利用上传图片的HttpClient

问题描述

我想用HttpClient的PHP脚本将其保存在一款Windows Phone 8.1应用程序的服务器上上传文件。

I want to upload a file using HttpClient to a php script to save it on a server in a Windows Phone 8.1 application.

下面是我的C＃代码我从。

Here is my C# code I got from this post.

private async Task<string> GetRawDataFromServer(byte[] data)
{
    //Debug.WriteLine("byte[] data length:" + Convert.ToBase64String(data).Length);
    var requestContent = new MultipartFormDataContent();
    //    here you can specify boundary if you need---^
    var imageContent = new ByteArrayContent(data);
    imageContent.Headers.ContentType =
        MediaTypeHeaderValue.Parse("image/jpeg");

    requestContent.Add(imageContent, "image", "image.jpg");
    using (var client = new HttpClient())
     {

         client.BaseAddress = new Uri("http://www.x.net/");

        var result = client.PostAsync("test/fileupload.php", requestContent).Result;

         return result.Content.ReadAsStringAsync().Result;

     }
}

和与此代码我检索在PHP脚本数据

And with this code I retrieve the data in the php script

<?
function base64_to_image( $imageData, $outputfile ) {
    /* encode & write data (binary) */
    $ifp = fopen( $outputfile, "wb" );
    fwrite( $ifp, base64_decode( $imageData ) );
    fclose( $ifp );
    /* return output filename */
    return( $outputfile );
}       

if (isset($_POST['image'])) {
    base64_to_jpeg($_POST['image'], "image.jpg");
}
else
    die("no image data found");
?>

但结果我总是得到的是没有内容，虽然有一个图像文件。难道我做错了什么把它当作一个POST参数？

But the result I always get is "No Data found" although there IS an image file. Am I doing something wrong passing it as a POST parameter?

推荐答案

好吧研究几个小时后，我来到了这一点，我应该从草稿重新开始。
我模拟HTML表单上传与下面的C＃代码：

Okay after hours of researching I came to the point that I should restart from draft.I simulate a Html form upload with following C# code:

  private async Task<string> UploadImage(StorageFile file)
        {
            HttpClient client = new HttpClient();
            client.BaseAddress = new Uri("http://your.url.com/");
            MultipartFormDataContent form = new MultipartFormDataContent();
            HttpContent content = new StringContent("fileToUpload");
            form.Add(content, "fileToUpload");
            var stream = await file.OpenStreamForReadAsync();
            content = new StreamContent(stream);
            content.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data")
            {
                Name = "fileToUpload",
                FileName = file.Name
            };
            form.Add(content);
            var response = await client.PostAsync("upload.php", form);
            return response.Content.ReadAsStringAsync().Result;
        }

和我的PHP文件接收数据看起来的follwing：

And my php file to receive the data looks as the follwing:

<?php 
$uploaddir = 'uploads/';
$uploadfile = $uploaddir . basename($_FILES['fileToUpload']['name']);
echo '<pre>';
if (move_uploaded_file($_FILES['fileToUpload']['tmp_name'], $uploadfile)) {
    echo "File is valid, and was successfully uploaded.\n";
} else {
    echo "Possible file upload attack!\n";
}

echo 'Here is some more debugging info:';
print_r($_FILES);
?>

现在它的作品，因为它应该，我希望有人可以重复使用我的代码。

Now it works as it should and I hope somebody can reuse my code.

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