Ruby on Rails MySQL query confusion

Question

I'd like to make my code much simply and efficient. My code goes like this (in MODEL)

class Employee < ApplicationRecord
  mount_uploader :image, AvatarUploader

  def self.search(search)
      if search
        where(('(name LIKE ? OR name = ?) AND (employee_type = "SE" OR employee_type = "OP")'), "%#{search}%", "")
    else
      unscoped
    end
  end
end

and my controller goes like this:

class EmployeesController < ApplicationController
  before_action :set_employee, only: [:show, :edit, :update, :destroy]
  # GET /employees
  # GET /employees.json
  helper_method :sort_column, :sort_direction

  def index
    @employees = Employee.all
    @employees = Employee.search(params[:employee_type])
    @employees = Employee.search(params[:search]).order(sort_column + " " + sort_direction).paginate(:per_page => 5, :page => params[:page])
  end 
end

it looks like you are over riding your variable in the controller @employees 3 times — MZaragoza
– MZaragoza, Commented Feb 20, 2017 at 2:27

Arun Kumar Mohan · Accepted Answer · 2017-02-20 02:37:05Z

1

Wou can start by doing something like this

class EmployeesController < ApplicationController
  before_action :set_employee, only: [:show, :edit, :update, :destroy]
  # GET /employees
  # GET /employees.json
  helper_method :sort_column, :sort_direction

  def index
    if params[:employee_type].present?
      @employees = Employee.search(params[:employee_type])
    elsif params[:search].present?
      @employees = Employee.search(params[:search]).order(sort_column + " " + sort_direction).paginate(:per_page => 5, :page => params[:page])
    else
      @employees = Employee.all
    end
  end 
end

now you wont over ride your variables

edited Feb 20, 2017 at 2:37

Arun Kumar Mohan

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answered Feb 20, 2017 at 2:29

MZaragoza

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2 Comments

Renz Daitol Over a year ago

Thanks @Mzaragoza! I didn't know we can put conditional statements in controller :D Thanks!

MZaragoza Over a year ago

@reynantedaitol I am glad to help :)

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