javascript - 如何避免这个很长的 if else 语句

Question

我仍在学习 javascript，并且一直在寻找使我的代码更好的方法。我写了这个很长的if else 语句。我想知道是否有更简单的方法来实现这一点。此代码使用jquery transit为旋转设置动画。基本上，当用户单击链接时，导航会将单击的链接旋转到顶部位置。导航定位在一个圆圈中。

var rotation = 0;
$('nav ul li a').bind('click', function () {
    'use strict';
    var item = $(this).parent();
    var itemI = item.index();
    var navAll = $(this).parents('ul');
    var pos0 = (rotation === 0);
    var pos1 = (rotation === 90);
    var pos2 = (rotation === 180);
    var pos3 = (rotation === 270);
    var item0 = (itemI === 0);
    var item1 = (itemI === 1);
    var item2 = (itemI === 2);
    var item3 = (itemI === 3);
    //------------------pos0 = 0----------------------------
    if (pos0 && item0) {
        //do nothing 0
    } else if (pos0 && item1) {
        navAll.transition({
            rotate: '-=90deg'
        });
        $('nav ul li a').transition({
            rotate: '90deg'
        });
        rotation = 270;
    } else if (pos0 && item2) {
        navAll.transition({
            rotate: '-=180deg'
        });
        $('nav ul li a').transition({
            rotate: '180deg'
        });
        rotation = 180;
    } else if (pos0 && item3) {
        navAll.transition({
            rotate: '+=90deg'
        });
        $('nav ul li a').transition({
            rotate: '-90deg'
        });
        rotation = 90;
    }
    //------------------pos1 = 90----------------------------
    else if (pos1 && item0) {
        navAll.transition({
            rotate: '0deg'
        });
        $('nav ul li a').transition({
            rotate: '0deg'
        });
        rotation = 0;
    } else if (pos1 && item1) {
        navAll.transition({
            rotate: '-90deg'
        });
        $('nav ul li a').transition({
            rotate: '90deg'
        });
        rotation = 270;
    } else if (pos1 && item2) {
        navAll.transition({
            rotate: '+=90deg'
        });
        $('nav ul li a').transition({
            rotate: '-180deg'
        });
        rotation = 180;
    } else if (pos1 && item3) {
        //do nothing 90
    }
    //------------------pos2 = 180----------------------------
    else if (pos2 && item0) {
        navAll.transition({
            rotate: '0deg'
        });
        $('nav ul li a').transition({
            rotate: '0deg'
        });
        rotation = 0;
    } else if (pos2 && item1) {
        navAll.transition({
            rotate: '-90deg'
        });
        $('nav ul li a').transition({
            rotate: '90deg'
        });
        rotation = 270;
    } else if (pos2 && item2) {
        //do nothing
        rotation = 180;
    } else if (pos2 && item3) {
        navAll.transition({
            rotate: '-=90deg'
        });
        $('nav ul li a').transition({
            rotate: '-90deg'
        });
        rotation = 90;
    }
    //------------------pos3 = 270----------------------------
    else if (pos3 && item0) {
        navAll.transition({
            rotate: '0deg'
        });
        $('nav ul li a').transition({
            rotate: '0deg'
        });
        rotation = 0;
    } else if (pos3 && item1) {
        //do nothing
    } else if (pos3 && item2) {
        navAll.transition({
            rotate: '-=90deg'
        });
        $('nav ul li a').transition({
            rotate: '180deg'
        });
        rotation = 180;
    } else if (pos3 && item3) {
        navAll.transition({
            rotate: '-=180deg'
        });
        $('nav ul li a').transition({
            rotate: '-90deg'
        });
        rotation = 90;
    }
});

Answer 1

你可以使用开关

现场演示

condition = rotation + "_" + itemI 

switch(condition)
{
   case "0_0":
        //your code
        break;
   case "90_0":
        //your code
        break;

   //Similarly you can add all cases.
}

Answer 2

这可能会缩短代码。

创建一个二维数组，其中包含旋转键和 itemI 以及结果 deg val 的组合值

然后只需访问数组并按如下方式运行您的命令

if(degVal[rotation][itemI]<0)
    rotateSign = '-';
else
    rotateSign = '+';

navAll.transition({rotate: rotateSign+'='+degVal[rotation][itemI]+'deg'});
$('nav ul li a').transition({rotate: degVal[rotation][itemI]+'deg'});
rotation = degVal[rotation][itemI];