Javascript 从参数扩展类

在理想的情况下，我将能够导入

google

，并对其进行扩展，同时在我的类中使用它

import google from 'google'

class GoogleMapsPopover extends google.maps.OverlayView{
  constructor ({point, map}) {
    this.bounds_ = new google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

但是，我想将

google

作为参数传递给

googlemapspover

类

如果不进行如下扩展，就无法拥有

super

：

class GoogleMapsPopover {
  constructor ({google, point, map}) {
    super(google.maps.OverlayView)
    this.google = google
    this.bounds_ = new this.google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

class GoogleMapsPopover {
  constructor ({google, point, map}) {
    this = google.maps.OverlayView
    this.google = google
    this.bounds_ = new this.google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

export const getGoogleMapsPopoverClass = (google) => {
  return class GoogleMapsPopover extends google.maps.OverlayView {
    constructor ({point, map}) {
      super()
      this.bounds_ = new google.maps.LatLngBounds(point, point)
      this.map_ = map
      this.div_ = null
      this.setMap(map)
    }
  }
}

const GoogleMapsPopover = getGoogleMapsPopoverClass(google)

您不能像这样设置

this

的值：

class GoogleMapsPopover {
  constructor ({google, point, map}) {
    super(google.maps.OverlayView)
    this.google = google
    this.bounds_ = new this.google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

class GoogleMapsPopover {
  constructor ({google, point, map}) {
    this = google.maps.OverlayView
    this.google = google
    this.bounds_ = new this.google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

export const getGoogleMapsPopoverClass = (google) => {
  return class GoogleMapsPopover extends google.maps.OverlayView {
    constructor ({point, map}) {
      super()
      this.bounds_ = new google.maps.LatLngBounds(point, point)
      this.map_ = map
      this.div_ = null
      this.setMap(map)
    }
  }
}

const GoogleMapsPopover = getGoogleMapsPopoverClass(google)

我能想到的唯一替代方法是将类包装在函数中，这有点粗糙

大概是这样的：

class GoogleMapsPopover {
  constructor ({google, point, map}) {
    super(google.maps.OverlayView)
    this.google = google
    this.bounds_ = new this.google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

class GoogleMapsPopover {
  constructor ({google, point, map}) {
    this = google.maps.OverlayView
    this.google = google
    this.bounds_ = new this.google.maps.LatLngBounds(point, point)
    this.map_ = map
    this.div_ = null
    this.setMap(map)
  }
}

export const getGoogleMapsPopoverClass = (google) => {
  return class GoogleMapsPopover extends google.maps.OverlayView {
    constructor ({point, map}) {
      super()
      this.bounds_ = new google.maps.LatLngBounds(point, point)
      this.map_ = map
      this.div_ = null
      this.setMap(map)
    }
  }
}

const GoogleMapsPopover = getGoogleMapsPopoverClass(google)

---

JS“类”看起来不太方便。为什么不使用

Object.create

？@evolutionxbox你能举个例子吗？到底是什么阻止你在理想世界中实现它？为什么你会“想把谷歌作为一个论据传进来”？不，将超类作为参数传递给构造函数是没有意义的。JS“classes”看起来不太方便。为什么不使用

Object.create

？@evolutionxbox你能举个例子吗？到底是什么阻止你在理想世界中实现它？为什么你会“想把谷歌作为一个论据传进来”？不，将超类作为参数传递给构造函数是没有意义的。