我试图了解属性在 ExtJs 类中是如何工作的。参考以下代码:
Ext.define('My.sample.Person', {
name: 'Unknown',
food: undefined,
foodList: [],
constructor: function(name) {
if (name) {
this.name = name;
}
},
eat: function(foodType) {
console.log(this.name + " is eating: " + foodType);
this.food = foodType;
this.foodList.push(foodType);
//this.foodList = [foodType]
},
showFood:function() {
console.log(this.name);
console.log(this.food);
},
showFoodList:function() {
console.log(this.name);
console.log(this.foodList);
}
});
var bob = Ext.create('My.sample.Person', 'Bob');
bob.eat("Salad");
bob.showFood();
bob.showFoodList();
console.log(bob)
var alan = Ext.create('My.sample.Person', 'alan');
console.log(alan)
alan.showFood();
alan.showFoodList();
<script src="https://cdnjs.cloudflare.com/ajax/libs/extjs/6.2.0/ext-all.js"></script>
如果您检查“alan”的结果,则 food = undefined 和 foodList = ['salad'] 因为以某种方式将 foodList 分配给了原型。
同时,如果您这样做,那么它将正常运行。知道为什么吗?背后的概念是什么?
结果: