Question

我有一些要上传的文件，有些文件失败了，因为帖子是异步的而不是同步的。

我试图将此通话设为同步通话..

我想等待回复。

如何将此调用设为同步？

static async Task<JObect> Upload(string key, string url, string 
                                 sourceFile, string targetFormat)
{ 
    using (HttpClientHandler handler = new HttpClientHandler { 
                                           Credentials = new NetworkCredential(key, "") 
                                       })
    using (HttpClient client = new HttpClient(handler))
    {
         var request = new MultipartFormDataContent();
         request.Add(new StringContent(targetFormat), "target_format");
         request.Add(new StreamContent(File.OpenRead(sourceFile)),
                                       "source_file",
                                        new FileInfo(sourceFile).Name);

        using (HttpResponseMessage response = await client.PostAsync(url,
                                                           request).ConfigureAwait(false))

        using (HttpContent content = response.Content)
        {
            string data = await content.ReadAsStringAsync().ConfigureAwait(false);
            return JsonObject.Parse(data);
        }
    }
}

任何帮助表示赞赏！

Answer 1

更改

await content.ReadAsStringAsync().ConfigureAwait(false)

到

content.ReadAsStringAsync().Result

ReadAsStringAsync返回一个Task对象。＆＃39; .Result＆＃39;在行的末尾告诉编译器返回内部字符串。

Answer 2

应该这样做：

static async Task<JObect> Upload(string key, string url, string 
                             sourceFile, string targetFormat)
{ 
    using (HttpClientHandler handler = new HttpClientHandler { 
                                           Credentials = new NetworkCredential(key, "") 
                                   })
    using (HttpClient client = new HttpClient(handler))
    {
         var request = new MultipartFormDataContent();
         request.Add(new StringContent(targetFormat), "target_format");
         request.Add(new StreamContent(File.OpenRead(sourceFile)),
                                   "source_file",
                                    new FileInfo(sourceFile).Name);

        using (HttpResponseMessage response = await client.PostAsync(url,request))

        using (HttpContent content = response.Content)
        {
            string data = await content.ReadAsStringAsync();
            return JsonObject.Parse(data);
        }
    }
}

使http客户端同步：等待响应

2 个答案: