我试图找出使用Task
和async/await
并行化HTTP请求的正确方法。我使用已经有异步方法检索数据的HttpClient
类。如果我只是在foreach循环中调用它并等待响应,则一次只发送一个请求(这是有道理的,因为在await
期间,控制返回到我们的事件循环,而不是下一次迭代foreach loop)。
围绕HttpClient
的我的包装看起来像
public sealed class RestClient
{
private readonly HttpClient client;
public RestClient(string baseUrl)
{
var baseUri = new Uri(baseUrl);
client = new HttpClient
{
BaseAddress = baseUri
};
}
public async Task<Stream> GetResponseStreamAsync(string uri)
{
var resp = await GetResponseAsync(uri);
return await resp.Content.ReadAsStreamAsync();
}
public async Task<HttpResponseMessage> GetResponseAsync(string uri)
{
var resp = await client.GetAsync(uri);
if (!resp.IsSuccessStatusCode)
{
// ...
}
return resp;
}
public async Task<T> GetResponseObjectAsync<T>(string uri)
{
using (var responseStream = await GetResponseStreamAsync(uri))
using (var sr = new StreamReader(responseStream))
using (var jr = new JsonTextReader(sr))
{
var serializer = new JsonSerializer {NullValueHandling = NullValueHandling.Ignore};
return serializer.Deserialize<T>(jr);
}
}
public async Task<string> GetResponseString(string uri)
{
using (var resp = await GetResponseStreamAsync(uri))
using (var sr = new StreamReader(resp))
{
return sr.ReadToEnd();
}
}
}
我们的事件循环调用的代码是
public async void DoWork(Action<bool> onComplete)
{
try
{
var restClient = new RestClient("https://example.com");
var ids = await restClient.GetResponseObjectAsync<IdListResponse>("/ids").Ids;
Log.Info("Downloading {0:D} items", ids.Count);
using (var fs = new FileStream(@"C:\test.json", FileMode.Create, FileAccess.Write, FileShare.Read))
using (var sw = new StreamWriter(fs))
{
sw.Write("[");
var first = true;
var numCompleted = 0;
foreach (var id in ids)
{
Log.Info("Downloading item {0:D}, completed {1:D}", id, numCompleted);
numCompleted += 1;
try
{
var str = await restClient.GetResponseString($"/info/{id}");
if (!first)
{
sw.Write(",");
}
sw.Write(str);
first = false;
}
catch (HttpException e)
{
if (e.StatusCode == HttpStatusCode.Forbidden)
{
Log.Warn(e.ResponseMessage);
}
else
{
throw;
}
}
}
sw.Write("]");
}
onComplete(true);
}
catch (Exception e)
{
Log.Error(e);
onComplete(false);
}
}
我尝试了一些涉及Parallel.ForEach
,Linq.AsParallel
的不同方法,并将整个循环内容包装在Task
中。
答案 0 :(得分:7)
基本思想是跟踪所有异步任务,并立即等待它们。最简单的方法是将foreach的主体提取到一个单独的异步方法,并执行以下操作:
var tasks = ids.Select(i => DoWorkAsync(i));
await Task.WhenAll(tasks);
这样,单独的任务是单独发布的(仍然按顺序发布,但不等待I / O完成),并且您可以同时等待它们。
请注意,您还需要进行一些配置 - 默认情况下,HTTP会受到限制,只允许两个同时连接到同一服务器。