如何在C＃中设置HttpWebRequest的内容？

Question

HttpWebRequest具有ContentLength和ContentType属性，但是如何实际设置请求的内容？

Answer 1

以下内容应该让您入门

byte[]  buffer = ...request data as bytes
var webReq = (HttpWebRequest) WebRequest.Create("http://127.0.0.1/target");

webReq.Method = "REQUIRED METHOD";
webReq.ContentType = "REQUIRED CONTENT TYPE";
webReq.ContentLength = buffer.Length;

var reqStream = webReq.GetRequestStream();
reqStream.Write(buffer, 0, buffer.Length);
reqStream.Close();

var webResp = (HttpWebResponse) webReq.GetResponse();

Answer 2

.NET 4.5（或通过添加NuGet的Microsoft.Net.Http包提供.NET 4.0）在设置请求内容时提供了很多额外的灵活性。 这是一个例子：

private System.IO.Stream Upload(string actionUrl, string paramString, Stream paramFileStream, byte [] paramFileBytes)
{
    HttpContent stringContent = new StringContent(paramString);
    HttpContent fileStreamContent = new StreamContent(paramFileStream);
    HttpContent bytesContent = new ByteArrayContent(paramFileBytes);
    using (var client = new HttpClient())
    using (var formData = new MultipartFormDataContent())
    {
        formData.Add(stringContent, "param1", "param1");
        formData.Add(fileStreamContent, "file1", "file1");
        formData.Add(bytesContent, "file2", "file2");
        var response = client.PostAsync(actionUrl, formData).Result;
        if (!response.IsSuccessStatusCode)
        {
            return null;
        }
        return response.Content.ReadAsStreamAsync().Result;
    }
}

Answer 3

这里有一个不同的选项，用于发布信息而不会弄乱Bytes和Streams。 我个人觉得更容易理解，阅读和调试。

// Convert Object to JSON
var requestMessage = JsonConvert.SerializeObject(requestObject);
var content = new StringContent(requestMessage, Encoding.UTF8, "application/json");

// Create the Client
var client = new HttpClient();
client.DefaultRequestHeaders.Add(AuthKey, AuthValue);

// Post the JSON
var responseMessage = client.PostAsync(requestEndPoint, content).Result;
var stringResult = responseMessage.Content.ReadAsStringAsync().Result;

// Convert JSON back to the Object
var responseObject = JsonConvert.DeserializeObject<ResponseObject>(stringResult);

Answer 4

HttpWebRequest的RequestStream是动作所在的位置 - 粗略代码......

//build the request object
HttpWebRequest request = (HttpWebRequest)WebRequest.Create(http://someapi.com/);
//write the input data (aka post) to a byte array
byte[] requestBytes = new ASCIIEncoding().GetBytes(inputData);
//get the request stream to write the post to
Stream requestStream = request.GetRequestStream();
//write the post to the request stream
requestStream.Write(requestBytes, 0, requestBytes.Length);

如果您要发送扩展字符，请使用UTF8Encoding，请确保您也设置了正确的content-type / charset标头。

Answer 5

HttpWebRequest.GetRequestStream()获取请求Stream。 设置标头后，使用GetRequestStream()并将内容写入流。

这篇文章解释了如何使用HttpWebRequest传输文件，这应该提供一个如何发送内容的好例子。

但是，基本上格式是

 var stream = request.GetRequestStream();
 stream.Write( stuff );
 stream.Close();
 var response = request.GetResponse();