Question

我做了一个httpclient postasync，我想将响应转换为我的模型对象。这是我的代码：

var response = client.PostAsync(TERMBASED_ENDPOINT,
    new StringContent(JsonConvert.SerializeObject(request).ToString(),
                      Encoding.UTF8, "application/json")).Result;

var result = await response.Content.ReadAsAsync<MyObject>();
//other code synchronously processed

因此，此代码是异步处理的。等待处理响应的最佳方法是什么，就在这种情况发生之后继续运行同步代码？

谢谢！

Answer 1

“等待”您的帖子调用以打开回复。

var response = await client.PostAsync(TERMBASED_ENDPOINT,
    new StringContent(JsonConvert.SerializeObject(request).ToString(),
                      Encoding.UTF8, "application/json"));

var result = JsonConvert.DeserializeObject<MyObject>(response.Content);

Answer 2

我认为您想要做的是将Task存储在变量中，做一些工作，然后等待响应。

var response = client.PostAsync(TERMBASED_ENDPOINT,
    new StringContent(JsonConvert.SerializeObject(request).ToString(),
                      Encoding.UTF8, "application/json")).Result;

var readTask = response.Content.ReadAsAsync<MyObject>();
//other code synchronously processed
var result = await readTask;

或者，如果您有多个异步任务，则可以等待它们，然后处理结果。

var response = client.PostAsync(TERMBASED_ENDPOINT,
    new StringContent(JsonConvert.SerializeObject(request).ToString(),
                      Encoding.UTF8, "application/json")).Result;

var readTask = response.Content.ReadAsAsync<MyObject>();
//other code synchronously processed
await Task.WhenAll(readTask);
var result = readTask.Result;

C＃async等待Task <t>处理

2 个答案: