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
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+)+$/; var segmentsA = a.replace(regExStrip0, '').split('.'); var segmentsB = b.replace(regExStrip0, '').split('.'); var l = Math.min(segmentsA.length, segmentsB.length); for (i = 0; i < l; i++) { diff = parseInt(segmentsA[i], 10) - parseInt(segmentsB[i], 10); if (diff) { return diff; } } return segmentsA.length - segmentsB.length; } // TEST console.log( ['2.5.10.4159', '1.0.0', '0.5', '0.4.1', '1', '1.1', '0.0.0', '2.5.0', '2', '0.0', '2.5.10', '10.5', '1.25.4', '1.2.15'].sort(cmpVersions)); // Result: // ["0.0.0", "0.0", "0.4.1", "0.5", "1.0.0", "1", "1.1", "1.2.15", "1.25.4", "2", "2.5.0", "2.5.10", "2.5.10.4159", "10.5"]
其他回答
下面是另一个简短的版本，适用于任何数量的子版本，填充零和偶数字母(1.0.0b3)
const compareVer = ((prep, repl) =>
{
prep = t => ("" + t)
//treat non-numerical characters as lower version
//replacing them with a negative number based on charcode of first character
.replace(/[^0-9\.]+/g, c => "." + (c.replace(/[\W_]+/, "").toLowerCase().charCodeAt(0) - 65536) + ".")
//remove trailing "." and "0" if followed by non-numerical characters (1.0.0b);
.replace(/(?:\.0+)*(\.-[0-9]+)(\.[0-9]+)?\.*$/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/
功能简单简短:
function isNewerVersion (oldVer, newVer) {
const oldParts = oldVer.split('.')
const newParts = newVer.split('.')
for (var i = 0; i < newParts.length; i++) {
const a = ~~newParts[i] // parse int
const b = ~~oldParts[i] // parse int
if (a > b) return true
if (a < b) return false
}
return false
}
测试:
isNewerVersion('1.0', '2.0') // true
isNewerVersion('1.0', '1.0.1') // true
isNewerVersion('1.0.1', '1.0.10') // true
isNewerVersion('1.0.1', '1.0.1') // false
isNewerVersion('2.0', '1.0') // false
isNewerVersion('2', '1.0') // false
isNewerVersion('2.0.0.0.0.1', '2.1') // true
isNewerVersion('2.0.0.0.0.1', '2.0') // false
function compare(versionA: string | undefined, versionB: string | undefined, operator: string = '>') {
if (versionA === undefined || versionB === undefined) {
return false
}
const listA = versionA.split('.')
const listB = versionB.split('.')
let a = []
let b = []
for (let i = 0; i < listA.length; i++) {
a.push(parseInt(listA[i].replace(/\D/g, ''), 10))
b.push(parseInt(listB[i].replace(/\D/g, ''), 10))
}
for (let i = 0; i < listA.length; i++) {
switch (operator) {
case '>':
case '>=':
if (a[i] === b[i]) {
continue
}
if (a[i] > b[i]) {
return true
}
if (a[i] < b[i]) {
return false
}
break
case '<':
case '<=':
if (a[i] === b[i]) {
continue
}
if (a[i] > b[i]) {
return false
}
if (a[i] < b[i]) {
return true
}
break
case '=':
if (a[i] > b[i]) {
return false
}
if (a[i] < b[i]) {
return false
}
break
}
}
switch (operator) {
case '>':
return false
case '<':
return false
case '=':
case '>=':
case '<=':
return true
}
}
比较不同条件下的功能:
const compareVer = (ver1, middle, ver2) => {
const res = new Intl.Collator("en").compare(ver1, ver2)
let comp
switch (middle) {
case "=":
comp = 0 === res
break
case ">":
comp = 1 === res
break
case ">=":
comp = 1 === res || 0 === res
break
case "<":
comp = -1 === res
break
case "<=":
comp = -1 === res || 0 === res
break
}
return comp
}
console.log(compareVer("1.0.2", "=", "1.0.2")) // true
console.log(compareVer("1.0.3", ">", "1.0.2")) // true
console.log(compareVer("1.0.1", ">=", "1.0.2")) // false
console.log(compareVer("1.0.3", ">=", "1.0.2")) // true
console.log(compareVer("1.0.1", "<", "1.0.2")) // true
console.log(compareVer("1.0.1", "<=", "1.0.2")) // true
我已经创建了这个解决方案，我希望你觉得它有用:
https://runkit.com/ecancino/5f3c6c59593d23001485992e
const quantify = max => (n, i) => n * (+max.slice(0, max.length - i))
const add = (a, b) => a + b
const calc = s => s.
split('.').
map(quantify('1000000')).
reduce(add, 0)
const sortVersions = unsortedVersions => unsortedVersions
.map(version => ({ version, order: calc(version) }))
.sort((a, b) => a.order - b.order)
.reverse()
.map(o => o.version)
