Question

我通过Azure Functions将webhook导入SQL数据库。

使用

导入时

{
        "remote_address": "1234",
}

这很有效。

但是当我使用

导入时

{
        "transaction": {
            "remote_address": "1234",
    }
}

它没有用。

我的Azure功能代码是：

using System.Net;
using Dapper;
using System.Data.SqlClient;
using System.Configuration;

public static async Task<HttpResponseMessage> Run(HttpRequestMessage req, TraceWriter log)
{
log.Info($"C# HTTP trigger function processed a request. RequestUri={req.RequestUri}");

var successful =true;
try
{
    var cnnString  = ConfigurationManager.ConnectionStrings["sqlConnection"].ConnectionString;

    using(var connection = new SqlConnection(cnnString))
    {
        connection.Open();

        var rLog = await req.Content.ReadAsAsync<LogRequest>();

        // insert a log to the database
        connection.Execute("INSERT INTO [dbo].[TABLE] ([COLUMN]) VALUES (@remote_address)", rLog);
        log.Info("Log added to database successfully!");
    }
}
catch
{
    successful=false;
}

return !successful
    ? req.CreateResponse(HttpStatusCode.BadRequest, "Unable to process your request!")
    : req.CreateResponse(HttpStatusCode.OK, "Data saved successfully!");
}

public class LogRequest
{
    public int Id{get;set;}
    public string remote_address{get;set;}
}

如何获取嵌套的JSON对象？

Answer 1

基本上，您可以嵌套C＃类，使它们对应于JSON结构：

public class Transaction
{
    public string remote_address { get; set; }
}

public class LogRequest
{
    public Transaction transaction { get; set; }
}

Answer 2

要扩展米哈伊尔的好答案，根据您的要求，有时最容易使用JObject/dynamic。

var jobjectBody = await req.Content.ReadAsAsync<JObject>();
dynamic dynamicBody = jobjectBody;
string remote_address = dynamicBody.transaction.remote_address;

将JSON导入SQL

2 个答案: