以下是软件版本号:

"1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"

我怎么比较呢?

假设正确的顺序是:

"1.0", "1.0.1", "2.0", "2.0.0.1", "2.0.1"

想法很简单… 读第一个数字,然后,第二个,第三个… 但是我不能将版本号转换为浮点数… 你也可以像这样看到版本号:

"1.0.0.0", "1.0.1.0", "2.0.0.0", "2.0.0.1", "2.0.1.0"

这样可以更清楚地看到背后的想法。 但是,我怎样才能把它转换成计算机程序呢?


当前回答

我找到了最简单的方法来比较它们,但我不确定这是否是你想要的。

当我在控制台中运行下面的代码时,它是有意义的,并且使用sort()方法,我可以获得版本字符串的排序数组。它是根据字母顺序排列的。

"1.0" < "1.0.1" //true
var arr = ["1.0.1", "1.0", "3.2.0", "1.3"]
arr.sort();     //["1.0", "1.0.1", "1.3", "3.2.0"]

其他回答

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。

这是另一种递归算法。

这段代码只使用了Array。shift和递归,这意味着它可以在Internet Explorer 6+中运行。如果你有任何疑问,你可以访问我的GitHub页面。

(function(root, factory) {
  if (typeof exports === 'object') {
    return module.exports = factory();
  } else if (typeof define === 'function' && define.amd) {
    return define(factory);
  } else {
    return root.compareVer = factory();
  }
})(this, function() {
  'use strict';
  var _compareVer;
  _compareVer = function(newVer, oldVer) {
    var VER_RE, compareNum, isTrue, maxLen, newArr, newLen, newMatch, oldArr, oldLen, oldMatch, zerofill;
    VER_RE = /(\d+\.){1,9}\d+/;
    if (arguments.length !== 2) {
      return -100;
    }
    if (typeof newVer !== 'string') {
      return -2;
    }
    if (typeof oldVer !== 'string') {
      return -3;
    }
    newMatch = newVer.match(VER_RE);
    if (!newMatch || newMatch[0] !== newVer) {
      return -4;
    }
    oldMatch = oldVer.match(VER_RE);
    if (!oldMatch || oldMatch[0] !== oldVer) {
      return -5;
    }
    newVer = newVer.replace(/^0/, '');
    oldVer = oldVer.replace(/^0/, '');
    if (newVer === oldVer) {
      return 0;
    } else {
      newArr = newVer.split('.');
      oldArr = oldVer.split('.');
      newLen = newArr.length;
      oldLen = oldArr.length;
      maxLen = Math.max(newLen, oldLen);
      zerofill = function() {
        newArr.length < maxLen && newArr.push('0');
        oldArr.length < maxLen && oldArr.push('0');
        return newArr.length !== oldArr.length && zerofill();
      };
      newLen !== oldLen && zerofill();
      if (newArr.toString() === oldArr.toString()) {
        if (newLen > oldLen) {
          return 1;
        } else {
          return -1;
        }
      } else {
        isTrue = -1;
        compareNum = function() {
          var _new, _old;
          _new = ~~newArr.shift();
          _old = ~~oldArr.shift();
          _new > _old && (isTrue = 1);
          return _new === _old && newArr.length > 0 && compareNum();
        };
        compareNum();
        return isTrue;
      }
    }
  };
  return _compareVer;
});

好吧,我希望这段代码能帮助到一些人。

下面是测试。

console.log(compareVer("0.0.2","0.0.1"));//1
console.log(compareVer("0.0.10","0.0.1")); //1
console.log(compareVer("0.0.10","0.0.2")); //1
console.log(compareVer("0.9.0","0.9")); //1
console.log(compareVer("0.10.0","0.9.0")); //1
console.log(compareVer("1.7", "1.07")); //1
console.log(compareVer("1.0.07", "1.0.007")); //1

console.log(compareVer("0.3","0.3")); //0
console.log(compareVer("0.0.3","0.0.3")); //0
console.log(compareVer("0.0.3.0","0.0.3.0")); //0
console.log(compareVer("00.3","0.3")); //0
console.log(compareVer("00.3","00.3")); //0
console.log(compareVer("01.0.3","1.0.3")); //0
console.log(compareVer("1.0.3","01.0.3")); //0

console.log(compareVer("0.2.0","1.0.0")); //-1
console.log(compareVer('0.0.2.2.0',"0.0.2.3")); //-1
console.log(compareVer('0.0.2.0',"0.0.2")); //-1
console.log(compareVer('0.0.2',"0.0.2.0")); //-1
console.log(compareVer("1.07", "1.7")); //-1
console.log(compareVer("1.0.007", "1.0.07")); //-1

console.log(compareVer()); //-100
console.log(compareVer("0.0.2")); //-100
console.log(compareVer("0.0.2","0.0.2","0.0.2")); //-100
console.log(compareVer(1212,"0.0.2")); //-2
console.log(compareVer("0.0.2",1212)); //-3
console.log(compareVer('1.abc.2',"1.0.2")); //-4
console.log(compareVer('1.0.2',"1.abc.2")); //-5

我的答案比这里的大多数答案都要简洁

/**
 * Compare two semver versions. Returns true if version A is greater than
 * version B
 * @param {string} versionA
 * @param {string} versionB
 * @returns {boolean}
 */
export const semverGreaterThan = function(versionA, versionB){
  var versionsA = versionA.split(/\./g),
    versionsB = versionB.split(/\./g)
  while (versionsA.length || versionsB.length) {
    var a = Number(versionsA.shift()), b = Number(versionsB.shift())
    if (a == b)
      continue
    return (a > b || isNaN(b))
  }
  return false
}

看看这篇博客文章。此函数适用于数字版本号。

function compVersions(strV1, strV2) {
  var nRes = 0
    , parts1 = strV1.split('.')
    , parts2 = strV2.split('.')
    , nLen = Math.max(parts1.length, parts2.length);

  for (var i = 0; i < nLen; i++) {
    var nP1 = (i < parts1.length) ? parseInt(parts1[i], 10) : 0
      , nP2 = (i < parts2.length) ? parseInt(parts2[i], 10) : 0;

    if (isNaN(nP1)) { nP1 = 0; }
    if (isNaN(nP2)) { nP2 = 0; }

    if (nP1 != nP2) {
      nRes = (nP1 > nP2) ? 1 : -1;
      break;
    }
  }

  return nRes;
};

compVersions('10', '10.0'); // 0
compVersions('10.1', '10.01.0'); // 0
compVersions('10.0.1', '10.0'); // 1
compVersions('10.0.1', '10.1'); // -1

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