Question

我正在尝试使用RestSharp发布以下JSON：

{"UserName":"UAT1206252627",
"SecurityQuestion":{
    "Id":"Q03",
    "Answer":"Business",
    "Hint":"The answer is Business"
},
}

我认为我很接近，但我似乎正在努力解决 SecurityQuestion （API正在抛出错误，说参数丢失了，但它没有说明哪一个）

这是我到目前为止的代码：

var request = new RestRequest("api/register", Method.POST);
request.RequestFormat = DataFormat.Json;

request.AddParameter("UserName", "UAT1206252627");

SecurityQuestion securityQuestion = new SecurityQuestion("Q03");
request.AddParameter("SecurityQuestion", request.JsonSerializer.Serialize(securityQuestion));

IRestResponse response = client.Execute(request);

我的安全问题类看起来像这样：

public class SecurityQuestion
{
    public string id {get; set;}
    public string answer {get; set;}
    public string hint {get; set;}

    public SecurityQuestion(string id)
    {
         this.id = id;
         answer = "Business";
         hint = "The answer is Business";
    }
}

谁能告诉我我做错了什么？有没有其他方式发布安全问题对象？

非常感谢。

Answer 1

您需要在标题中指定内容类型：

String tag = (String) checkBox.getTag()

此外，if ( grep { $_ eq $in }@array ){ print "available\n"; }else{ print "not available\n"; }根据方法

添加到POST或URL查询字符串

我认为您需要将它添加到正文中：

request.AddHeader("Content-type", "application/json");

Answer 2

再次感谢您的帮助。为了实现这一点，我必须将所有内容作为单个参数提交。这是我最后使用的代码。

首先，我创建了几个名为Request Object和Security Question的类：

public class SecurityQuestion
{
    public string Id { get; set; }
    public string Answer { get; set; }
    public string Hint { get; set; }
}

public class RequestObject
{
    public string UserName { get; set; }
    public SecurityQuestion SecurityQuestion { get; set; }
}

然后我将它作为单个参数添加，并在发布之前将其序列化为JSON，如下所示：

var yourobject = new RequestObject
            {
                UserName = "UAT1206252627",
                SecurityQuestion = new SecurityQuestion
                {
                    Id = "Q03",
                    Answer = "Business",
                    Hint = "The answer is Business"
                },
            };
var json = request.JsonSerializer.Serialize(yourobject);

request.AddParameter("application/json; charset=utf-8", json, ParameterType.RequestBody);

IRestResponse response = client.Execute(request);

它有效！

Answer 3

要发布原始json正文字符串，AddBody（）或AddJsonBody（）方法将无效。请改用以下内容

request.AddParameter(
   "application/json",
   "{ \"username\": \"johndoe\", \"password\": \"secretpassword\" }", // <- your JSON string
   ParameterType.RequestBody);

Answer 4

通过RestSharp方法

从对象支持

AddObject

request.AddObject(securityQuestion);

Answer 5

看起来最简单的方法是让RestSharp处理所有的序列化。您只需要像这样指定RequestFormat。这就是我为自己的工作提出的建议。

    public List<YourReturnType> Get(RestRequest request)
    {
        var request = new RestRequest
        {
            Resource = "YourResource",
            RequestFormat = DataFormat.Json,
            Method = Method.POST
        };
        request.AddBody(new YourRequestType());

        var response = Execute<List<YourReturnType>>(request);
        return response.Data;
    }

    public T Execute<T>(RestRequest request) where T : new()
    {
        var client = new RestClient(_baseUrl);

        var response = client.Execute<T>(request);
        return response.Data;
    }

RestSharp发布一个JSON对象

5 个答案: