使用C#中的Json.net格式化Json

Question

我试图将一堆XML文件解析为一个已经正常工作的JSON文件.

最终的JSON文件如下所示:

{
"items": [
    {
        "subItems": [
            {
                "Name": "Name",
                "Value": "Value",
                "Type": "text"
            },
            {
                "Name": "Name",
                "Value": "Value",
                "Type": "text"
            }
        ]
    },
    {
        "subItems": [
            {
                "Name": "Name",
                "Value": "Value",
                "Type": "text"
            },
            {
                "Name": "Name",
                "Value": "Value",
                "Type": "text"
            },
...

相反,我想实现以下结构:

{
"items": [
    [   
        {
            "Name": "Name",
            "Value": "Value",
            "Type": "text"
        },
        {
            "Name": "Name",
            "Value": "Value",
            "Type": "text"
        }

    ],
    [
        {
            "Name": "Name",
            "Value": "Value",
            "Type": "text"
        },
        {
            "Name": "Name",
            "Value": "Value",
            "Type": "text"
        }
    ]
]
}

但我不知道如何定义我的对象才能这样做,我目前的结构如下:

public class Items
{
    public List<Item> items;
}

public class Item
{
    public List<SubItem> subItems;
}

public class SubItem
{
    public string Name { get; set; }
    public string Value { get; set; }
    public string Type { get; set; }
}

我该怎么做？

Answer 1

答案很简单:将对象转换为列表:这将删除道具名称(以及json中的对象表示法).

public class Items
{
    public List<Item> items; //list with prop name 'items'
}

public class Item : List<SubItem> // list in list like json notation
{
}

public class SubItem // Object in the list in list
{
    public string Name { get; set; }
    public string Value { get; set; }
    public string Type { get; set; }
}

正如@FlilipCordas所说,从列表继承是不好的做法(有充分理由)你这样做更好:

public class Items
{
    public List<List<SubItem>> items; //list with list with prop name 'items'
}

public class SubItem // Object in the list in list
{
    public string Name { get; set; }
    public string Value { get; set; }
    public string Type { get; set; }
}