将curl命令转换为restsharp

Question

如何将这个curl命令转换为restsharp？

curl -X POST \
--header "Authorization: Bearer SomeToken" \
--header "_Id: SomeId" \    
--data-urlencode 'keys: name, age' \
https://api.SomeDomain.com/object-storage/classes/query/Player

我试过了：

RestClient client = new RestClient("https://api.SomeDomain.com");
RestRequest request = new RestRequest("/object-storage/classes/query/hotels", Method.POST);
request.AddHeader("_Id", "SomeId");
request.AddHeader("Authorization", "Bearer " + SomeToken);
request.AddParameter("application/json", "{\"keys\" : \"name\" ,\"age\"}");

var response = client.Execute(request);

我不知道如何转换这行代码：

--data-urlencode 'keys: name, age' \

编辑1：

好的，就像@Evk所说的，我试着使用curl commond：

curl -X POST \
--header "Authorization: Bearer SomeToken" \
 --header "X-Backtory-Object-Storage-Id: SomeId"  \
 --data-urlencode 'keys:Name' \
 --trace-ascii /dev/stdout 
 https://api.backtory.com/object-storage/classes/query/hotels

这是跟踪打印：

 => Send header, 908 bytes (0x38c)
 0000: POST /object-storage/classes/query/hotels HTTP/1.1
 0034: Host: api.backtory.com
 004c: User-Agent: curl/7.52.1
 0065: Accept: */*
 0072: Authorization: Bearer SomeToken
 030d: X-Backtory-Object-Storage-Id: SomeId
 0345: Content-Length: 17
 0359: Content-Type: application/x-www-form-urlencoded
 038a:
 => Send data, 17 bytes (0x11)
 0000: %27keys%3AName%27

以及错误：

 {
 "timestamp" : "2017-01-13T13:59:29.883UTC",
 "status" : 500,
 "error" : "Internal Server Error",
 "exception" : "com.google.gson.JsonSyntaxException",
 "message" : "org.springframework.web.util.NestedServletException: Request processing failed; nested exception is com.google.gson.JsonSyntaxException: com.google.gson.stream.MalformedJsonException: Use JsonReader.setLenient(true) to accept malformed JSON at line 1 column 19 path $",
 "path" : "/classes/query/hotels"
 }

对于更多细节，我在服务器端没有控制权！