Jquery插件将属性传递给另一个对象并不起作用

Question

(function( $, window ) {

    $.fn.slideInterval = function( options ) {


    return this.each(function() {

        $(this).css({overflow: "hidden"})

        var slider = {
            self: this,
        init:function( options, elem ) {                
            self.elem = elem;
            self.$elem = $( elem );
            slider.action()

        },
        action:function() {
            slider.moved.call(slider.self)
        },

        moved:function() {

                console.log(options.fade) // options.fade is undefined

        }

    }

    slider.init( options, this );
    });
 }; 

    $.fn.slideInterval.options = function() {
    var options = $.extend({
        fade: 5,
        ImgSlide: null,
        interval: null,
        bullet: 'default',
        context: 'Your Title'
    })( options );

    };


})( jQuery, window);

我有一些问题，当我在slider方法中调用options.fade但不起作用。如何在滑块对象方法中获取选项对象属性..... ??

............................................... .................................................. .................................................. ......

Answer 1

我认为您希望在调用options时扩展传递的slideInterval对象，但我不确定，因为您没有正确解释您的代码和问题：

(function ($, window) {

    $.fn.slideInterval = function(options) {
        // argument default options with passed options
        options = $.extend({}, $.fn.slideInterval.options, options);

        return this.each(function () {
            // ...
        });
    };

    $.fn.slideInterval.options = {
        fade: 5,
        ImgSlide: null,
        interval: null,
        bullet: 'default',
        context: 'Your Title'
    };

})(jQuery, window);