function versionCompare(version1, version2){
                var a = version1.split('.');
                var b = version2.split('.');
                for (var i = 0; i < a.length; ++i) {
                    a[i] = Number(a[i]);
                }
                for (var i = 0; i < b.length; ++i) {
                    b[i] = Number(b[i]);
                }
                var length=a.length;

                for(j=0; j<length; j++){
                    if(typeof b[j]=='undefined')b[j]=0;
                    if (a[j] > b[j]) return true;
                    else if(a[j] < b[j])return false;
                    if(j==length-1 && a[j] >= b[j])return true;
                }             

                return false;
            },

我喜欢@mar10的版本，尽管从我的角度来看，有误用的可能(如果版本与Semantic Versioning文档兼容，似乎不是这样，但如果使用了一些“构建号”，则可能是这样):

versionCompare( '1.09', '1.1');  // returns 1, which is wrong:  1.09 < 1.1
versionCompare('1.702', '1.8');  // returns 1, which is wrong: 1.702 < 1.8

这里的问题是，在某些情况下，版本号的子数字被删除了后面的零(至少我最近在使用不同的软件时看到的)，这类似于数字的有理数部分，因此:

5.17.2054 > 5.17.2
5.17.2 == 5.17.20 == 5.17.200 == ... 
5.17.2054 > 5.17.20
5.17.2054 > 5.17.200
5.17.2054 > 5.17.2000
5.17.2054 > 5.17.20000
5.17.2054 < 5.17.20001
5.17.2054 < 5.17.3
5.17.2054 < 5.17.30

但是，第一个(或第一个和第二个)版本子号始终被视为它实际等于的整数值。

如果你使用这种版本控制，你可以只改变例子中的几行:

// replace this:
p1 = parseInt(v1parts[i], 10);
p2 = parseInt(v2parts[i], 10);
// with this:
p1 = i/* > 0 */ ? parseFloat('0.' + v1parts[i], 10) : parseInt(v1parts[i], 10);
p2 = i/* > 0 */ ? parseFloat('0.' + v2parts[i], 10) : parseInt(v2parts[i], 10);

因此，除了第一个子数字外，每个子数字都将作为浮点数进行比较，因此09和1将相应地变成0.09和0.1，并以这种方式进行正确比较。2054和3将变成0.2054和0.3。

那么，完整的版本是(归功于@mar10):

/** Compare two dotted version strings (like '10.2.3').
 * @returns {Integer} 0: v1 == v2, -1: v1 < v2, 1: v1 > v2
 */
function versionCompare(v1, v2) {
    var v1parts = ("" + v1).split("."),
        v2parts = ("" + v2).split("."),
        minLength = Math.min(v1parts.length, v2parts.length),
        p1, p2, i;
    // Compare tuple pair-by-pair. 
    for(i = 0; i < minLength; i++) {
        // Convert to integer if possible, because "8" > "10".
        p1 = i/* > 0 */ ? parseFloat('0.' + v1parts[i], 10) : parseInt(v1parts[i], 10);;
        p2 = i/* > 0 */ ? parseFloat('0.' + v2parts[i], 10) : parseInt(v2parts[i], 10);
        if (isNaN(p1)){ p1 = v1parts[i]; } 
        if (isNaN(p2)){ p2 = v2parts[i]; } 
        if (p1 == p2) {
            continue;
        }else if (p1 > p2) {
            return 1;
        }else if (p1 < p2) {
            return -1;
        }
        // one operand is NaN
        return NaN;
    }
    // The longer tuple is always considered 'greater'
    if (v1parts.length === v2parts.length) {
        return 0;
    }
    return (v1parts.length < v2parts.length) ? -1 : 1;
}

注:这是比较慢的，但也可以考虑重用相同的比较函数来操作字符串实际上是字符数组的事实:

 function cmp_ver(arr1, arr2) {
     // fill the tail of the array with smaller length with zeroes, to make both array have the same length
     while (min_arr.length < max_arr.length) {
         min_arr[min_arr.lentgh] = '0';
     }
     // compare every element in arr1 with corresponding element from arr2, 
     // but pass them into the same function, so string '2054' will act as
     // ['2','0','5','4'] and string '19', in this case, will become ['1', '9', '0', '0']
     for (i: 0 -> max_length) {
         var res = cmp_ver(arr1[i], arr2[i]);
         if (res !== 0) return res;
     }
 }

这就是我的解。它已经在leetcode上被接受。我在今天的面试中遇到了问题。但我当时没有解决它。 我又想了想。加0使两个数组的长度相等。然后比较。

var compareVersion = function(version1, version2) { let arr1 = version1.split('.').map(Number); let arr2 = version2.split('.').map(Number); let diff = 0; if (arr1.length > arr2.length){ diff = arr1.length - arr2.length; while (diff > 0){ arr2.push(0); diff--; } } else if (arr1.length < arr2.length){ diff = arr2.length - arr1.length; while (diff > 0){ arr1.push(0); diff--; } } let i = 0; while (i < arr1.length){ if (arr1[i] > arr2[i]){ return 1; } else if (arr1[i] < arr2[i]){ return -1; } i++; } return 0; };

你可以使用JavaScript的localeCompare方法:

a.localeCompare(b, undefined， {numeric: true})

这里有一个例子:

"1.1".localeCompare("2.1.1"， undefined， {numeric: true}) => -1

"1.0.0".localeCompare("1.0"， undefined， {numeric: true}) =>

"1.0.0".localeCompare("1.0.0"， undefined， {numeric: true}) => 0

我写了一个排序版本的节点模块，你可以在这里找到它:version-sort

特点:

序列“1.0.1.5.53.54654.114.1.154.45”不受限制 不限制序列长度:'1.1546515465451654654654654138754431574364321353734'工作 可以根据版本对对象进行排序(参见README) 阶段(如alpha, beta, rc1, rc2)

如果您需要其他功能，请不要犹豫，打开一个问题。

你可以使用带有选项的String#localeCompare

sensitivity Which differences in the strings should lead to non-zero result values. Possible values are: "base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A. "accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A. "case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A. "variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A. The default is "variant" for usage "sort"; it's locale dependent for usage "search". numeric Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.

var版本=[" 2.0.1”、“2.0”、“1.0”、“1.0.1”,“2.0.0.1”); 版本。sort((a, b) => a.localeCompare(b, undefined， {numeric: true，灵敏度:'base'})); console.log(版本);