出错时重试异步文件上传答案

【问题标题】：Retrying asynchronous file upload on error出错时重试异步文件上传
【发布时间】：2013-01-12 19:35:43
【问题描述】：

我正在尝试在 WPF 应用程序中上传文件。如果服务器响应，一切正常，但应用程序将在“不安全”互联网连接的环境中使用。因此，如果第一次尝试失败，我想在短暂休息后重试上传。

我用 async/await 尝试了几件事，最后得到了以下代码。如果服务器正在运行，一切正常，但如果不是，则程序在 while 循环的第二次迭代中失败并返回 ObjectDisposedException。

有什么想法吗？

private void UploadButton_Click(object sender, RoutedEventArgs e)
{
    // build content to send
    content = new MultipartFormDataContent();
    var filestream = new FileStream(filePath, FileMode.Open);
    var fileName = System.IO.Path.GetFileName(filePath);
    content.Add(new StreamContent(filestream), "file", fileName);
    content.Add(new StringContent(terminal_id.ToString()), "terminal_id");

    UploadTask(content);
    /*var task_a = new Task(() => UploadTask(content));
    task_a.Start();*/
}

private async void UploadTask(HttpContent content)
{
    bool success = false;
    int counter = 0;

    while (counter < 3 && !success)
    {
        Debug.WriteLine("starting upload");
        success = await UploadFileAsync(content);
        Debug.WriteLine("finished upload. result " + success.ToString());
        //if (!success) System.Threading.Thread.Sleep(5000);
        counter++;
    }
}

private async Task<bool> UploadFileAsync(HttpContent content)
{
    var message = new HttpRequestMessage();
    message.Method = HttpMethod.Post;
    message.Content = content;
    message.RequestUri = new Uri(target_url);

    using (HttpClient client = new HttpClient())
    {
        try
        {
            HttpResponseMessage res = await client.SendAsync(message);
            if (res.IsSuccessStatusCode) return true;
        }
        catch (HttpRequestException hre)
        {
            Debug.WriteLine(hre.ToString());
        }
        return false;
    }
}

【问题讨论】：

标签： c# .net wpf asynchronous async-await

【解决方案1】：

我认为问题可能是content 超出范围？尝试在UploadTask 方法中创建content。此外，可能值得从 UploadTask 返回 Task<bool> 并将其缓存为类级别变量（因此您不必返回 void）。

例如：

Task<bool> newTask;

private void UploadButton_Click(object sender, RoutedEventArgs e)
{
    newTask = UploadTask();

}

private async Task<bool> UploadTask()
{
bool success = false;
int counter = 0;

// build content to send
HttpContent content = new MultipartFormDataContent();
var filestream = new FileStream(filePath, FileMode.Open);
var fileName = System.IO.Path.GetFileName(filePath);
content.Add(new StreamContent(filestream), "file", fileName);
content.Add(new StringContent(terminal_id.ToString()), "terminal_id");

while (counter < 3 && !success)
{
    Debug.WriteLine("starting upload");
    success = await UploadFileAsync(content);
    Debug.WriteLine("finished upload. result " + success.ToString());

    counter++;
}

return success;
}

【讨论】：

恐怕（几乎）没有任何改变。异常在后台任务中引发，但结果保持不变。但是有了这个建议，我将内容的创建移到 UploadFileAsync() 并且它可以工作。谢谢！

【解决方案2】：

您的文件流似乎已被处置/关闭。您需要从头开始重试（content = new MultipartFormDataContent(); 等）。

【讨论】：

【解决方案3】：

将内容的创建移到 UploadFileAsync() 之后，它就可以工作了。结果：

Task<bool> newTask;

private void UploadButton_Click(object sender, RoutedEventArgs e)
{
    newTask = UploadTask();
}

private async Task<bool> UploadTask()
{
    bool success = false;
    int counter = 0;

    while (counter < 3 && !success)
    {
        Debug.WriteLine("starting upload");
        success = await UploadFileAsync();
        Debug.WriteLine("finished upload. result " + success.ToString());
        if (!success) System.Threading.Thread.Sleep(5000);
        counter++;
    }
    return success;
}

private async Task<bool> UploadFileAsync()
{
    MultipartFormDataContent content = new MultipartFormDataContent();
    var filestream = new FileStream(filePath, FileMode.Open);
    var fileName = System.IO.Path.GetFileName(filePath);
    content.Add(new StreamContent(filestream), "file", fileName);
    content.Add(new StringContent(terminal_id.ToString()), "terminal_id");

    var message = new HttpRequestMessage();
    message.Method = HttpMethod.Post;
    message.Content = content;
    message.RequestUri = new Uri(target_url);

    using (HttpClient client = new HttpClient())
    {
        try
        {
            HttpResponseMessage res = await client.SendAsync(message);
            if (res.IsSuccessStatusCode) return true;
        }
        catch (HttpRequestException hre)
        {
            Debug.WriteLine(hre.ToString());
        }
        return false;
    }
}

【讨论】：