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
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;
},
其他回答
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);
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
例如，如果我们想检查当前jQuery版本是否小于1.8，如果version是"1.10.1"，parseFloat($.ui.version) < 1.8)将会给出错误的结果，因为parseFloat("1.10.1")返回1.1。 字符串比较也会出错，因为"1.8" < "1.10"的结果为false。
所以我们需要一个这样的测试
if(versionCompare($.ui.version, "1.8") < 0){
alert("please update jQuery");
}
下面的函数可以正确地处理这个问题:
/** 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 = parseInt(v1parts[i], 10);
p2 = 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;
}
下面是一些例子:
// compare dotted version strings
console.assert(versionCompare("1.8", "1.8.1") < 0);
console.assert(versionCompare("1.8.3", "1.8.1") > 0);
console.assert(versionCompare("1.8", "1.10") < 0);
console.assert(versionCompare("1.10.1", "1.10.1") === 0);
// Longer is considered 'greater'
console.assert(versionCompare("1.10.1.0", "1.10.1") > 0);
console.assert(versionCompare("1.10.1", "1.10.1.0") < 0);
// Strings pairs are accepted
console.assert(versionCompare("1.x", "1.x") === 0);
// Mixed int/string pairs return NaN
console.assert(isNaN(versionCompare("1.8", "1.x")));
//works with plain numbers
console.assert(versionCompare("4", 3) > 0);
看到这里的现场示例和测试套件: http://jsfiddle.net/mar10/8KjvP/
最简单的方法是使用localeCompare:
a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' })
这将返回:
0:版本字符串相等 1: a版本大于b版本 -1:版本b大于版本a
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"));
