评论区都看了一下，应该还是map遍历方法最简单

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#1 Created at ... [Delete] [Delete and Lock User]

var arr_list=Array.from(document.getElementById('test-list').children)

var arr_lang=Array.from(document.getElementsByClassName("lang"),item=>item.innerText)

arr_lang.sort();

arr_list.map((x,i)=>{x.innerText=arr_lang[i]})

//Array.from可以用扩展运算符[...]代替

#2 Created at ... [Delete] [Delete and Lock User]

太强了你，感觉每章都可以看到你

#3 Created at ... [Delete] [Delete and Lock User]

强无敌啊！

#4 Created at ... [Delete] [Delete and Lock User]

大佬，这个map的arr_list.map((x,i)=>{x.innerText=arr_lang[i]}) 看不懂了

应该去看哪一部分的内容啊...

#5 Created at ... [Delete] [Delete and Lock User]

去看map啊

#6 Created at ... [Delete] [Delete and Lock User]

这样结果会出现class名称和textContent不对应的情况吧🤯

#7 Created at ... [Delete] [Delete and Lock User]

arr_list.map((x,i)=>{x.innerText=arr_lang[i]})

这里的i为什么能自增啊

#8 Created at ... [Delete] [Delete and Lock User]

arr_list.map((x,i)=>{x.innerText=arr_lang[i]})

懂了，回调函数可以有三个参数，第二个参数是数组中正在处理的当前元素的索引。