我也遇到了版本比较的问题，但是版本可能包含任何内容(例如:不是点的分隔符，像rc1, rc2…)

我使用了这个方法，它基本上将版本字符串分为数字和非数字，并尝试根据类型进行比较。

function versionCompare(a,b) {
  av = a.match(/([0-9]+|[^0-9]+)/g)
  bv = b.match(/([0-9]+|[^0-9]+)/g)
  for (;;) {
    ia = av.shift();
    ib = bv.shift();
    if ( (typeof ia === 'undefined') && (typeof ib === 'undefined') ) { return 0; }
    if (typeof ia === 'undefined') { ia = '' }
    if (typeof ib === 'undefined') { ib = '' }

    ian = parseInt(ia);
    ibn = parseInt(ib);
    if ( isNaN(ian) || isNaN(ibn) ) {
      // non-numeric comparison
      if (ia < ib) { return -1;}
      if (ia > ib) { return 1;}
    } else {
      if (ian < ibn) { return -1;}
      if (ian > ibn) { return 1;}
    }
  }
}

对于某些情况，这里有一些假设，例如:"1.01" === "1.1"，或"1.8" < "1.71"。它无法管理“1.0.0-rc”。1" < "1.0.0"，由语义版本2.0.0指定

2017答:

v1 = '20.0.12'; 
v2 = '3.123.12';

compareVersions(v1,v2) 
// return positive: v1 > v2, zero:v1 == v2, negative: v1 < v2 
function compareVersions(v1, v2) {
        v1= v1.split('.')
        v2= v2.split('.')
        var len = Math.max(v1.length,v2.length)
        /*default is true*/
        for( let i=0; i < len; i++)
            v1 = Number(v1[i] || 0);
            v2 = Number(v2[i] || 0);
            if (v1 !== v2) return v1 - v2 ;
            i++;
        }
        return 0;
    }

最简单的现代浏览器代码:

 function compareVersion2(ver1, ver2) {
      ver1 = ver1.split('.').map( s => s.padStart(10) ).join('.');
      ver2 = ver2.split('.').map( s => s.padStart(10) ).join('.');
      return ver1 <= ver2;
 }

这里的想法是比较数字，但以字符串的形式。为了使比较工作，两个字符串必须在相同的长度。所以:

"123" > "99"变成"123" > "099" 填充短数字“修复”比较

这里我用0填充每个部分，长度为10。然后使用简单的字符串比较来得到答案

例子:

var ver1 = '0.2.10', ver2=`0.10.2`
//become 
ver1 = '0000000000.0000000002.0000000010'
ver2 = '0000000000.0000000010.0000000002'
// then it easy to see that
ver1 <= ver2 // true

这个非常小，但非常快的比较函数接受每个段的任何长度和任何数字大小的版本号。

返回值: -如果a < b，则数字< 0 -如果是> b，则为> 0 如果a = b - 0

所以你可以使用它作为array。sort()的比较函数;

编辑:修正了版本剥离尾随零识别“1”和“1.0.0”相等的错误

function cmpVersions (a, b) { var i, diff; var regExStrip0 = /(\.0+)+$/; var segmentsA = a.replace(regExStrip0, '').split('.'); var segmentsB = b.replace(regExStrip0, '').split('.'); var l = Math.min(segmentsA.length, segmentsB.length); for (i = 0; i < l; i++) { diff = parseInt(segmentsA[i], 10) - parseInt(segmentsB[i], 10); if (diff) { return diff; } } return segmentsA.length - segmentsB.length; } // TEST console.log( ['2.5.10.4159', '1.0.0', '0.5', '0.4.1', '1', '1.1', '0.0.0', '2.5.0', '2', '0.0', '2.5.10', '10.5', '1.25.4', '1.2.15'].sort(cmpVersions)); // Result: // ["0.0.0", "0.0", "0.4.1", "0.5", "1.0.0", "1", "1.1", "1.2.15", "1.25.4", "2", "2.5.0", "2.5.10", "2.5.10.4159", "10.5"]

下面是另一个简短的版本，适用于任何数量的子版本，填充零和偶数字母(1.0.0b3)

const compareVer = ((prep, repl) =>
{
  prep = t => ("" + t)
      //treat non-numerical characters as lower version
      //replacing them with a negative number based on charcode of first character
    .replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".")
      //remove trailing "." and "0" if followed by non-numerical characters (1.0.0b);
    .replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/g, "$1$2")
    .split('.');

  return (a, b, c, i, r) =>
  {
    a = prep(a);
    b = prep(b);
    for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;)
    {
      r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]);
    }
    return r;
  }
})();

