你可以做這樣的事情在客戶端:
HttpClient client = new HttpClient();
var imageStream = File.OpenRead(@"C:\p1.jpg");
var content = new StreamContent(imageStream);
content.Headers.ContentType = new MediaTypeHeaderValue("image/jpeg");
var response = await client.PostAsync("URL", content);
您可以從NugetPackage Microsoft.Net.Http
得到
HttpClient的
在REST API端(接收端),您可以從Request.Content對象中獲取它,如下所示:
public void Post()
{
using (var fileStream = File.Create("C:\\NewFile.jpg"))
{
using (MemoryStream tempStream = new MemoryStream())
{
var task = this.Request.Content.CopyToAsync(tempStream);
task.Wait();
tempStream.Seek(0, SeekOrigin.Begin);
tempStream.CopyTo(fileStream);
tempStream.Close();
}
}
}
http://stackoverflow.com/questions/10320232/how-to-accept-a-file-post-asp-net-mvc-4-webapi –