Drahcir
Drahcir

Reputation: 11972

How To Call XML POST API

I need to call an API with POST arguments, for example:

I've had a look at XDocument but not sure how to send params in the request. I'm also not sure how I would call asynchronously, and even if I should or if it's better/easier to run in another thread.

I'd be calling this from my windows based C# application.

Upvotes: 2

Views: 6418

Answers (4)

Er. Binod Mehta
Er. Binod Mehta

Reputation: 63

in asp .NET

var client = new RestClient("www.api.url");
client.Timeout = -1;
var request = new RestRequest(Method.POST);
request.AddHeader("id", "givenid");
request.AddHeader("HASH", "generatedHash");
request.AddParameter("text/plain", "fullxml or body",  ParameterType.RequestBody);
IRestResponse response = client.Execute(request);
Console.WriteLine(response.Content);

In HTTP

POST /api/somename/1.1
Host: google.com
id: merchantid
HASH: generatedshah
<node1><element>Individual</element></node1>

Phython - http.client

import http.client
import mimetypes
conn = http.client.HTTPSConnection("url")
payload = "<Node><Element>Individual</Element></Node>"
headers = {
  'id': 'givenid',
  'HASH': 'generatedhash'
}
conn.request("POST", "/api/blacklist/verify", payload, headers)
res = conn.getresponse()
data = res.read()
print(data.decode("utf-8"))

Upvotes: 1

Steven Edison
Steven Edison

Reputation: 527

Call it from where? Javascript? If so you can use JQuery:

http://api.jquery.com/jQuery.post/

$.post('http://localhost/myAPI/', { options: "blue", type="car"}, function(data) {
  $('.result').html(data);
});

data will contain the result of your post.

If from server-side you can use HttpWebRequest and write to it's stream with your params.

// Create a request using a URL that can receive a post. 
        WebRequest request = WebRequest.Create ("http://localhost/myAPI/");
        // Set the Method property of the request to POST.
        request.Method = "POST";
        // Create POST data and convert it to a byte array.
        string postData = "options=blue&type=car";
        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 ();

Upvotes: 0

Ross
Ross

Reputation: 2468

To throw another option into the ring, you may want to consider using the PostAsync method of HttpClient in .NET 4.5. An admittedly untested stab:

        HttpClient client = new HttpClient();
        var task = client.PostAsync(string.Format("{0}{1}", "http://localhost/myAPI", "?options=blue&type=car"), null);
        Car car = task.ContinueWith(
            t =>
            {
                return t.Result.Content.ReadAsAsync<Car>();
            }).Unwrap().Result;

Upvotes: 0

jrummell
jrummell

Reputation: 43077

You can use one of the Upload methods of WebClient.

WebClient client = new WebClient();
string response = client.UploadString(
                  "http://localhost/myAPI/?options=blue&type=car", 
                  "POST data");

Upvotes: 3

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