如何在javascript中对混合的数字/字母数字数组进行排序 [英] how to sort mixed numeric/alphanumeric array in javascript

问题描述

我有一个混合数组，需要按数字，字母然后按数字排序-

I have a mixed array that I need to sort by number, alphabet and then by digit-

['A1', 'A10', 'A11', 'A12', 'A3A', 'A3B', 'A3', 'A4', 'B10', 'B2', 'F1', '1', '2', 'F3']

我如何将其排序为:

['1', '2', 'A1', 'A2', 'A3', 'A3A', 'A3B', 'A4', 'A10', 'A11', 'A12', 'B2', 'B10', 'F1', 'F3']

这是我尝试过的:

var reA = /[^a-zA-Z]/g;
var reN = /[^0-9]/g;
function sortAlphaNum(a, b) {
    var AInt = parseInt(a.Field, 10);
    var BInt = parseInt(b.Field, 10);

    if (isNaN(AInt) && isNaN(BInt)) {
        var aA = (a.Field).replace(reA, "");
        var bA = (b.Field).replace(reA, "");
        if (aA === bA) {
            var aN = parseInt((a.Field).replace(reN, ""), 10);
            var bN = parseInt((b.Field).replace(reN, ""), 10);
            return aN === bN ? 0 : aN > bN ? 1 : -1;
        } else {
            return aA > bA ? 1 : -1;
        }
    } else if (isNaN(AInt)) {//A is not an Int
        return 1;//to make alphanumeric sort first return -1 here
    } else if (isNaN(BInt)) {//B is not an Int
        return -1;//to make alphanumeric sort first return 1 here
    } else {
        return AInt > BInt ? 1 : -1;
    }
}

fieldselecteddata.sort(sortAlphaNum);

，但是只能按字母/数字排序，直到1个数字和1个字符(如 A1 ， A2 ， A10 )组合为止.但是，在这种情况下，如果存在诸如 A3A ， A3B 之类的值，它将无法正确排序.可以使用直接的JavaScript或jQuery完成此操作吗?

but that only sorts it alphabetically/numeric till combination of 1 numeric and 1 character like A1, A2, A10. But if there will be values like A3A, A3B in that case it wont sort properly. Can this be done with either straight JavaScript or jQuery?

推荐答案

var arr = ['A1', 'A10', 'A11', 'A12', 'A3A', 'A3B', 'A3', 'A4', 'B10', 'B2', 'F1', '1', '2', 'F3'];

// regular expression to get the alphabetic and the number parts, if any
var regex = /^([a-z]*)(\d*)/i;

function sortFn(a, b) {
  var _a = a.match(regex);
  var _b = b.match(regex);

  // if the alphabetic part of a is less than that of b => -1
  if (_a[1] < _b[1]) return -1;
  // if the alphabetic part of a is greater than that of b => 1
  if (_a[1] > _b[1]) return 1;

  // if the alphabetic parts are equal, check the number parts
  var _n = parseInt(_a[2]) - parseInt(_b[2]);
  if(_n == 0) // if the number parts are equal start a recursive test on the rest
      return sortFn(a.substr(_a[0].length), b.substr(_b[0].length));
  // else, just sort using the numbers parts
  return _n;
}

console.log(arr.sort(sortFn));

注意:正则表达式(/.../i )中的 i 修饰符表示不区分大小写(同时查找小写字母和大写字母).

Note: the i modifier in the regular expression (/.../i) means case-insensitive (looks for both lowercases and uppercases).

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