任务不包含内容的定义

Question

如何克服下面的错误？我已经按照以下教程进行操作，但我得到了不包含Content. https://www.youtube.com/watch?v=IVvJX4CoLUY

private void UploadFile_Cliked(object sender, EventArgs e)
{
    var content = new MultipartFormDataContent();
    content.Add(new StreamContent(_mediaFile.GetStream()),
        "\"file\"",
        $"\"{_mediaFile.Path}\"");
    var httpClient = new HttpClient();

    var uploadServiceBaseAddress = "http://192.168.137.1/pic/";
    var httpResponseMessage = httpClient.PostAsync(uploadServiceBaseAddress, content);
    RemotePathLabel.Text = httpResponseMessage.Content.ReadAsStringAsync();
}

“Task”不包含“Content”的定义，并且找不到接受“Task”类型的第一个参数的扩展方法“Content”（您是否缺少 using 指令或程序集引用？）

这是错误，但有解决它的想法吗？

我已经添加了这个Microsoft.AspNet.WebApi.Client包，但仍然失败

Answer 1

使事件处理程序异步，然后等待返回 Task 派生结果的函数，因为您在调用时会遇到相同的问题ReadAsStringAsync

private HttpClient httpClient = new HttpClient();

private async void UploadFile_Cliked(object sender, EventArgs e) {
    var content = new MultipartFormDataContent();
    content.Add(new StreamContent(_mediaFile.GetStream()),
        "\"file\"",
        $"\"{_mediaFile.Path}\"");        

    var uploadServiceBaseAddress = "http://192.168.137.1/pic/";
    var httpResponseMessage = await httpClient.PostAsync(uploadServiceBaseAddress, content);
    RemotePathLabel.Text = await httpResponseMessage.Content.ReadAsStringAsync();
}

async void请注意，事件处理程序是允许的规则的例外

参考Async/Await - 异步编程的最佳实践