Question

我的JSON对象中的属性名称需要通过字符串分配，如何使用JObject执行此操作？或者我应该从字符串中只是JObject.parse？我知道.Net Json文档，但它非常基础，并没有显示如下的任何示例。

以下是我现在所拥有的：

return JObject.FromObject(new { 
attachment = new { 
    type = "template",
    payload = new {
        template_type = "button",
        text = Title,
        buttons = new {
            type = type,
            Variable1 = Value,
            Variable2 = Payload
        }
    }               
}
});

另外，如果我按照下面的方式执行string.parse方式，这是格式化字符串的最佳方法吗？

JObject.Parse(@"{
    attachment : { 
        type : 'template',
        payload : {
            template_type : 'button',
            text : '"+Title+@"',
                buttons : [{
                  type : '"+Type+@"',
                  "+Variable1+" : '"+Value+@"',
                  "+Variable2+" : '"+Payload+@"'
                }]      
        }
    }
}"

Answer 1

您可以使用外部匿名类型对象中包含的Creating JSON: Creating JSON with LINQ模式手动构造内部var obj = JObject.FromObject(new { attachment = new { type = "template", payload = new { template_type = "button", text = Title, buttons = new JArray( new JObject( new JProperty("type", Type), new JProperty(Variable1, Value), new JProperty(Variable2, Payload))) } } });，然后将整个匿名对象序列化为JSON：

Business =  {
  1: { Country: "France" },
  2: { Country: "France" },
  3: { Country: "UnitedKingdom" },
  4: { Country: "France" }
}

示例fiddle。

Answer 2

Dictionary<string,object>使用buttons，例如：

var obj = JObject.FromObject(new
{
    attachment = new
    {
        type = "template", 
        payload = new
        {
            template_type = "button", 
            text = Title, 
            buttons = new object[] 
            { 
                new Dictionary<string,object>() 
                {
                    { "type", "type" },
                    { Variable1, Value },
                    { Variable2, Payload }
                }
            }
        }
    }
});

JSON使用字符串变量

2 个答案: