Question

我想让我的代码简单有效。我的代码是这样的（在MODEL中）

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

我的控制器是这样的：

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

Answer 1

你可以从做这样的事情开始

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

现在你不会超越你的变量

Ruby on Rails MySQL查询混乱

1 个答案: