RESTsharp使用＆获取命令

Question

Hey all this is the first time I am trying out RESTsharp . I am trying to create a GET call that looks like this:

http://labs.bible.org/api/?passage=random&type=json

I've tried the following with looking at some online examples:

var client = new RestClient("http://labs.bible.org/");
var request = new RestRequest("api/?{0}&{1}", Method.GET);

request.AddParameter("passage", "random");
request.AddParameter("type", "json");

var queryResult = client.Execute<List<quotesAPI>>(request).Data;

When I put a stop on the queryResult it just says NULL.

quotesAPI looks like this:

public class qAPI
{
   public string bookname { get; set; }
   public string chapter { get; set; }
   public string verse { get; set; }
   public string text { get; set; }
}

So how do I need to format the call in order for it to work as it should be?

update 1

 var client = new RestClient("http://labs.bible.org/");
 var request = new RestRequest("api", Method.GET);

 request.AddParameter("passage", "random");
 request.AddParameter("type", "json");
 client.AddHandler("application/x-javascript", new RestSharp.Deserializers.JsonDeserializer());

 var queryResult = client.Execute<List<quotesAPI>>(request).Data;

Answer 1

First, there is no need to create rest client like this:

new RestRequest("api/?{0}&{1}", Method.GET);

This will result in query to http://labs.bible.org/api/?{0}&{1}&passage=random&type=json . In this specific case it might still "work", but in general you should of course avoid this. Instead, create it like this:

new RestRequest("api", Method.GET);

For GET methods, parameters you then create will be appended to query string for you.

Another problem here is unusual content type of response. Response for your query has content type application/x-javascript . RestSharp has no idea what to do with such content type, so you should tell it:

var client = new RestClient("http://labs.bible.org/");
client.AddHandler("application/x-javascript", new RestSharp.Deserializers.JsonDeserializer());

Here you say that it should deserialize response with such content type as json. After that you should receive expected result.