所以我需要将一个图像上传到我的Mysql数据库以及一些其他字符串,例如名称... 我能够将名称添加到Mysql DB中,但我无法为图像执行此操作。 我在一个字节[]中转换了图像,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);
}
答案 0 :(得分:1)
您可以使用HttpStreamContent class将流直接上传到您的网络服务器。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);
}
正如我们在您的另一个问题中所讨论的那样,您可以参考官方HttpClient sample,方案5是关于发布流。
对于客户端应用,我们可以做的只是正确上传文件流,重要的是,您需要实现图像解码功能并保存到服务器中的mysql。
顺便说一句,你曾经问过同一个问题upload an image into mysql database using windows universal app,而且我已经看到了你对我的回答的最新评论,在那种情况下我不会更新我的答案,希望这是你要求的权利。
答案 1 :(得分:1)
尝试一下对我有用:
私有静态异步任务上传(字符串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();
返回响应; }
答案 2 :(得分:0)
@semwal
我不记得具体的解决方法,但这是我的解决方案。希望对您有帮助
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 = "" };
}
}
}