如何在C#的API Post方法中传递JSON请求体

下面是两种常用的实现方案，针对你提供的JSON请求体，分别给出代码示例：

方案一：使用强类型模型（推荐，利于维护和类型检查） #

首先根据你的JSON结构定义对应的C#类：

public class PaymentRequest
{
    public string intent { get; set; }
    public List<PurchaseUnit> purchase_units { get; set; }
    public ApplicationContext application_context { get; set; }
}

public class PurchaseUnit
{
    public Amount amount { get; set; }
}

public class Amount
{
    public string currency_code { get; set; }
    public string value { get; set; }
}

public class ApplicationContext
{
    public string return_url { get; set; }
    public string cancel_url { get; set; }
}

然后使用HttpClient发送Post请求（.NET Core/.NET 5+推荐使用）：

using System;
using System.Net.Http;
using System.Text;
using System.Text.Json;
using System.Threading.Tasks;
using System.Collections.Generic;

class PaymentSender
{
    static async Task Main(string[] args)
    {
        // 构造请求数据
        var requestData = new PaymentRequest
        {
            intent = "CAPTURE",
            purchase_units = new List<PurchaseUnit>
            {
                new PurchaseUnit
                {
                    amount = new Amount
                    {
                        currency_code = "USD",
                        value = "10"
                    }
                }
            },
            application_context = new ApplicationContext
            {
                return_url = "https://example.com/return",
                cancel_url = "https://example.com/cancel"
            }
        };

        // 将对象序列化为JSON字符串
        string json = JsonSerializer.Serialize(requestData);
        var content = new StringContent(json, Encoding.UTF8, "application/json");

        // 注意：实际项目中HttpClient应复用单例，避免频繁创建
        using var client = new HttpClient();
        try
        {
            HttpResponseMessage response = await client.PostAsync("你的API接口地址", content);
            response.EnsureSuccessStatusCode(); // 若请求失败会抛出异常

            string responseBody = await response.Content.ReadAsStringAsync();
            Console.WriteLine("API响应：" + responseBody);
        }
        catch (HttpRequestException ex)
        {
            Console.WriteLine("请求失败：" + ex.Message);
        }
    }
}

方案二：直接使用匿名对象（快速实现，无需定义模型） #

如果不想定义实体类，可以直接构造匿名对象并序列化，这里以Newtonsoft.Json为例（需先安装Newtonsoft.Json NuGet包）：

using System;
using System.Net.Http;
using System.Text;
using System.Threading.Tasks;
using Newtonsoft.Json;

class PaymentSender
{
    static async Task Main(string[] args)
    {
        var requestData = new
        {
            intent = "CAPTURE",
            purchase_units = new[]
            {
                new
                {
                    amount = new
                    {
                        currency_code = "USD",
                        value = "10"
                    }
                }
            },
            application_context = new
            {
                return_url = "https://example.com/return",
                cancel_url = "https://example.com/cancel"
            }
        };

        string json = JsonConvert.SerializeObject(requestData);
        var content = new StringContent(json, Encoding.UTF8, "application/json");

        using var client = new HttpClient();
        try
        {
            HttpResponseMessage response = await client.PostAsync("你的API接口地址", content);
            response.EnsureSuccessStatusCode();

            string responseBody = await response.Content.ReadAsStringAsync();
            Console.WriteLine("API响应：" + responseBody);
        }
        catch (HttpRequestException ex)
        {
            Console.WriteLine("请求失败：" + ex.Message);
        }
    }
}

关键注意事项 #

• 务必设置请求内容的Content-Type为application/json，否则API可能无法正确解析请求体
• 生产环境中，HttpClient建议复用单例实例，避免频繁创建导致的资源泄漏
• 可根据API要求调整序列化配置（比如大小写、忽略空值等）

内容的提问来源于stack exchange，提问作者Nikhil Jose