Javascript jQuery函数交互

我的编码有问题，请检查以下代码

var sliderFunction = {
  init: function(sliderSettings) {
    sliderFunction.sliderSettings();
  },
  sliderSettings: function() {
    var sliderDiv = $('.slider');
    var ulDiv = $('.slider ul');
    var liDiv = $('.slider ul li');

    var sliderWidth = sliderDiv.width();
    var countLi = liDiv.length;
    var ulTotalWidth = sliderWidth * countLi;

    ulDiv.outerWidth(ulTotalWidth);
  },
  nextContent: function() {
  }
}

sliderFunction.init();

我想在nextContent函数中访问sliderSettings函数的var。我该怎么做

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提前感谢

在全局范围内定义它们，它们将随时可访问

 var sliderDiv,ulDiv,liDiv,sliderWidth,countLi,ulTotalWidth,ulDiv;
var sliderFunction = {
  init: function(sliderSettings) {
  sliderFunction.sliderSettings();
},
sliderSettings: function() {
 sliderDiv = $('.slider');
 ulDiv = $('.slider ul');
 liDiv = $('.slider ul li');

 sliderWidth = sliderDiv.width();
 countLi = liDiv.length;
 ulTotalWidth = sliderWidth * countLi;

 ulDiv.outerWidth(ulTotalWidth);
},

nextContent:function(){
    console.log(sliderDiv,ulDiv,liDiv,sliderWidth,countLi,ulTotalWidth,ulDiv);
}
}

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你不能，范围问题。