

EDIT Didn't see the mention about defining class in your question, but I'll leave the example here for others who may come here with the same issue.

有没有简单的解决方法。不幸的解决方案是定义明确的具有属性的类：

There is no easy workaround. The unfortunate solution is to define a class with properties explicitly:

type News() =
  let mutable id = 0
  let mutable author = ""
  let mutable title = ""
  let mutable content = ""
  member x.ID with get() = id and set(v) = id <- v
  member x.Author with get() = author and set(v) = author <- v
  member x.Title with get() = title and set(v) = title <- v
  member x.Content with get() = content and set(v) = content <- v

相比，记录这是非常丑陋的，但它是做在F＃的当前版本的唯一途径。这是一些在F＃团队知道，所以有可能会在下一版本的一些更好的解决方案。

That's very ugly compared to records, but it's the only way to do it in the current version of F#. It is something that the F# team is aware of, so there may be some better solution in the next version.

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