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WebRequestを使用してPOSTいくつかのデータを読み取り、応答を読み取りますか？

サーバーにAPIにPOSTを作成させる必要があります。WebRequestオブジェクトにPOST値を追加する方法と、それを送信して応答を取得する方法（文字列になります）

私はPOST 2つの値、さらにもっと多くの場合、これらの例では文字列postData = "投稿する文字列"が表示されていますが、どのようにPOSTしていることを知らせますか？複数のフォーム値があることを？

c#asp.netwebrequest
19
MetaGuru

[〜＃〜] msdn [〜＃〜] から

// Create a request using a URL that can receive a post. 
WebRequest request = WebRequest.Create ("http://contoso.com/PostAccepter.aspx ");
// Set the Method property of the request to POST.
request.Method = "POST";
// Create POST data and convert it to a byte array.
string postData = "This is a test that posts this string to a Web server.";
byte[] byteArray = Encoding.UTF8.GetBytes (postData);
// Set the ContentType property of the WebRequest.
request.ContentType = "application/x-www-form-urlencoded";
// Set the ContentLength property of the WebRequest.
request.ContentLength = byteArray.Length;
// Get the request stream.
Stream dataStream = request.GetRequestStream ();
// Write the data to the request stream.
dataStream.Write (byteArray, 0, byteArray.Length);
// Close the Stream object.
dataStream.Close ();
// Get the response.
WebResponse response = request.GetResponse ();
// Display the status.
Console.WriteLine (((HttpWebResponse)response).StatusDescription);
// Get the stream containing content returned by the server.
dataStream = response.GetResponseStream ();
// Open the stream using a StreamReader for easy access.
StreamReader reader = new StreamReader (dataStream);
// Read the content.
string responseFromServer = reader.ReadToEnd ();
// Display the content.
Console.WriteLine (responseFromServer);
// Clean up the streams.
reader.Close ();
dataStream.Close ();
response.Close ();

Key1 = value1＆key2 = value2の形式で情報を送信する必要があることを考慮してください

30
Claudio Redi

ここに私のために働くものがあります。改善できると確信していますので、お気軽に提案や編集を行って改善してください。

const string WEBSERVICE_URL = "http://localhost/projectname/ServiceName.svc/ServiceMethod";
//This string is untested, but I think it's ok.
string jsonData = "{ \"key1\" : \"value1\", \"key2\":\"value2\"  }"; 
try
{       
    var webRequest = System.Net.WebRequest.Create(WEBSERVICE_URL);
    if (webRequest != null)
    {
        webRequest.Method = "POST";
        webRequest.Timeout = 20000;
        webRequest.ContentType = "application/json";

    using (System.IO.Stream s = webRequest.GetRequestStream())
    {
        using (System.IO.StreamWriter sw = new System.IO.StreamWriter(s))
            sw.Write(jsonData);
    }

    using (System.IO.Stream s = webRequest.GetResponse().GetResponseStream())
    {
        using (System.IO.StreamReader sr = new System.IO.StreamReader(s))
        {
            var jsonResponse = sr.ReadToEnd();
            System.Diagnostics.Debug.WriteLine(String.Format("Response: {0}", jsonResponse));
        }
    }
}
}
catch (Exception ex)
{
    System.Diagnostics.Debug.WriteLine(ex.ToString());
}
21
Adrian Carr

HttpWebRequestおよびHttpWebResponseオブジェクトを使用してWebサービスに投稿する例を次に示します。

StringBuilder sb = new StringBuilder();
    string query = "?q=" + latitude + "%2C" + longitude + "&format=xml&key=xxxxxxxxxxxxxxxxxxxxxxxx";
    string weatherservice = "http://api.worldweatheronline.com/free/v1/marine.ashx" + query;
    HttpWebRequest request = (HttpWebRequest)WebRequest.Create(weatherservice);
    request.Referer = "http://www.yourdomain.com";
    HttpWebResponse response = (HttpWebResponse)request.GetResponse();
    Stream stream = response.GetResponseStream();
    StreamReader reader = new StreamReader(stream);
    Char[] readBuffer = new Char[256];
    int count = reader.Read(readBuffer, 0, 256);

    while (count > 0)
    {
        String output = new String(readBuffer, 0, count);
        sb.Append(output);
        count = reader.Read(readBuffer, 0, 256);
    }
    string xml = sb.ToString();
2
Johnson Gibbs

以下は、テキストファイルからデータを読み取り、ハンドラーに送信して処理し、ハンドラーから応答データを受信し、読み取ってデータを文字列ビルダークラスに格納するコードです。

 //Get the data from text file that needs to be sent.
                FileStream fileStream = new FileStream(@"G:\Papertest.txt", FileMode.OpenOrCreate, FileAccess.ReadWrite);
                byte[] buffer = new byte[fileStream.Length];
                int count = fileStream.Read(buffer, 0, buffer.Length);

                //This is a handler would recieve the data and process it and sends back response.
                WebRequest myWebRequest = WebRequest.Create(@"http://localhost/Provider/ProcessorHandler.ashx");

                myWebRequest.ContentLength = buffer.Length;
                myWebRequest.ContentType = "application/octet-stream";
                myWebRequest.Method = "POST";
                // get the stream object that holds request stream.
                Stream stream = myWebRequest.GetRequestStream();
                       stream.Write(buffer, 0, buffer.Length);
                       stream.Close();

                //Sends a web request and wait for response.
                try
                {
                    WebResponse webResponse = myWebRequest.GetResponse();
                    //get Stream Data from the response
                    Stream respData = webResponse.GetResponseStream();
                    //read the response from stream.
                    StreamReader streamReader = new StreamReader(respData);
                    string name;
                    StringBuilder str = new StringBuilder();
                    while ((name = streamReader.ReadLine()) != null)
                    {
                        str.Append(name); // Add to stringbuider when response contains multple lines data
                   }
                }
                catch (Exception ex)
                {
                    throw ex;
                }
0
Sheo Dayal Singh

より強力で柔軟な例は、ここにあります： フォームフィールド、Cookie、ヘッダーを使用したC＃ファイルのアップロード

0
Sky Sanders