Question

我想将参数传递给Azure函数2中的HttpTrigger类型。我在下面的链接中看到Azure函数1的答案。

How to pass parameters by POST to an Azure function?

以上链接中的答案为File cacheDirectory = context.getCacheDir(); //App crashes here since there is no external storage String cachePath = ""; if(cacheDirectory != null){ cachePath = cacheDirectory.toPath().normalize().toString(); } 我正在寻找与Azure Function 2类似的答案。

在这里，我们使用的是HttpRequest类而不是HttpRequestMessage类。

下面是代码。

await req.Content.ReadAsAsync<>();

Answer 1

public static async void Run(
            [HttpTrigger(AuthorizationLevel.Function, "post", Route = null)]
            HttpRequest req,
            ILogger log)
        {
            var content = await new StreamReader(req.Body).ReadToEndAsync();

            MyClass myClass = JsonConvert.DeserializeObject<MyClass>(content);

        }

您可以拥有一个自定义类，该类可以从发布请求中以JSON的形式发送。

Answer 2

似乎您正在尝试向POST发送Azure Function V2请求。请参见下面的代码段。

自定义请求类别：

public class Users
    {

        public string Name { get; set; }
        public string Email { get; set; }

    }

Azure Function V2：

在此示例中，我使用自定义类获取两个参数，并返回该两个类属性作为响应。

public static async Task<IActionResult> Run(
            [HttpTrigger(AuthorizationLevel.Anonymous, "get", "post", Route = null)] HttpRequest req,
            ILogger log)
        {
            log.LogInformation("C# HTTP trigger function processed a request.");

            //Read Request Body
            var content = await new StreamReader(req.Body).ReadToEndAsync();

            //Extract Request Body and Parse To Class
            Users objUsers = JsonConvert.DeserializeObject<Users>(content);

            //As we have to return IAction Type So converting to IAction Class Using OkObjectResult We Even Can Use OkResult
            var result = new OkObjectResult(objUsers);
            return (IActionResult)result;
        }

索取样品：

{
   "Name": "Kiron" ,
   "Email": "kiron@email.com"
}

PostMan测试：

注意： 您实际上是在寻找await req.Content.ReadAsAsync<>(); 从您的函数发送POST请求中需要。并从中读取该服务器的响应。但是请记住，req.Content不支持Azure Function V2作为阅读帖子的请求，它已经显示了Function V1示例here

另一个示例：

请参阅以下代码段：

public static async Task<IActionResult> Run(
            [HttpTrigger(AuthorizationLevel.Anonymous, "get", "post", Route = null)] HttpRequest req,
            ILogger log)
        {
            log.LogInformation("C# HTTP trigger function processed a request.");

            //Read Request Body
            var content = await new StreamReader(req.Body).ReadToEndAsync();

            //Extract Request Body and Parse To Class
            Users objUsers = JsonConvert.DeserializeObject<Users>(content);

            //Post Reuqest to another API 
            HttpClient client = new HttpClient();
            var json = JsonConvert.SerializeObject(objUsers);
            //Parsing json to post request content
            var stringContent = new StringContent(json, UnicodeEncoding.UTF8, "application/json");
            //Posting data to remote API
            HttpResponseMessage responseFromApi = await client.PostAsync("YourRequstURL", stringContent);

            //Variable for next use to bind remote API response
            var remoteApiResponse = "";
            if (responseFromApi.IsSuccessStatusCode)
            {
               remoteApiResponse = responseFromApi.Content.ReadAsStringAsync().Result; // According to your sample, When you read from server response
            }

            //As we have to return IAction Type So converting to IAction Class Using OkObjectResult We Even Can Use OkResult
            var result = new OkObjectResult(remoteApiResponse);
            return (IActionResult)result;
        }

希望您能理解，如果您还有任何疑问，请随时与我们分享。谢谢，祝您编程愉快！

Answer 3

首先，基于HTTP请求创建Azure Function Project（或添加到您的解决方案中）。触发。在创建的项目的单个类中，您可以看到示例。您从请求正文传递了JSON：

{"name":"John"}

，然后将其转换为代码中的变量：

 public static async Task<IActionResult> Run(
        [HttpTrigger(AuthorizationLevel.Function, "get", "post", Route = null)] HttpRequest req,
        ILogger log)
    {
        log.LogInformation("C# HTTP trigger function processed a request.");

       

        string requestBody = await new StreamReader(req.Body).ReadToEndAsync();
        dynamic data = JsonConvert.DeserializeObject(requestBody);
        string name = data.name;
        ...

如何通过POST将参数传递给Azure Functions 2

3 个答案: