Question

我想问邮递员的param帖子，所以我可以在rails中的控制器中获取它。

这是我的控制器

class Api::V1::UsersController < ApplicationController
  before_action :get_user, only: [:show, :fullname]
  #get all user object
  def index
    render :json => User.all
  end

  # get user object with id
  def show
    render :json => @user
  end

  # get full name of user object with id
  def fullname
    @hasil = @user.firstname + ' ' + @user.lastname
    render :json => @hasil
  end

  # post object into model
  def create
    @user = User.new(user_params)
    @user.save
    if @user.save
        render :json => @user
    else
        render :json => @user.errors.full_messages
    end
  end

  private
    def user_params
      params.permit(:username, :firstname, :lastname, :age)
    end

    def get_user
      @user = User.find(params[:id])
    end
end

当我在邮递员中尝试时，我的用户名，名字，姓氏和年龄都是空的。

我像这样发布

[
  {
    "username": "admin2",
    "firstname": "Reza Adha",
    "lastname": "Hamonangan",
    "age": 23,
  }
]

当我尝试发布

时，我的rails服务器日志

Started POST "/api/v1/users" for ::1 at 2016-11-18 18:04:24 +0700
  ActiveRecord::SchemaMigration Load (0.0ms)  SELECT `schema_migrations`.* FROM `schema_migrations`
Processing by Api::V1::UsersController#create as */* 
   (0.0ms)  BEGIN
  SQL (1.0ms)  INSERT INTO `users` (`created_at`, `updated_at`) VALUES ('2016-11-18 11:04:25', '2016-11-18 11:04:25')
   (4.2ms)  COMMIT
   (0.0ms)  BEGIN
   (0.0ms)  COMMIT
Completed 200 OK in 22ms (Views: 0.3ms | ActiveRecord: 7.2ms)

我认为我在控制器中做错了从邮递员处获取POST数据。请帮我.. 谢谢你的关注。

Answer 1

你需要在params上使用user键：

[
  "user": {
    "username": "admin2",
    "firstname": "Reza Adha",
    "lastname": "Hamonangan",
    "age": 23,
  }
]

允许user使用强大的参数：

params(:user).permit(:username, :firstname, :lastname, :age)

Answer 2

好的，所以我从here找到答案。我在邮递员T_T中插错了

它看起来应该是这样的

{
    "user" : {
      "username": "admin2",
      "firstname": "Reza Adha",
      "lastname": "Hamonangan",
      "age": 23
    }
}

并且我没有给出头文件aplication / json。我很抱歉这个大惊小怪。谢谢@ user100693和@MurifoX

如何从Rails控制器中的Postman获取数据？

2 个答案: