如何在 UWP 中使用 httpClient.postAsync 上傳圖像或字節 []

Question

所以我需要將圖像與其他一些字符串（如名稱）一起上傳到我的 Mysql 數據庫中，例如……我能夠將名稱添加到 Mysql DB 中，但我無法為圖像執行此操作。 我將圖像轉換為一個字節 [] 並且我現在卡住了..這是我使用的代碼

private Stream stream = new MemoryStream();
    private CancellationTokenSource cts;

    public MainPage()
    {
        this.InitializeComponent();
    }

    private async void buttonUpload_Click(object sender, RoutedEventArgs e)
    {
        FileOpenPicker open = new FileOpenPicker();
        open.SuggestedStartLocation = PickerLocationId.PicturesLibrary;
        open.ViewMode = PickerViewMode.Thumbnail;

        // Filter to include a sample subset of file types
        open.FileTypeFilter.Clear();
        open.FileTypeFilter.Add(".bmp");
        open.FileTypeFilter.Add(".png");
        open.FileTypeFilter.Add(".jpeg");
        open.FileTypeFilter.Add(".jpg");

        // Open a stream for the selected file
        StorageFile file = await open.PickSingleFileAsync();

        // Ensure a file was selected
        if (file != null)
        {
            // Ensure the stream is disposed once the image is loaded
            using (IRandomAccessStream fileStream = await file.OpenAsync(FileAccessMode.Read))
            {
                BitmapImage bitmapImage = new BitmapImage();
                await bitmapImage.SetSourceAsync(fileStream);
                fileStream.AsStream().CopyTo(stream);
                img.Source = bitmapImage;
            }
        }
    }

    private async void submit_Click(object sender, RoutedEventArgs e)
    {


        Uri uri = new Uri("http://localhost/mydatabase/add.php");
        HttpClient client = new HttpClient();
        HttpStreamContent streamContent = new HttpStreamContent(stream.AsInputStream());
        HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, uri);
        request.Content = streamContent;
        HttpResponseMessage response = await client.PostAsync(uri, streamContent).AsTask(cts.Token);



    }

Answer 1

試試這個對我有用：

private static async Task Upload(string actionUrl)
{
    Image newImage = Image.FromFile(@"Absolute Path of image");
    ImageConverter _imageConverter = new ImageConverter();
    byte[] paramFileStream= (byte[])_imageConverter.ConvertTo(newImage, typeof(byte[]));

    var formContent = new MultipartFormDataContent
    {
        //send form text values here
        {new StringContent("value1"), "key1"},
        {new StringContent("value2"), "key2" },
        // send Image Here
        {new StreamContent(new MemoryStream(paramFileStream)), "imagekey", "filename.jpg"}
    };

    var myHttpClient = new HttpClient();
    var response = await myHttpClient.PostAsync(actionUrl.ToString(), formContent);
    string stringContent = await response.Content.ReadAsStringAsync();

    return response;
}

Answer 2

您可以使用HttpStreamContent 類將流直接上傳到您的 Web 服務器。 例如：

private Stream stream = new MemoryStream();
private CancellationTokenSource cts;

private async void SelectImage(object sender, RoutedEventArgs e)
{
    FileOpenPicker open = new FileOpenPicker();
    open.SuggestedStartLocation = PickerLocationId.PicturesLibrary;
    open.ViewMode = PickerViewMode.Thumbnail;

    // Filter to include a sample subset of file types
    open.FileTypeFilter.Clear();
    open.FileTypeFilter.Add(".bmp");
    open.FileTypeFilter.Add(".png");
    open.FileTypeFilter.Add(".jpeg");
    open.FileTypeFilter.Add(".jpg");

    // Open a stream for the selected file
    StorageFile file = await open.PickSingleFileAsync();

    // Ensure a file was selected
    if (file != null)
    {
        // Ensure the stream is disposed once the image is loaded
        using (IRandomAccessStream fileStream = await file.OpenAsync(FileAccessMode.Read))
        {
            BitmapImage bitmapImage = new BitmapImage();
            await bitmapImage.SetSourceAsync(fileStream);
            fileStream.AsStream().CopyTo(stream);
            img.Source = bitmapImage;
        }
    }
}

private async void UploadImage(object sender, RoutedEventArgs e)
{
    Uri uri = new Uri("you web uri");
    HttpClient client = new HttpClient();
    HttpStreamContent streamContent = new HttpStreamContent(stream.AsInputStream());
    HttpRequestMessage request = new HttpRequestMessage(HttpMethod.Post, uri);
    request.Content = streamContent;
    HttpResponseMessage response = await client.PostAsync(uri, streamContent).AsTask(cts.Token);
}

正如我們在您的另一個問題中所討論的，您可以參考官方的HttpClient 示例，場景 5 是關於發布流的。

對於客戶端應用程序，我們能做的只是正確上傳文件流，重要的是，您需要實現圖像解碼功能並保存到服務器中的mysql。

順便說一句，你曾經問過同樣的問題， 使用 windows 通用應用程序將圖像上傳到 mysql 數據庫，我已經看到你對我的答案的最新評論，在這種情況下我不會更新我的答案，希望這是正確的你問的。

Answer 3

@semwal

我不記得具體的實際修復是什么，但這是我的解決方案。 希望能幫助到你

   public async Task<JsonApiResult> SendHttpData(string file, string token, string claimid, string serviceMethod)
    {
        serviceMethod = $"{serviceMethod}/?claimid={claimid}&token={token}";

        HttpClient _httpClient = new HttpClient();

        _httpClient.Timeout = TimeSpan.FromMinutes(10);

        _httpClient.BaseAddress = new Uri(_url);

        try
        {
            string filename = Path.GetFileName(file);

            MultipartFormDataContent content = new MultipartFormDataContent();

            var fileContent = new StreamContent(File.OpenRead(file));
            fileContent.Headers.ContentDisposition = new ContentDispositionHeaderValue("form-data") { Name = "result", FileName = $"\"{filename}\"" };
            fileContent.Headers.ContentType = new MediaTypeHeaderValue("multipart/form-data");

            content.Add(fileContent);

            HttpResponseMessage response = await _httpClient.PostAsync(serviceMethod, content);

            if (response.IsSuccessStatusCode)
            {
                return new JsonApiResult { Result = "success", Message = "File Sent", Data = "" };
            }
            else
            {
                return new JsonApiResult { Result = "fail", Message = response.ToString(), Data = "" };
            }
        }
        catch (Exception e)
        {
            return new JsonApiResult { Result = "fail", Message = e.Message, Data = "" };
        }
    }
}