SyntaxError：＆＃x27; super＆＃x27;关键字在扩展超级类时出乎意料

Question

我有一所超级班级学校，具有属性名称，级别和学生人数

class School{
      constructor(name,level,numberOfStudents){
        this._name = name;
        this._level = level
        this._numberOfStudents = numberOfStudents
      }
      get name(){
        return this._name;
      }
      get level(){
        return this._level
      }
      get numberOfStudents(){
        return this._numberOfStudents
      }
    }// end School() class 
    

但是，当我将其扩展到子类小学时...

    class PrimarySchool extends School{
      constructror(name,numberOfStudents,pickupPolicy){
        super(name,'primary',numberOfStudents)
        this._pickupPolicy = pickupPolicy
      }
      get pickupPolicy(){
        return this._pickupPolicy
      }
    }//end PrimarySchool

我在这里得到SyntaxError：“超级”关键字出乎意料。但是我应该能够将“主要”作为父级属性的值 level 。与名称和数字学生相同。即使我在子级构造函数中声明级别，甚至不通过一个值，我也会得到相同的错误：

class PrimarySchool extends School{
  constructror(name,level,numberOfStudents,pickupPolicy){
    super(name,level,numberOfStudents)
    this._pickupPolicy = pickupPolicy
  }
  get pickupPolicy(){
    return this._pickupPolicy
  }
}//end PrimarySchool 

Answer 1

conductror中的priendarschool类中有一个错字。正确的单词是构造函数。这是一个有效的示例代码：

class School{
      constructor(name,level,numberOfStudents){
        this._name = name;
        this._level = level
        this._numberOfStudents = numberOfStudents
      }
      get name(){
        return this._name;
      }
      get level(){
        return this._level
      }
      get numberOfStudents(){
        return this._numberOfStudents
      }
    }// end School() class 
    

   class PrimarySchool extends School{
      constructor(name,numberOfStudents,pickupPolicy){
        super(name,'primary',numberOfStudents)
        this._pickupPolicy = pickupPolicy
      }
      get pickupPolicy(){
        return this._pickupPolicy
      }
    }//end PrimarySchool


const school = new PrimarySchool( 'NY School', 2137, true );

console.log( school );

您是否将任何IDE用于语法突出显示或linter（例如ESLINT）？