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Topic source拉了一泡屎回来,发现应该可以省略一次遍历,下面是改进版:
// sort list:
let thatOl = document.getElementById('test-list');
[].slice.call(thatOl.querySelectorAll('li')).sort(function(x, y){
x.innerText > y.innerText && (thatOl.insertBefore(y, x));
});
// sort list: var langList = document.getElementById('test-list'); var langs = document.getElementsByClassName('lang'); var i; var arr = []; for (i = 0; i < langs.length; i++) { arr.push(langs[i]); } arr .sort(function(x, y) { var s1 = x.innerText; var s2 = y.innerText; if (s1 < s2) { return -1; } else if(s1 > s2) { return 1; } else { return 0; } });
for (i = 0; i < arr.length; i++) { langList.appendChild(arr[i]); }
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