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如何按名称或类型创建类的实例?

typescript

POV · 技术社区 · 6 年前

我的课程结构如下:

class FilterWeekScheduleClass {

}

class FilterClassJournal { 

}

const registryFilterClasses = {
  FilterWeekScheduleClass,
  FilterClassJournal
};

class SingletonClassRegister {
  public registeredClasses = {};

  public constructor() {
    for (let obj in registryFilterClasses) {
      // CREATE INSTANCE OF obj LIKE new obj();
    }
  }
}

let a = new SingletonClassRegister();

在的构造函数中 SingletonClassRegister 类,我尝试根据创建实例 registryFilterClasses . 如何正确创建实例?

我试着这样做:

class SingletonClassRegister {
  public registeredClasses = {};

  public constructor() {
    for (let obj of registryFilterClasses) {
      registryFilterClasses[obj.constructor.name] = new obj();
    }
  }
}

对吗?

1 回复 | 直到 6 年前

Christoph Lütjen 6 年前

首先,我认为您需要一个数组,所以让我们修复registryfilterclasses:

const registryFilterClasses = [
  FilterWeekScheduleClass,
  FilterClassJournal
];

然后简单地创建实例:

for (let cls of registryFilterClasses) {
  let myInstance = new cls();
}

或与数据源对象相同:

const registryFilterClasses = {
  FilterWeekScheduleClass,
  FilterClassJournal
};
// see https://eslint.org/docs/rules/object-shorthand
// same as
//    const registryFilterClasses = {
//      FilterWeekScheduleClass: FilterWeekScheduleClass,
//      FilterClassJournal: FilterClassJournal
//    };

for (let key in registryFilterClasses) {
  // key is now the name of the class as string
  let myInstance = new registryFilterClasses[key]();
}

更新:获取类名:

class Test {}
Test.name => "Test"

const t = Test;
t.name => "Test"

const i = new t();
i.constructor.name => "Test"

注册表的可能实现:

class SingletonClassRegister {
  public registeredClasses = {
    MyClass
  };

  private instances = {};

  getInstanceByClassName(name) {
    if(this.instances[name] == undefined)
      this.instances[name] = new this.registeredClasses[name]();
    return this.instances[name];
  }

  getInstancesForAllRegisteredClasses() {
    return Object.keys(this.registeredClasses)
       .map(name => this.getInstanceByClassName(name));
  }
}

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