我想并行调用我的API次数,以便可以快速完成处理。我下面有三种方法必须并行调用API。我试图了解哪种是执行此操作的最佳方法。
基本代码
var client = new System.Net.Http.HttpClient();
client.DefaultRequestHeaders.Add("Accept", "application/json");
client.BaseAddress = new Uri("https://jsonplaceholder.typicode.com");
var list = new List<int>();
var listResults = new List<string>();
for (int i = 1; i < 5; i++)
{
list.Add(i);
}
使用Parallel.ForEach的第一种方法
Parallel.ForEach(list,new ParallelOptions() { MaxDegreeOfParallelism = 3 }, index =>
{
var response = client.GetAsync("posts/" + index).Result;
var contents = response.Content.ReadAsStringAsync().Result;
listResults.Add(contents);
Console.WriteLine(contents);
});
Console.WriteLine("After all parallel tasks are done with Parallel for each");
带有任务的第二种方法。我不确定这是否并行。让我知道是否可以
var loadPosts = new List<Task<string>>();
foreach(var post in list)
{
var response = await client.GetAsync("posts/" + post);
var contents = response.Content.ReadAsStringAsync();
loadPosts.Add(contents);
Console.WriteLine(contents.Result);
}
await Task.WhenAll(loadPosts);
Console.WriteLine("After all parallel tasks are done with Task When All");
使用动作阻止的第三种方法-这是我应该一直做的事情,但我希望听到社区的意见
var responses = new List<string>();
var block = new ActionBlock<int>(
async x => {
var response = await client.GetAsync("posts/" + x);
var contents = await response.Content.ReadAsStringAsync();
Console.WriteLine(contents);
responses.Add(contents);
},
new ExecutionDataflowBlockOptions
{
MaxDegreeOfParallelism = 6, // Parallelize on all cores
});
for (int i = 1; i < 5; i++)
{
block.Post(i);
}
block.Complete();
await block.Completion;
Console.WriteLine("After all parallel tasks are done with Action block");
方法2接近。这是一条经验法则:I / O绑定操作=>使用Tasks / WhenAll(异步),计算绑定操作=>使用并行性。Http请求是网络I / O。
foreach (var post in list)
{
async Task<string> func()
{
var response = await client.GetAsync("posts/" + post);
return await response.Content.ReadAsStringAsync();
}
tasks.Add(func());
}
await Task.WhenAll(tasks);
var postResponses = new List<string>();
foreach (var t in tasks) {
var postResponse = await t; //t.Result would be okay too.
postResponses.Add(postResponse);
Console.WriteLine(postResponse);
}
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