How to set the content of an HttpWebRequest in C#?

问题:

An HttpWebRequest has the properties ContentLength and ContentType, but how do you actually set the content of the request?

回答1:

The following should get you started

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();

回答2:

.NET 4.5 (or .NET 4.0 by adding the Microsoft.Net.Http package from NuGet) provides a lot of additional flexibility in setting the request content. Here is an example:

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;
    }
}

回答3:

HttpWebRequest's RequestStream is where the action is at - rough code...

//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);

If you're sending extended chars, use UTF8Encoding, make sure you set the right content-type/charset header too.

回答4:

Here's a different option for posting info without messing with Bytes and Streams. I personally find it easier to follow, read, and debug.

// 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);

回答5:

HttpWebRequest.GetRequestStream() gets the request Stream. After you have set the headers, use GetRequestStream() and write the content to the stream.

This post explains how to transmit files using HttpWebRequest, which should provide a good example of how to send content.

But, basically the format would be

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