JavaScript localeCompare 排序不好 [ "aaa-1.0"、 "aaa-1.1"、 "aaa-1.01"、 "aaa-11","aaa-2.01"]

Question

我想按字符串的“十进制”数值以及字母顺序对字符串进行排序。保持自然顺序。

var strArray = ["aaa-0", "aaa-0.01", "aaa-1.0", "aaa-1.1", "aaa-1.01", "aaa-2.01", "bbb-2.01", "aaa-11", "aaa-2.01"]

strArray.sort(function (a, b) {
    return a.localeCompare(b,undefined, {numeric: true, sensitivity: 'base'});
});

返回:

["aaa-0", "aaa-0.01", "aaa-1.0", "aaa-1.1", "aaa-1.01", "aaa-2.01", "aaa-2.01", "aaa-11", "bbb-2.01"]

虽然我想要:

["aaa-0", "aaa-0.01", "aaa-1.0", "aaa-1.01", "aaa-1.1", "aaa-2.01", "aaa-2.01", "aaa-11", "bbb-2.01"]

如何实现这一点？即使我使用“，”而不是“。”排序是一样的。

Answer 1

您需要分别对数字和字母进行排序。

演示

var strArray = ["aaa-0", "aaa-0.01", "aaa-1.0", "aaa-1.1", "aaa-1.01", "aaa-2.01", "bbb-2.01", "aaa-11", "aaa-2.01"]

strArray.sort(function(a, b) {
  var splitA = a.split("-");
  var splitB = b.split("-");
  if (splitA[0] == splitB[0]) {
    return (Number(splitA[1]) - Number(splitB[1]));
  } else {
    return splitA[0].localeCompare(splitB[0], undefined, {
      numeric: true,
      sensitivity: 'base'
    });
  }
});
console.log(strArray);