将StorageFile上传到PHP文件

Question

我一直在尝试将StorageFile（主要是图像文件）上传到PHP文件，以便它可以保存到服务器中。

ViewModel.cs

public async Task<bool> uploadFile(StorageFile file)
{
    try
    {
        using (HttpMultipartFormDataContent form = new HttpMultipartFormDataContent())
            {
                using (IInputStream fileStream = await file.OpenSequentialReadAsync())
                {
                    HttpStreamContent streamContent = new HttpStreamContent(fileStream);
                    form.Add(streamContent, "file", file.Name);

                    using (HttpClient client = new HttpClient())
                    {
                        using (HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, new Uri("localhost/uploadFile.php")))
                        {
                            request.Content = form;
                            HttpResponseMessage response =  await client.SendRequestAsync(request);
                            Debug.WriteLine("\nRequest: " + request.ToString());
                            Debug.WriteLine("\n\nResponse: " + response.ToString());
                        }
                    }
                }
            }
            return true;
   }
   catch (Exception e)
   {
       Debug.WriteLine(e.Message);
       return false;
   }
}

uploadFile.php

<?php
$uploaddir = 'uploads/';
$uploadedFile = $uploaddir . basename($_FILES['file']['name']);

if (move_uploaded_file($_FILES['file']['tmp_name'], $uploadedFile)){
    echo 'File upload success!';
} else {
    echo 'Possible file upload attack!';
}
?>

问题是，当我尝试上传文件时，它向我发送了一个错误The object has been closed. (Exception from HRESULT: 0X80000013)和Exception thrown: 'System.ObjectDisposedException' in mscorlib.ni.dll ..我不明白，我在{{1}内部上传文件声明，怎么处理？我做错了吗？

调试向我展示了这个

  

请求：方法：POST，RequestUri：＆＃39; http://localhost/uploadFile.php＆＃39;，内容：Windows.Web.Http.HttpMultipartFormDataContent，TransportInformation：ServerCertificate：＆＃39;＆＃39;，ServerCertificateErrorSeverity：无，ServerCertificateErrors：{}，ServerIntermediateCertificates：{}，标题：{Accept-Encoding：gzip，deflate} {Content-Length：27749，Content-Type：multipart / form-data; boundary = 9955f08b-e82d-428b-82e1-3197e5011ccd}

     

响应：StatusCode：200，ReasonPhrase：＆＃39; OK＆＃39;，版本：2，内容：Windows.Web.Http.HttpStreamContent，Headers：{Connection：Keep-Alive，Server：Apache / 2.4.18 （Ubuntu），Keep-Alive：timeout = 5，max = 100，Date：Sun，2016年11月6日04:02:40 GMT} {Content-Length：28，Content-Type：text / html;字符集= UTF-8}

Answer 1

Using statement提供了一种方便的语法，可确保正确使用IDisposable对象。简单地说，它可以帮助您执行Dispose()方法。所以你的代码等于：

 HttpMultipartFormDataContent form = new HttpMultipartFormDataContent();               
 IInputStream fileStream = await file.OpenSequentialReadAsync();
 HttpStreamContent streamContent = new HttpStreamContent(fileStream);
 form.Add(streamContent, "file", file.Name);    
 HttpClient client = new HttpClient();             
 HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, new Uri("http://127.0.0.1:9096/hello.php"));
 request.Content = form;      
 HttpResponseMessage response = await client.SendRequestAsync(request);
 Debug.WriteLine("\nRequest: " + request.ToString());
 Debug.WriteLine("\n\nResponse: " + response.ToString());
 request.Dispose();
 client.Dispose();
 fileStream.Dispose();
 form.Dispose();

您将在form.Dispose();代码行获得例外。原因是HttpMultipartFormDataContent.Add方法不需要处理。在我看来，并非需要处理的非托管资源，HttpMultipartFormDataContent类ReadAsBufferAsync之类的其他方法可能需要处理。

按如下方式更新您的代码，这不会引发closed异常：

HttpMultipartFormDataContent form = new HttpMultipartFormDataContent();
using (IInputStream fileStream = await file.OpenSequentialReadAsync())
{
    HttpStreamContent streamContent = new HttpStreamContent(fileStream);
    form.Add(streamContent, "file", file.Name);

    using (HttpClient client = new HttpClient())
    {
        using (HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, new Uri("http://127.0.0.1:9096/hello.php")))
        {
            request.Content = form;
            HttpResponseMessage response = await client.SendRequestAsync(request);
            Debug.WriteLine("\nRequest: " + request.ToString());
            Debug.WriteLine("\n\nResponse: " + response.ToString());
        }
    }
}