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

我不喜欢任何一个解决方案,所以我根据自己的编码偏好重新编写了它。请注意,最后四个检查结果与接受的答案略有不同。对我有用。

function v_check(version_a, version_b) {
    // compares version_a as it relates to version_b
    // a = b => "same"
    // a > b => "larger"
    // a < b => "smaller"
    // NaN   => "invalid"

    const arr_a = version_a.split('.');
    const arr_b = version_b.split('.');

    let result = "same"; // initialize to same // loop tries to disprove

    // loop through a and check each number against the same position in b
    for (let i = 0; i < arr_a.length; i++) {
        let a = arr_a[i];
        let b = arr_b[i];

        // same up to this point so if a is not there, a is smaller
        if (typeof a === 'undefined') {
            result = "smaller";
            break;

        // same up to this point so if b is not there, a is larger
        } else if (typeof b === 'undefined') {
            result = "larger";
            break;

        // otherwise, compare the two numbers
        } else {

            // non-positive numbers are invalid
            if (a >= 0 && b >= 0) {

                if (a < b) {
                    result = "smaller";
                    break;
                }
                else if (a > b) {
                    result = "larger";
                    break;
                }

            } else {
                result = "invalid";
                break;
            }
        }
    }

    // account for the case where the loop ended but there was still a position in b to evaluate
    if (result == "same" && arr_b.length > arr_a.length) result = "smaller";

    return result;
}


console.log(v_check("1.7.1", "1.7.10"));  // smaller
console.log(v_check("1.6.1", "1.7.10"));  // smaller
console.log(v_check("1.6.20", "1.7.10")); // smaller
console.log(v_check("1.7.1", "1.7.10"));  // smaller
console.log(v_check("1.7", "1.7.0"));     // smaller
console.log(v_check("1.7", "1.8.0"));     // smaller

console.log(v_check("1.7.10", "1.7.1"));  // larger
console.log(v_check("1.7.10", "1.6.1"));  // larger
console.log(v_check("1.7.10", "1.6.20")); // larger
console.log(v_check("1.7.0", "1.7"));     // larger
console.log(v_check("1.8.0", "1.7"));     // larger

console.log(v_check("1.7.10", "1.7.10")); // same
console.log(v_check("1.7", "1.7"));       // same

console.log(v_check("1.7", "1..7")); // larger
console.log(v_check("1.7", "Bad"));  // invalid
console.log(v_check("1..7", "1.7")); // smaller
console.log(v_check("Bad", "1.7"));  // invalid

其他回答

replace()函数只替换字符串中的第一个出现项。我们来替换。与,。然后全部删除。然后做,to。再次将其解析为float。

for(i=0; i<versions.length; i++) {
    v = versions[i].replace('.', ',');
    v = v.replace(/\./g, '');
    versions[i] = parseFloat(v.replace(',', '.'));
}

最后,排序:

versions.sort();

semver

npm使用的语义版本解析器。

$ npm install semver
var semver = require('semver');

semver.diff('3.4.5', '4.3.7') //'major'
semver.diff('3.4.5', '3.3.7') //'minor'
semver.gte('3.4.8', '3.4.7') //true
semver.ltr('3.4.8', '3.4.7') //false

semver.valid('1.2.3') // '1.2.3'
semver.valid('a.b.c') // null
semver.clean(' =v1.2.3 ') // '1.2.3'
semver.satisfies('1.2.3', '1.x || >=2.5.0 || 5.0.0 - 7.2.3') // true
semver.gt('1.2.3', '9.8.7') // false
semver.lt('1.2.3', '9.8.7') // true

var versions = [ '1.2.3', '3.4.5', '1.0.2' ]
var max = versions.sort(semver.rcompare)[0]
var min = versions.sort(semver.compare)[0]
var max = semver.maxSatisfying(versions, '*')

语义版本控制链接:https://www.npmjs.com/package/semver#prerelease-identifiers

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

两个版本比较

const val = '1.2.3 5.4.3';
const arr = val.split(' ');
let obj = {};
for(let i = 0; i<2; i++) {
    const splitArr = arr[i].split('.')
    const reduced = splitArr.reduce((pre, 
    curr)=>parseInt(pre)+parseInt(curr));
    obj[i] = reduced;
}
if(obj[0]>obj[1]) {
    console.log(arr[0]);
} else {
    console.log(arr[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