我正在使用在此处找到的一些代码将图像发布到服务器。问题是我没有收到我应该收到的 Json 响应，而是在我的“ResponseReady”回调中收到 OL'SERVER NOT FOUND 响应。（编辑：原来这只是我的参数，这段代码工作得很好。）
这是我用来制作 POST 的课程
public class PostSubmitter
{
public string url { get; set; }
public Dictionary<string, object> parameters { get; set; }
string boundary = "----------" + DateTime.Now.Ticks.ToString();
HttpWebRequest webRequest;
public void Submit()
{
// Prepare web request...
webRequest = WebRequest.CreateHttp(url);
webRequest.Method = "POST";
webRequest.ContentType = string.Format("multipart/form-data; boundary={0}", boundary);
webRequest.BeginGetRequestStream(new AsyncCallback(RequestReady), webRequest);
}
private void RequestReady(IAsyncResult asynchronousResult)
{
using (Stream postStream = webRequest.EndGetRequestStream(asynchronousResult))
{
writeMultipartObject(postStream, parameters);
}
webRequest.BeginGetResponse(new AsyncCallback(ResponseReady), webRequest);
}
private void ResponseReady(IAsyncResult asynchronousResult)
{
try
{
using (var response =
(HttpWebResponse)webRequest.EndGetResponse(asynchronousResult))
using (var streamResponse = response.GetResponseStream())
using (var streamRead = new StreamReader(streamResponse))
{
var responseString = streamRead.ReadToEnd();
var success = response.StatusCode == HttpStatusCode.OK;
if (responseString != null)
{
//JObject comes from Newtonsoft.Json ddl. This is a good one if your working with json
JObject jsonResponse = JObject.Parse(responseString);
//Do stuff with json.....
}
}
}
catch (Exception e)
{
if (e.Message == "The remote server returned an error: NotFound.")
{
webRequest.Abort();
Deployment.Current.Dispatcher.BeginInvoke(delegate() { MessageBox.Show("Unable to connect to server at this time, please try again later"); });
}
else
Deployment.Current.Dispatcher.BeginInvoke(delegate() { MessageBox.Show("Unable to upload photo at this time, please try again later"); });
return;
}
}
public void writeMultipartObject(Stream stream, object data)
{
using (StreamWriter writer = new StreamWriter(stream))
{
if (data != null)
{
foreach (var entry in data as Dictionary<string, object>)
{
WriteEntry(writer, entry.Key, entry.Value);
}
}
writer.Write("--");
writer.Write(boundary);
writer.WriteLine("--");
writer.Flush();
}
}
private void WriteEntry(StreamWriter writer, string key, object value)
{
if (value != null)
{
writer.Write("--");
writer.WriteLine(boundary);
if (value is byte[])
{
byte[] ba = value as byte[];
writer.WriteLine(@"Content-Disposition: form-data; name=""{0}""; filename=""{1}""", key, "sentPhoto.jpg");
writer.WriteLine(@"Content-Type: application/octet-stream");
writer.WriteLine(@"Content-Type: image / jpeg");
writer.WriteLine(@"Content-Length: " + ba.Length);
writer.WriteLine();
writer.Flush();
Stream output = writer.BaseStream;
output.Write(ba, 0, ba.Length);
output.Flush();
writer.WriteLine();
}
else
{
writer.WriteLine(@"Content-Disposition: form-data; name=""{0}""", key);
writer.WriteLine();
writer.WriteLine(value.ToString());
}
}
}
}
使用这个类，我们可以使用以下代码行对服务器进行简单的 POST：
Dictionary<string, object> postData = new Dictionary<string, object>()
{
{"file", byteArrayOfImage}
//You can add other parameters here
};
PostSubmitter postToServer = new PostSubmitter() { url = getPicturePostUrl(), parameters = postData };
postToServer.Submit();
这方面有很多问题......你会认为它们会让复杂的网络请求变得更容易......
提前感谢您的有用评论或随时提出问题。