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 期望得到理想的结果。 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

正如评论中所讨论的，这看起来像是规范或供应商实现中的不一致。
由于先前关于同一主题的问题没有任何答案，我很乐意 go 使用，这里是一个手动实现：

 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));