Google Map 类实例在类式继承中的实现
众所周知,程序的实现不可能会是完美的。下面是google Map类在继承实现的写法。首先是照抄《JavaScript设计模式》中的类式继承:
function extend(subClass, superClass) { function F() {} F.prototype = superClass.prototype; subClass.prototype = new F(); subClass.prototype.constructor = subClass; subClass.superclass = superClass.prototype; if (superClass.prototype.constructor == Object.prototype.constructor) { superClass.prototype.constructor = superClass; } } function SubMap(elm, config) { console.log(SubMap.superclass.constructor); SubMap.superclass.constructor.call(this, elm, config); } extend(SubMap, google.maps.Map); var mapObj1 = new SubMap($('#map_Box')[0], { zoom: 13, center: new google.maps.LatLng(31.227, 121.519), mapTypeId: google.maps.MapTypeId.ROADMAP });
上面的google地图在页面中显示不了,说明代码是有问题的。下面是对上面实现的改写:
function extend(subClass, superClass) { function F() {} F.prototype = superClass.prototype; subClass.prototype = new F(); subClass.prototype.constructor = subClass; subClass.superclass = superClass; } function SubMap(elm, config) { SubMap.superclass.call(this, elm, config); } extend(SubMap, google.maps.Map); var mapObj1 = new SubMap($('#map_Box')[0], { zoom: 13, center: new google.maps.LatLng(31.227, 121.519), mapTypeId: google.maps.MapTypeId.ROADMAP });
嗯,这种继承实现的写法google地图在页面中显示得很好。
(完)