java Rails/设计不可处理的实体

Question

我正在尝试使用post请求（JSON）来创建用户。使用curl的示例效果很好。下面是curl命令：

curl -X POST -H "Content-Type: application/json" 'http://localhost:3000/users.json' -d '{ "user": {"email": "e@f.com", "password": "foobar", "password_confirmation": "foobar"}}'

卷曲输出：

Started POST "/users.json" for 127.0.0.1 at 2012-01-28 16:19:10 -0800
  Processing by Devise::RegistrationsController#create as JSON
  Parameters: {"user"=>{"email"=>"e@f.com", "password"=>"[FILTERED]", "password_confirmation"=>"[FILTERED]"}}

但是，当我使用java中的http库尝试请求时，我得到一个错误：

Started POST "/users.json" for 192.168.1.88 at 2012-01-28 16:47:56 -0800
  Processing by Devise::RegistrationsController#create as JSON
  Parameters: {"user"=>"{\"password_confirmation\":\"secret\",\"password\":\"secret\",\"email\":\"a@foo.com\"}"}
Completed 422 Unprocessable Entity in 73ms (Views: 3.0ms | ActiveRecord: 0.0ms)

用于创建请求的代码：

            RestClient client = new RestClient(SIGNUP_URL);
            JSONObject jObj = new JSONObject();
            JSONObject jsonUserObj = new JSONObject();

            try {
                jObj.put("email", txtUserName.getText().toString());
                jObj.put("password", txtPassword.getText().toString());
                jObj.put("password_confirmation", txtPassword.getText().toString());

                jsonUserObj.put("user", jObj.toString());
            } catch (JSONException e1) {
                e1.printStackTrace();
            }

            client.setJSONParams(jsonUserObj);
            try {
                client.Execute(RequestMethod.JSON_POST);
            } catch (IOException e) {
                e.printStackTrace();
            } catch (JSONException e) {
                    e.printStackTrace();
            } catch (Exception e) {
                    e.printStackTrace();
            }   
            if (client.getResponseCode() == HttpStatus.SC_OK) {
               // Good response
               try {
                   jObj = new JSONObject(client.getResponse());
                   System.out.println("Signup Successful");
               } catch (JSONException e) {
                   // TODO Auto-generated catch block
                   e.printStackTrace();
               }
            }
        }

以下是为执行调用的函数：

    HttpPost request = new HttpPost(url);
            request.addHeader("Content-Type", "application/json");
            request.addHeader("Accept", "application/json");
            StringEntity s = new StringEntity(jsonParams.toString(), HTTP.UTF_8);
            s.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, "application/json"));
            request.setEntity(s);
            executeRequest(request, url);
            break;

Answer 1

对我来说，同样的问题来自于RegistrationsController中的一些错误，因此我将与您分享我的问题，您可以进行必要的更改以满足您的需要

class RegistrationsController < Devise::RegistrationsController
  skip_before_filter :verify_authenticity_token,
                     :if => Proc.new { |c| c.request.format == 'application/json' }

  respond_to :json

  def create
    build_resource
    resource = User.new(params[:user])
    if resource.save
      sign_in resource
      render :status => 200,
           :json => { :success => true,
                      :info => "Registered",
                      :data => { :user => resource,
                                 :auth_token => current_user.authentication_token } }
    else
      logger.info("current user passed from json: #{resource.to_yaml}")
      render :status => :unprocessable_entity,
             :json => { :success => false,
                        :info => resource.errors,
                        :data => {} }
    end
  end
end

Answer 2

如果查看Java库请求的日志，您会发现用户的内容只是一个大的转义字符串，而不是嵌套的哈希

当您查看创建请求的代码时，您有：

jsonUserObj.put("user", jObj.toString());

我猜如果你把它改成：

jsonUserObj.put("user", jObj);

它将起作用，因为这样它将把作为jObj的散列与“user”关联起来，而不是将该对象的字符串与“user”关联起来