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
const compareAppVersions = (firstVersion, secondVersion) => { const firstVersionArray = firstVersion.split(".").map(Number); const secondVersionArray = secondVersion.split(".").map(Number); const loopLength = Math.max( firstVersionArray.length, secondVersionArray.length ）; 对于(设I = 0;i < loopLength;+ + i) { const a = firstVersionArray[i] || 0; const b = secondVersionArray[i] || 0; If (a !== b) { 返回> b ?1: -1; ｝ ｝ 返回0; };
其他回答
我认为这是一个值得分享的实现，因为它简短，简单，但功能强大。请注意，它只使用数字比较。通常它会检查version2是否比version1晚，如果是，则返回true。假设您有version1: 1.1.1和version2: 1.1.2。它遍历两个版本的每个部分，将它们的部分相加如下:对于版本1(1 + 0.1)然后(1.1 + 0.01)，对于版本2(1 + 0.1)然后(1.1 + 0.02)。
function compareVersions(version1, version2) {
version1 = version1.split('.');
version2 = version2.split('.');
var maxSubVersionLength = String(Math.max.apply(undefined, version1.concat(version2))).length;
var reduce = function(prev, current, index) {
return parseFloat(prev) + parseFloat('0.' + Array(index + (maxSubVersionLength - String(current).length)).join('0') + current);
};
return version1.reduce(reduce) < version2.reduce(reduce);
}
如果你想从版本列表中找到最新的版本，那么这可能是有用的:
function findLatestVersion(versions) {
if (!(versions instanceof Array)) {
versions = Array.prototype.slice.apply(arguments, [0]);
}
versions = versions.map(function(version) { return version.split('.'); });
var maxSubVersionLength = String(Math.max.apply(undefined, Array.prototype.concat.apply([], versions))).length;
var reduce = function(prev, current, index) {
return parseFloat(prev) + parseFloat('0.' + Array(index + (maxSubVersionLength - String(current).length)).join('0') + current);
};
var sums = [];
for (var i = 0; i < versions.length; i++) {
sums.push(parseFloat(versions[i].reduce(reduce)));
}
return versions[sums.indexOf(Math.max.apply(undefined, sums))].join('.');
}
console.log(findLatestVersion('0.1000000.1', '2.0.0.10', '1.6.10', '1.4.3', '2', '2.0.0.1')); // 2.0.0.10
console.log(findLatestVersion(['0.1000000.1', '2.0.0.10', '1.6.10', '1.4.3', '2', '2.0.0.1'])); // 2.0.0.10
我必须比较我的扩展版本，但我没有 在这里找到一个可行的解决方案。在比较1.89 > 1.9或1.24.1 == 1.240.1时，几乎所有提议的期权都被打破了
这里，我从仅在最后的记录1.1 == 1.10和1.10.1 > 1.1.1中0下降的事实开始
compare_version = (new_version, old_version) => {
new_version = new_version.split('.');
old_version = old_version.split('.');
for(let i = 0, m = Math.max(new_version.length, old_version.length); i<m; i++){
//compare text
let new_part = (i<m-1?'':'.') + (new_version[i] || 0)
, old_part = (i<m-1?'':'.') + (old_version[i] || 0);
//compare number (I don’t know what better)
//let new_part = +((i<m-1?0:'.') + new_version[i]) || 0
//, old_part = +((i<m-1?0:'.') + old_version[i]) || 0;
//console.log(new_part, old_part);
if(old_part > new_part)return 0; //change to -1 for sort the array
if(new_part > old_part)return 1
}
return 0
};
compare_version('1.0.240.1','1.0.240.1'); //0
compare_version('1.0.24.1','1.0.240.1'); //0
compare_version('1.0.240.89','1.0.240.9'); //0
compare_version('1.0.24.1','1.0.24'); //1
我不是一个大专家，但我构建了简单的代码来比较两个版本，将第一个返回值更改为-1以对版本数组进行排序
['1.0.240', '1.0.24', '1.0.240.9', '1.0.240.89'].sort(compare_version)
//results ["1.0.24", "1.0.240", "1.0.240.89", "1.0.240.9"]
和短版本的比较全字符串
c=e=>e.split('.').map((e,i,a)=>e[i<a.length-1?'padStart':'padEnd'](5)).join('');
//results " 1 0 2409 " > " 1 0 24089 "
c('1.0.240.9')>c('1.0.240.89') //true
如果您有意见或改进，请不要犹豫提出建议。
// Returns true if v1 is bigger than v2, and false if otherwise.
function isNewerThan(v1, v2) {
v1=v1.split('.');
v2=v2.split('.');
for(var i = 0; i<Math.max(v1.length,v2.length); i++){
if(v1[i] == undefined) return false; // If there is no digit, v2 is automatically bigger
if(v2[i] == undefined) return true; // if there is no digit, v1 is automatically bigger
if(v1[i] > v2[i]) return true;
if(v1[i] < v2[i]) return false;
}
return false; // Returns false if they are equal
}
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$2") .split('.'); return (a, b, c, i, r) => { a = prep(a); b = prep(b); for (i = 0, r = 0, c = Math.max(a.length, b.length); !r && i++ < c;) { r = -1 * ((a[i] = ~~a[i]) < (b[i] = ~~b[i])) + (a[i] > b[i]); } return r; } })(); //examples let list = [ ["1.0", "1.0.0.0.0.0"], ["1.0", "1.0.1"], ["1.0b1", "1.0"], ["1.0b", "1.0b"], ["1.1", "1.0.1b"], ["1.1alpha", "1.1beta"], ["1.1rc1", "1.1beta"], ["1.1rc1", "1.1rc2"], ["1.1.0a1", "1.1a2"], ["1.1.0a10", "1.1.0a1"], ["1.1.0alpha", "1.1a"], ["1.1.0alpha2", "1.1b1"], ["1.0001", "1.00000.1.0.0.0.01"] ] for(let i = 0; i < list.length; i++) { console.log( list[i][0] + " " + "<=>"[compareVer(list[i][0], list[i][1]) + 1] + " " + list[i][1] ); }
https://jsfiddle.net/vanowm/p7uvtbor/
这个非常小，但非常快的比较函数接受每个段的任何长度和任何数字大小的版本号。
返回值: -如果a < b，则数字< 0 -如果是> b，则为> 0 如果a = b - 0
所以你可以使用它作为array。sort()的比较函数;
编辑:修正了版本剥离尾随零识别“1”和“1.0.0”相等的错误
function cmpVersions (a, b) { var i, diff; var regExStrip0 = /(\.0+)+$/; 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"]
