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
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
其他回答
这里有一个面向对象的有趣方法:
function versionString(str) {
var parts = str.split('.');
this.product = parts.length > 0 ? parts[0] * 1 : 0;
this.major = parts.length > 1 ? parts[1] * 1 : 0;
this.minor = parts.length > 2 ? parts[2] * 1 : 0;
this.build = parts.length > 3 ? parts[3] * 1 : 0;
this.compareTo = function(vStr){
vStr = this._isVersionString(vStr) ? vStr : new versionString(vStr);
return this.compare(this, vStr);
};
this.toString = function(){
return this.product + "." + this.major + "." + this.minor + "." + this.build;
}
this.compare = function (str1, str2) {
var vs1 = this._isVersionString(str1) ? str1 : new versionString(str1);
var vs2 = this._isVersionString(str2) ? str2 : new versionString(str2);
if (this._compareNumbers(vs1.product, vs2.product) == 0) {
if (this._compareNumbers(vs1.major, vs2.major) == 0) {
if (this._compareNumbers(vs1.minor, vs2.minor) == 0) {
return this._compareNumbers(vs1.build, vs2.build);
} else {
return this._compareNumbers(vs1.minor, vs2.minor);
}
} else {
return this._compareNumbers(vs1.major, vs2.major);
}
} else {
return this._compareNumbers(vs1.product, vs2.product);
}
};
this._isVersionString = function (str) {
return str !== undefined && str.build !== undefined;
};
this._compareNumbers = function (n1, n2) {
if (n1 > n2) {
return 1;
} else if (n1 < n2) {
return -1;
} else {
return 0;
}
};
}
还有一些测试:
var v1 = new versionString("1.0");
var v2 = new versionString("1.0.1");
var v3 = new versionString("2.0");
var v4 = new versionString("2.0.0.1");
var v5 = new versionString("2.0.1");
alert(v1.compareTo("1.4.2"));
alert(v3.compareTo(v1));
alert(v5.compareTo(v4));
alert(v4.compareTo(v5));
alert(v5.compareTo(v5));
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
这不是一个很好的解决问题的方法，但它非常相似。
这个排序函数是针对语义版本的，它处理的是解析版本，所以它不能处理像x或*这样的通配符。
它适用于正则表达式匹配的版本:/\d+\.\d+\.\d+.*$/。它与这个答案非常相似，除了它也适用于像1.2.3-dev这样的版本。 与另一个答案的比较:我删除了一些我不需要的检查，但我的解决方案可以与另一个相结合。
semVerSort = function(v1, v2) {
var v1Array = v1.split('.');
var v2Array = v2.split('.');
for (var i=0; i<v1Array.length; ++i) {
var a = v1Array[i];
var b = v2Array[i];
var aInt = parseInt(a, 10);
var bInt = parseInt(b, 10);
if (aInt === bInt) {
var aLex = a.substr((""+aInt).length);
var bLex = b.substr((""+bInt).length);
if (aLex === '' && bLex !== '') return 1;
if (aLex !== '' && bLex === '') return -1;
if (aLex !== '' && bLex !== '') return aLex > bLex ? 1 : -1;
continue;
} else if (aInt > bInt) {
return 1;
} else {
return -1;
}
}
return 0;
}
合并后的解为:
function versionCompare(v1, v2, options) {
var zeroExtend = options && options.zeroExtend,
v1parts = v1.split('.'),
v2parts = v2.split('.');
if (zeroExtend) {
while (v1parts.length < v2parts.length) v1parts.push("0");
while (v2parts.length < v1parts.length) v2parts.push("0");
}
for (var i = 0; i < v1parts.length; ++i) {
if (v2parts.length == i) {
return 1;
}
var v1Int = parseInt(v1parts[i], 10);
var v2Int = parseInt(v2parts[i], 10);
if (v1Int == v2Int) {
var v1Lex = v1parts[i].substr((""+v1Int).length);
var v2Lex = v2parts[i].substr((""+v2Int).length);
if (v1Lex === '' && v2Lex !== '') return 1;
if (v1Lex !== '' && v2Lex === '') return -1;
if (v1Lex !== '' && v2Lex !== '') return v1Lex > v2Lex ? 1 : -1;
continue;
}
else if (v1Int > v2Int) {
return 1;
}
else {
return -1;
}
}
if (v1parts.length != v2parts.length) {
return -1;
}
return 0;
}
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
在排序之前对版本进行预处理意味着parseInt不会被不必要地多次调用。 使用array# map类似于Michael Deal的建议，下面是我用来查找标准3部分semver的最新版本的排序:
var semvers = ["0.1.0", "1.0.0", "1.1.0", "1.0.5"]; var versions = semvers.map(function(semver) { return semver.split(".").map(function(part) { return parseInt(part); }); }); versions.sort(function(a, b) { if (a[0] < b[0]) return 1; else if (a[0] > b[0]) return -1; else if (a[1] < b[1]) return 1; else if (a[1] > b[1]) return -1; else if (a[2] < b[2]) return 1; else if (a[2] > b[2]) return -1; return 0; }); var newest = versions[0].join("."); console.log(newest); // "1.1.0"
