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我有一个原型，它有一个可以添加回调的方法：

/*
 * Add a callback function that is invoked on every element submitted and must return a data object.
 * May be used as well for transmitting static data.
 * 
 * The callback function is supposed to expect a jQuery element as single parameter 
 * and must return a data object (for additional data to be sent along with the one already given upon initialization).
 * Adding multiple callback functions results in those functions being invoked in the same order as they were added.
 * 
 * 1) Therefore subsequent assignments to the same key in the data array override those that were performed by a previous callback function.
 * 2) If data_arg = true is given, the data returned by the callback function that was previously called is given to the new_callback as 2nd argument, so it can be manipulated.
 *    However, even if it isn't, the unchanged data must be returned anyway to have any effect.
 */

this.add_data_callback = function(new_callback, data_arg) {

    if(this.data_callback) {
        old_callback = this.data_callback;

        if(!data_arg) {
            //alert('add it');
            //alert(old_callback);
            //alert(new_callback);

            this.data_callback = function(element) {
                //alert('do it');
                //alert(old_callback);
                //alert(new_callback);
                return jQuery.extend(old_callback(element), new_callback(element));
            };
        }
        else {
            this.data_callback = function(element, data) {
                return new_callback(element, old_callback(element));
            };
        }
    }
    else {
        //alert('add the first');
        //alert(new_callback);
        this.data_callback = new_callback;
    }
};

（请忽略 data_arg = true 的 else 部分，因为它不相关。）

在我的具体情况下，我添加了三个回调函数。但是，如果最终为元素调用 this.data_callback() ，则会导致整个事物无限循环。在我试图追踪错误的过程中，上面的警报（是的，我知道有调试工具，但这样更舒服）对问题提供了以下见解：

  1. 匿名函数/闭包（包含 jQuery.extend() 的那个）被调用。new_callback 包含第三个回调函数。old_callback 包含另一个...
  2. 被调用的匿名函数。new_callback 包含第二个回调函数。old_callback 应该包含第一个回调函数，但实际上它是另一个......
  3. 被调用的匿名回调。new_callback 再次包含第二个回调函数。old_callback 包含匿名函数
  4. ...

现在，我想念什么？是一些奇怪的关闭魔法还是只是一些我显然看不到的明显错误？

提前致谢！

4

1 回答 1

2

为什么没有old_callback用 定义var？

var old_callback = this.data_callback;

事实上，它是一个全局变量。也许它已在其他地方宣布，但对我来说仍然很可疑。

于 2010-03-29T13:46:23.797 回答