Question

当我发送请求时，服务器将发送如下响应：

{
    "id":"YWauEwCUIe",
    "name":"User 1",
    "status":"DONE"
}

为此，我正在使用以下代码：

var client = new RestClient(mServer);
var request = new RestRequest(Method.POST);
request.AddQueryParameter("action", "GET");
IRestResponse response = client.Execute(request);
string mResponse = response.Content;

mUser = new JObject();
mUser = JObject.Parse(mResponse); // Error Line
string mStatus = mUser["status"].ToString();
if (mStatus.Equals("DONE")
{
    MessageBox.Show("Status: DONE");
}

错误：Newtonsoft.Json.JsonReaderException：'从JsonReader读取JObject时出错。当前的JsonReader项不是对象：空。路径”，第1行，位置4。'

Answer 1

我的回答是null导致MySQL错误的原因，现在它已解决。

PHP代码：

<?php
    $Message['id'] = "YWauEwCUIe";
    $Message['name'] = "User 1";
    $Message['status'] = "DONE";
    echo json_encode($Message);
?>

，这将显示此result：

{
    "id":"YWauEwCUIe",
    "name":"User 1",
    "status":"DONE"
}

此C#代码将起作用：

var client = new RestClient(mServer);
var request = new RestRequest(Method.POST);
request.AddQueryParameter("action", "GET");
IRestResponse response = client.Execute(request);
string mResponse = response.Content;

mUser = new JObject();
mUser = JObject.Parse(mResponse);
string mStatus = mUser["status"].ToString();
if (mStatus.Equals("DONE")
{
    MessageBox.Show("Status: DONE");
}

Answer 2

我会使用非常受欢迎的NewtonSoft库：

采用从服务器获得的响应形状（假设JSON对象不变并且您具有所有字段），可以在此处生成响应类：https://app.quicktype.io/#l=cs&r=json2csharp

对我来说，它生成了以下代码：

    using System;
    using System.Collections.Generic;

    using System.Globalization;
    using Newtonsoft.Json;
    using Newtonsoft.Json.Converters;

    public partial class Welcome
    {
        [JsonProperty("id")]
        public string Id { get; set; }

        [JsonProperty("name")]
        public string Name { get; set; }

        [JsonProperty("status")]
        public string Status { get; set; }
    }

    public partial class Welcome
    {
        public static Welcome FromJson(string json) => JsonConvert.DeserializeObject<Welcome>(json, QuickType.Converter.Settings);
    }

    public static class Serialize
    {
        public static string ToJson(this Welcome self) => JsonConvert.SerializeObject(self, QuickType.Converter.Settings);
    }

    internal static class Converter
    {
        public static readonly JsonSerializerSettings Settings = new JsonSerializerSettings
        {
            MetadataPropertyHandling = MetadataPropertyHandling.Ignore,
            DateParseHandling = DateParseHandling.None,
            Converters =
            {
                new IsoDateTimeConverter { DateTimeStyles = DateTimeStyles.AssumeUniversal }
            },
        };
    }

这是我的示例代码：

using System;
using System.Globalization;
using Newtonsoft.Json;
using Newtonsoft.Json.Converters;

namespace JsonParse
{
    public partial class Welcome
    {
        [JsonProperty("id")]
        public string Id { get; set; }

        [JsonProperty("name")]
        public string Name { get; set; }

        [JsonProperty("status")]
        public string Status { get; set; }
    }

    public partial class Welcome
    {
        public static Welcome FromJson(string json) => JsonConvert.DeserializeObject<Welcome>(json, Converter.Settings);
    }

    public static class Serialize
    {
        public static string ToJson(this Welcome self) => JsonConvert.SerializeObject(self, Converter.Settings);
    }

    internal static class Converter
    {
        public static readonly JsonSerializerSettings Settings = new JsonSerializerSettings
        {
            MetadataPropertyHandling = MetadataPropertyHandling.Ignore,
            DateParseHandling = DateParseHandling.None,
            Converters =
            {
                new IsoDateTimeConverter { DateTimeStyles = DateTimeStyles.AssumeUniversal }
            },
        };
    }

    class Program
    {
        static void Main(string[] args)
        {

            var jsonPayload = "{\"id\":\"YWauEwCUIe\",\"name\":\"User 1\",\"status\":\"DONE\"}";
            var jsonAsObject = Newtonsoft.Json.JsonConvert.DeserializeObject<Welcome>(jsonPayload);
            Console.WriteLine($"{jsonAsObject.Id} {jsonAsObject.Name} {jsonAsObject.Status}");
            Console.ReadKey();
        }
    }
}

在C＃中从服务器获取JSON响应

2 个答案: