hil*_*uff 3 php c# mysql image win-universal-app
所以我需要将图像以及其他一些字符串(例如名称)上传到我的 Mysql 数据库中...我能够将名称添加到 Mysql DB 中,但我无法对图像执行此操作。我将图像转换为字节 [],现在我被卡住了..这是我使用的代码
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);
}
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小智 5
试试这个它对我有用:
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;
}
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