我有一个用C#写的WebForm应用程序,我想做的是用户提交带有图像的HTML表单后,我将该图像发送到WCF REST服务也在C#中编写的WCF REST服务。。问题是当我在Web服务中获取图像时,这将被损坏。
我想问题是我无法正确编码文件,但是在互联网上阅读几天后,我没有找到线索。
WebForm代码:
protected void Page_Load(object sender, EventArgs e)
{
HttpPostedFile image = Request.Files["imagen"];
string serverResponse = Send("mywebservice/postimage", "POST", Encoding.UTF8.GetBytes(StreamToString(image.InputStream)));
}
编辑(这种方式工作)
protected void Page_Load(object sender, EventArgs e)
{
HttpPostedFile image = Request.Files["imagen"];
MemoryStream ms = new MemoryStream();
image.InputStream.CopyTo(ms);
byte[] bytes = ms.ToArray();
string serverResponse = Send("mywebservice/postimage", "POST", bytes);
}
public int Send(string url, string method, byte[] data)
{
string serverResponse = "";
HttpWebRequest newRequest = (HttpWebRequest)WebRequest.Create(url);
newRequest.ContentType = "image/jpeg";
newRequest.Method = method;
newRequest.Timeout = 10000;
if (newRequest.Method == "POST" || newRequest.Method == "PUT")
{
Stream reqStream = newRequest.GetRequestStream();
reqStream.Write(data, 0, data.Length);
reqStream.Close();
}
WSMessageEnt wsMessageEnt = new WSMessageEnt();
try
{
HttpWebResponse webResponse;
webResponse = (HttpWebResponse)newRequest.GetResponse();
Stream dataStream = webResponse.GetResponseStream();
serverResponse = new StreamReader(dataStream).ReadToEnd();
}
catch (WebException we)
{
}
return serverResponse;
}
public static string StreamToString(Stream data)
{
StreamReader reader = new StreamReader(data);
string body = reader.ReadToEnd();
reader.Close();
reader.Dispose();
return body;
}
WebService代码:
[WebInvoke(UriTemplate = "upload-user-image", Method = "POST")]
public Stream UploadUserImage(Stream streamdata)
{
System.Drawing.Image img = System.Drawing.Image.FromStream(streamImagen, true);
// here I get a format error
}
可能是一个问题:您将字节流转换为字符串,然后返回到字节流。图像字节是任意的,可以或可能不会映射到字符串,并且您在一个转换中使用的编码(Encoding.default)可能与另一个转换(encoding.default)不同(Encoding.utf8)。
而不是这样做:
Encoding.UTF8.GetBytes(StreamToString(image.InputStream))
尝试做
之类的事情MemoryStream ms = new MemoryStream();
image.InputStream.CopyTo(ms);
byte[] bytes = ms.ToArray();
或只是将流传递给Send
并将其复制到请求流。