创建具有通用类作为列表的对象

Question

我有这种情况：

public class ExtResult<T>
{
    public bool Success { get; set; }
    public string Msg { get; set; }
    public int Total { get; set; }
    public T Data { get; set; }
}

//create list object:
List<ProductPreview> gridLines;
...
...
//At the end i would like to create object
ExtResult<gridLines> result = new ExtResult<gridLines>() { 
    Success = true, Msg = "", 
    Total=0, 
    Data = gridLines 
}

但是我得到一个错误：

错误：“无法解析gridLines”

我该怎么做才能解决此问题？

Answer 1

gridLines是一个变量，其类型为List<ProductPreview> ，应将其用作ExtResult<T>的类型参数：

ExtResult<List<ProductPreview>> result = new ExtResult<List<ProductPreview>>() { 
    Success = true, 
    Msg = "", 
    Total=0, 
    Data = gridLines 
};

Answer 2

您应该将类​​型作为通用参数而不是变量传递：

var result = new ExtResult<List<ProductPreview>> // not gridLines, but it's type
{ 
    Success = true,
    Msg = "",
    Total=0,
    Data = gridLines
}