为什么我会收到“错误:...必须是引用类型"

问题描述

在各种数据库表中，我有一个属性列和一个值列.我正在使用 Linq to SQL 访问数据库.

In various database tables I have both a property and a value column. I'm using Linq to SQL to access the database.

我正在编写一个方法，它返回一个包含从给定数据库表中检索到的属性/值的字典:

I'm writing a method which returns a dictionary containing the properties/values retrieved from the given database table:

private static Dictionary<string, string> GetProperties<T>(Table<T> table)
{
    Dictionary<string, string> properties = new Dictionary<string, string>();

    foreach (var row in table)
    {
        properties[row.Property]=row.Value;
    }

    return properties;
}

编译后，我得到:

错误 1 ​​类型T"必须是引用类型，才能将其用作泛型类型或方法System.Data.Linq.Table"中的参数TEntity"

我尝试搜索此错误消息，但没有成功.

I've tried searching for this error message without luck.

搜索 StackOverflow，这个问题似乎很相似，但关于参数列表:Generic List<＞作为方法的参数 - 尽管该参数在该问题的答案中仍然不是引用类型.

Searching StackOverflow, this question seems similar, though regarding a parameter List: Generic List<T> as parameter on method - though the parameter still isn't a reference type in the answers to that question, either.

阅读 MSDN 上的 C# 编程指南:http://msdn.microsoft.com/en-us/library/twcad0zb(VS.80).aspx 我看他们的例子都是通过引用传递参数的.但是，在我的特定情况下，我看不到如何通过引用传递，因为泛型类型仅用于指定 Table 的泛型类型.

Reading the C# Programming Guide on MSDN: http://msdn.microsoft.com/en-us/library/twcad0zb(VS.80).aspx I see their examples all pass the parameters by reference. However, I can't see how to pass by reference in my particular case, since the generic type is just for specifying the generic type of Table.

任何指针将不胜感激.

PS:抱歉，如果我需要时间来接受答案，因为此功能目前无法使用(我是盲人，使用屏幕阅读器).

PS: Appologies if it takes time for me to accept an answer, as this feature is currently not accessible (I'm blind and use a screen reader).

推荐答案

发生这种情况的原因是 Table 的声明方式:

This happens because of how Table<T> is declared:

public sealed class Table<TEntity> : IQueryable<TEntity>, 
    IQueryProvider, IEnumerable<TEntity>, ITable, IQueryable, IEnumerable, 
    IListSource
where TEntity : class  // <-- T must be a reference type!

编译器抱怨，因为你的方法对T没有约束，这意味着你可以接受一个不符合规范的T表.

The compiler is complaining because your method has no constraints on T, which means that you could accept a T which doesn't conform to the specification of Table<T>.

因此，您的方法至少需要对其接受的内容同样严格.试试这个:

Thus, your method needs to be at least as strict about what it accepts. Try this instead:

private static Dictionary<string, string> GetProperties<T>(Table<T> table) where T : class

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