函数返回:

如果a = b则为0

1如果a > b

-1如果a < b

1.0         = 1.0.0.0.0.0
1.0         < 1.0.1
1.0b1       < 1.0
1.0b        = 1.0b
1.1         > 1.0.1b
1.1alpha    < 1.1beta
1.1rc1      > 1.1beta
1.1rc1      < 1.1rc2
1.1.0a1     < 1.1a2
1.1.0a10    > 1.1.0a1
1.1.0alpha  = 1.1a
1.1.0alpha2 < 1.1b1
1.0001      > 1.00000.1.0.0.0.01

/*use strict*/ const compareVer = ((prep, repl) => { prep = t => ("" + t) //treat non-numerical characters as lower version //replacing them with a negative number based on charcode of first character .replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".") //remove trailing "." and "0" if followed by non-numerical characters (1.0.0b); .replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/g, "$1$2") .split('.'); return (a, b, c, i, r) => { a = prep(a); b = prep(b); for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;) { r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]); } return r; } })(); //examples let list = [ ["1.0", "1.0.0.0.0.0"], ["1.0", "1.0.1"], ["1.0b1", "1.0"], ["1.0b", "1.0b"], ["1.1", "1.0.1b"], ["1.1alpha", "1.1beta"], ["1.1rc1", "1.1beta"], ["1.1rc1", "1.1rc2"], ["1.1.0a1", "1.1a2"], ["1.1.0a10", "1.1.0a1"], ["1.1.0alpha", "1.1a"], ["1.1.0alpha2", "1.1b1"], ["1.0001", "1.00000.1.0.0.0.01"] ] for(let i = 0; i < list.length; i++) { console.log( list[i][0] + " " + "<=>"[compareVer(list[i][0], list[i][1]) + 1] + " " + list[i][1] ); }

https://jsfiddle.net/vanowm/p7uvtbor/

我也遇到过类似的问题，而且我已经为它创建了一个解决方案。你可以试一试。

如果等于则返回0，如果版本号大于则返回1，如果版本号小于则返回-1

function compareVersion(currentVersion, minVersion) { let current = currentVersion.replace(/\./g," .").split(' ').map(x=>parseFloat(x,10)) let min = minVersion.replace(/\./g," .").split(' ').map(x=>parseFloat(x,10)) for(let i = 0; i < Math.max(current.length, min.length); i++) { if((current[i] || 0) < (min[i] || 0)) { return -1 } else if ((current[i] || 0) > (min[i] || 0)) { return 1 } } return 0 } console.log(compareVersion("81.0.1212.121","80.4.1121.121")); console.log(compareVersion("81.0.1212.121","80.4.9921.121")); console.log(compareVersion("80.0.1212.121","80.4.9921.121")); console.log(compareVersion("4.4.0","4.4.1")); console.log(compareVersion("5.24","5.2")); console.log(compareVersion("4.1","4.1.2")); console.log(compareVersion("4.1.2","4.1")); console.log(compareVersion("4.4.4.4","4.4.4.4.4")); console.log(compareVersion("4.4.4.4.4.4","4.4.4.4.4")); console.log(compareVersion("0","1")); console.log(compareVersion("1","1")); console.log(compareVersion("1","1.0.00000.0000")); console.log(compareVersion("","1")); console.log(compareVersion("10.0.1","10.1"));

我根据Kons的想法做了这个，并针对Java版本“1.7.0_45”进行了优化。它只是一个将版本字符串转换为浮点数的函数。这是函数:

function parseVersionFloat(versionString) {
    var versionArray = ("" + versionString)
            .replace("_", ".")
            .replace(/[^0-9.]/g, "")
            .split("."),
        sum = 0;
    for (var i = 0; i < versionArray.length; ++i) {
        sum += Number(versionArray[i]) / Math.pow(10, i * 3);
    }
    console.log(versionString + " -> " + sum);
    return sum;
}

字符串“1.7.0_45”被转换为1.0070000450000001，这足以进行正常的比较。这里解释的错误:如何处理JavaScript中的浮点数精度?如果需要超过3个数字在任何部分，你可以改变除法数学。Pow (10, I * 3);;

输出如下所示:

1.7.0_45         > 1.007000045
ver 1.7.build_45 > 1.007000045
1.234.567.890    > 1.23456789