Question

在下面的代码中，

body.on(event, bindSelector, callback);

我将回调作为对此函数(kit.callback.renderPage)的引用传递。

我希望(kit.callback.renderPage)也接受参数。如何从下面的init()函数传递一个对象作为参数？

jQuery(document).ready(function($) { 

    (function(send, config) {

        var kit = {
            selectors: {
                "body": "body",
                "start": "[data-ckit-toggle-on]"
            },

            callback: {

                 renderPage : function() {
                    console.debug('opt');
                 }
            },


            getSelectors: function(selector) {
                return $(selector);     
            },

            init: function() {

                kit.bindNow(
                    kit.events.chatToolbar,
                    kit.selectors.body, 
                    kit.selectors.start, 
                    kit.callback.renderPage
                );


            },  

            bindNow: function(event, bindTo, bindSelector, callback) {
                var body = kit.getSelectors(bindTo);
                body.on(event, bindSelector, callback);
            }
        };

        //starting point
        kit.init(); 
    })(interface, settings);
});

Answer 1

您可以将回调包装在内联/匿名函数中，例如：

bindNow: function(event, bindTo, bindSelector, callback) {
            var body = kit.getSelectors(bindTo);
            body.on(event, bindSelector, function() { 
                callback.call(this, param1, param2);
            });
         }

或

kit.bindNow(
            kit.events.chatToolbar,
            kit.selectors.body, 
            kit.selectors.start, 
            function() { 
                kit.callback.renderPage.call(this, param1, param2);
            }
        );

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Function/call

如何将参数传递给作为引用传递的函数？

1 个答案: