使用today.getDay()的代码仅周日显示Undefined的原因咨询

正常运行的代码示例（指定周六日期） #

weeks=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"];
var today = new Date("September 24, 2022 01:15:00"); 
document.getElementById("current-day").innerHTML=weeks[today.getDay()-1];

<h1 id="current-day"></h1>

这段代码在周六（2022年9月24日）及除周日外的其他日期都能正常显示对应星期，但到周日就会输出Undefined，下面是周日的测试代码：

周日测试代码 #

weeks=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"];
var today = new Date("September 25, 2022 01:15:00");
document.getElementById("current-day").innerHTML=weeks[today.getDay()-1];

<h1 id="current-day"></h1>

问题根源 #

问题出在JavaScript的getDay()方法上：

• getDay()返回0到6的数字，0对应周日，1对应周一，以此类推，6对应周六
• 你的weeks数组是从Monday（周一）开始的，索引0是周一、索引6是周日
• 当日期是周日时，getDay()返回0，执行0-1=-1，数组取-1索引会返回undefined（JavaScript数组不支持负索引访问有效元素）

修复方案 #

有几种简单的修复方式：

1. 调整数组顺序：把周日放在数组第一个位置，让索引和getDay()返回值对应

weeks=["Sunday","Monday","Tuesday","Wednesday","Thursday","Friday","Saturday"];
var today = new Date("September 25, 2022 01:15:00");
document.getElementById("current-day").innerHTML=weeks[today.getDay()];

2. 处理索引计算逻辑：当getDay()返回0时，让它对应数组的最后一个索引（6）

weeks=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"];
var today = new Date("September 25, 2022 01:15:00");
// 用(today.getDay() || 7)把0转换成7，再减1得到6，对应数组里的周日
document.getElementById("current-day").innerHTML=weeks[(today.getDay() || 7) - 1];

3. 使用模运算：通过模7来处理负数索引的问题

weeks=["Monday","Tuesday","Wednesday","Thursday","Friday","Saturday","Sunday"];
var today = new Date("September 25, 2022 01:15:00");
document.getElementById("current-day").innerHTML=weeks[(today.getDay() - 1 + 7) % 7];

内容的提问来源于stack exchange，提问作者Haruu