javascript - localeCompare区分大小写的字母顺序问题

Question

我正在尝试在这里实现 aAbBcC。我尝试了 en-US 希望得到想要的结果。在“Acuityads”之前期待“alpha”。

var array = [{name:"Acuityads"},{name:"alpha"}];

console.log(array.sort(function(a,b){return a.name.localeCompare(b.name, 
'en-US-u-kf-lower'); }));


console.log(array.sort(function(a,b){return a.name.localeCompare(b.name, 
'en-US'); }));

"alpha".localeCompare("Acuityads", 'en-US') // output as 1

Answer 1

讨论 in the comments ，这看起来像是规范或 vendor 实现中的不一致。
自 an earlier question on the same topic没有任何我很乐意接受的答案，这里是 a manual implementation :

const input = ["aaaaa", "aaa", "aa", "aaaa", "aA", "Aa", "AA", "ab", "aB", "Ab", "AB"];

function caseSensitiveCompare(a, b) {
  // Sort character by character, return early if possible
  for (let ii = 0; ii < Math.max(a.length, b.length); ii++) {
    const aChar = a.charAt(ii);
    const bChar = b.charAt(ii);

    // If inputs match up to here, but lengths don't match, sort by length
    if (!(aChar && bChar)) {
      return a.length - b.length;
    }

    // If we meet a differing character, return early
    const comp = aChar.localeCompare(bChar);
    if (comp !== 0) {
      return comp;
    }
  }
  // If we found nothing to do, the strings are equal
  return 0;
}

console.log(input.sort(caseSensitiveCompare));