使用 HttpClient 上传图片

Question

我想使用 HttpClient 将文件上传到 php 脚本，以将其保存在 Windows Phone 8.1 应用程序的服务器上。

这是我从 this post 获得的 C# 代码。

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;

     }
}

通过这段代码，我在 php 脚本中检索数据

<?
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");
?>

但我总是得到的结果是“找不到数据”，尽管有一个图像文件。将其作为 POST 参数传递是不是我做错了什么？

Answer 1

好吧，经过数小时的研究，我发现我应该从草稿中重新开始。 我使用以下 C# 代码模拟 Html 表单上传：

  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;
        }

我接收数据的 php 文件如下所示：

<?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);
?>

现在它可以正常工作了，我希望有人可以重用我的代码。