Question

真的很难达到我想要的效果。我基本上需要创建一个对象链接，这些对象将包含对HTMLElements的引用。在此对象中更改状态值时，应添加/删除类。

但是，当我创建此对象的副本时，我需要保持指向此状态的链接，因此所有其他属性都是唯一的，并且在此副本中设置了新图像，但是如果状态为要更改第一个对象，那么它也应该在这个新对象中更改，但引用新图像。

我有以下脚本示例，但是当在第一个对象中更改状态时，第二个对象不会更改。我相信我没有保留对旧对象属性的引用。

<img src="https://mgreviewsblog.files.wordpress.com/2014/08/2014-ducati-899-panigale-white-16.jpg" alt="Image 1" class="responsive-img" data-lightbox-group="detail_gallery" data-lightbox-name="Gallery">
<img src="https://mgreviewsblog.files.wordpress.com/2014/08/2014-ducati-899-panigale-white-1.jpg" alt="Image 2" class="responsive-img" data-lightbox-group="detail_gallery" data-lightbox-name="Gallery">
<img src="https://mgreviewsblog.files.wordpress.com/2014/08/2014-ducati-899-panigale-white-3.jpg" alt="Image 3" class="responsive-img" data-lightbox-group="detail_gallery" data-lightbox-name="Gallery">

<script>
   let images = querySelectorAll('img');

   images = Array.from(images, function (elm) {
      var activeState = {
         _state: 0,
         object: null,
         set State(val) {
            if (this._state !== val) {
            switch(val) {
               case 0:
                  elm.classList.remove('active');
               break;
               case 1:
                  this.object.classList.add('active'); // Needs to reference image within object
               break;
            }
            this._state = val;
            };
         }
      };

      return {
         object: elm, // Store a reference to the image
         _active: activeState,
         get activeState() {
            this._active.object = this.object;
            return this._active;
         }
      };
   });


   // Duplicate the above object, but retain the state object,
   // however the object within the state needs to be different. 
   let clone = [];
   for (let i = 0; i < _.images.length; i++) {
      var img = document.createElement('img');
      img.src = _.images[i].object.src;
      img.alt = _.images[i].alt;

      newList.push({
         object: img,
         index: i,
         src: _.images[i].src,
         alt: _.images[i].alt,
         _active: _.images[i].activeState, // Retain reference to active state across objects
         get activeState() {
            this._active.object = this.object;
            return this._active;
         }
      });
   }
</script>

修改简化版：

var test = {
    object: 'test1', 
    _active: { val: 0 },
    active: function(x) {
        this._active.val = x;
        console.log(this._active);
    }
  };
var test2 = {
    object: 'test1', 
    _active: test._value,
    active: test.value
  };

_active.val仍然在对象之间保持链接，但如果我更改它test2.active(2)只执行 test2 值， test 应该执行为好。

Answer 1

我认为你的整个代码过于复杂。我会这样做：

function statify(el,state){
  var obj={
   get state(){
     return state;
   },
   set state(v){
    state=v;
    if(v){
      el.classList.add("active");
    }else{
      el.classList.remove("active");
    }
  },
  get el(){return el;},
  cloneTo(el){
   return statify(el,state);
  }
 };
 obj.state=state;
 return obj;
}

所以你可以像这样使用它：

var states=Array.from(images).map(image=>statify(image));

states[0].state=1;

var clone=states.map(function(state){
 var el=state.el;
 var img=document.createElement("img");
 img.src=el.src;
 return state.cloneTo(img);
});

http://jsbin.com/tuhehihade/edit?console

复制对象，保留一个属性链接

1 个答案: