如何使用 Newtonsoft.Json 讀取 C# 中的 json 輸入

Question

我試圖讀取進入 AWS lambda function 的 json，下面是初始代碼的一部分

using System;
using System.Collections.Generic;
using System.Threading.Tasks;
using Amazon.Lambda.Core;
using Newtonsoft.Json;

// Assembly attribute to enable the Lambda function's JSON input to be converted into a .NET class.
[assembly: LambdaSerializer(typeof(Amazon.Lambda.Serialization.SystemTextJson.DefaultLambdaJsonSerializer))]

namespace DemoApp
{
    public class Function
    {
        public string FunctionHandler(dynamic input,ILambdaContext context)
        {
            dynamic results = JsonConvert.DeserializeObject<dynamic>(input);
            String customer = results.customer;
             .......

進來的json是

{'customer': 'Test Customer',
 'phone': '0435 434 544',
 'asset': 'ASSET',
 'po': 'PO09876',
 'location': 'XYZ',
 'items': {'hose': [[{'goods': 'A231',
     'qty': '0.23',
     'backorder': '0.0',
     'unitprice': '0.0',
     'total': '0.0'},
    {'goods': 'B564',
     'qty': '1.0',
     'backorder': '0.0',
     'unitprice': '0.0',
     'total': '0.0'},
    {'goods': 'C544',
     'qty': '1.0',
     'backorder': '0.0',
     'unitprice': '0.0',
     'total': '0.0'}]],
  'items': []}}

我收到一個錯誤The best overloaded method match for 'Newtonsoft.Json.JsonConvert.DeserializeObject<object>(string)' has some invalid arguments", 。如何正確讀取 json？

Answer 1

JsonConvert.DeserializeObject需要一個字符串，所以我將 json 轉換為字符串，一切正常。

using System;
using System.Collections.Generic;
using System.Threading.Tasks;
using Amazon.Lambda.Core;
using Newtonsoft.Json;

// Assembly attribute to enable the Lambda function's JSON input to be converted into a .NET class.
[assembly: LambdaSerializer(typeof(Amazon.Lambda.Serialization.SystemTextJson.DefaultLambdaJsonSerializer))]

namespace DemoApp
{
    public class Function
    {
        public string FunctionHandler(Object input,ILambdaContext context)
        {
            string jsonString = System.Text.Json.JsonSerializer.Serialize(input);
            dynamic results = JsonConvert.DeserializeObject(jsonString);
            String customer = results.customer;
             .......