我有一个以下字符串的数组：
['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0']
...等等
我需要一个能够给我以下订购结果的解决方案
['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0'].
我尝试实现一种排序，因此它首先按第一个位置中的数字排序，而不是在相等的情况下，按第二个位置中的数字（第一个点之后）排序，依此类推...
我尝试使用 sort() 和 localeCompare()，但是如果我有元素
'4.5.0''4.11.0'['4.11.0','4.5.0']['4.5.0','4.11.0']我怎样才能实现这个目标？
您可以将所有部分添加到固定大小的字符串中，然后对其进行排序，最后再次删除填充。
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.map( a => a.split('.').map( n => +n+100000 ).join('.') ).sort()
.map( a => a.split('.').map( n => +n-100000 ).join('.') );
console.log(arr)显然，您必须明智地选择数字 100000 的大小：它应该比您最大的数字部分至少多一位数字。
replacevar arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.map( a => a.replace(/\d+/g, n => +n+100000 ) ).sort()
.map( a => a.replace(/\d+/g, n => +n-100000 ) );
console.log(arr)由于填充及其反向函数非常相似，因此对两者使用一个函数 f 似乎是一个很好的练习，并使用一个额外的参数定义“方向”（1 = 填充，-1 = 取消填充）。这导致了这段相当晦涩且极端的代码。考虑这只是为了好玩，而不是真正的用途：
var arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = (f=>f(f(arr,1).sort(),-1)) ((arr,v)=>arr.map(a=>a.replace(/\d+/g,n=>+n+v*100000)));
console.log(arr);sortsortarr.sort( (a, b) => a.replace(/\d+/g, n => +n+100000 )
.localeCompare(b.replace(/\d+/g, n => +n+100000 )) );
但对于较大的数组，这会导致性能下降。这是因为排序算法通常需要对某个值进行多次比较，每次都与数组中的不同值进行比较。这意味着必须对同一数字执行多次填充。因此，对于较大的数组，首先在整个数组中应用填充，然后使用标准排序，然后再次删除填充会更快。
但对于较短的数组，这种方法可能仍然是最快的。在这种情况下，所谓的 natural 排序选项（可以通过
localeComparevar arr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
arr = arr.sort( (a, b) => a.localeCompare(b, undefined, { numeric:true }) );
console.log(arr);要了解填充的工作原理，请查看它生成的中间结果：
[ "100005.100005.100001", "100004.100021.100000", "100004.100022.100000",
"100006.100001.100000", "100005.100001.100000" ]
关于表达式
+n+100000+以上只是填充字符串的一种方法。请参阅此Q&A了解其他一些替代方案。
如果您正在寻找一个 npm 包来比较两个 semver 版本，https://www.npmjs.com/package/compare-versions 就是最佳选择。
然后你可以像这样对版本进行排序：
// ES6/TypeScript
import compareVersions from 'compare-versions';
var versions = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
var sorted = versions.sort(compareVersions);
您可以拆分字符串并比较各个部分。
function customSort(data, order) {
function isNumber(v) {
return (+v).toString() === v;
}
var sort = {
asc: function (a, b) {
var i = 0,
l = Math.min(a.value.length, b.value.length);
while (i < l && a.value[i] === b.value[i]) {
i++;
}
if (i === l) {
return a.value.length - b.value.length;
}
if (isNumber(a.value[i]) && isNumber(b.value[i])) {
return a.value[i] - b.value[i];
}
return a.value[i].localeCompare(b.value[i]);
},
desc: function (a, b) {
return sort.asc(b, a);
}
}
var mapped = data.map(function (el, i) {
return {
index: i,
value: el.split('.')
};
});
console.log(mapped);
mapped.sort(sort[order] || sort.asc);
return mapped.map(function (el) {
return data[el.index];
});
}
var array = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0'];
console.log('sorted array asc', customSort(array));
console.log('sorted array desc ', customSort(array, 'desc'));
console.log('original array ', array);.as-console-wrapper { max-height: 100% !important; top: 0; }您可以在循环中检查值是否不同，返回差异，否则继续
var a=['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
a.sort(function(a,b){
var a1 = a.split('.');
var b1 = b.split('.');
var len = Math.max(a1.length, b1.length);
for(var i = 0; i< len; i++){
var _a = +a1[i] || 0;
var _b = +b1[i] || 0;
if(_a === _b) continue;
else return _a > _b ? 1 : -1
}
return 0;
})
console.log(a)虽然有点晚了，这将是我的解决方案；
var arr = ["5.1.1","5.1.12","5.1.2","3.7.6","2.11.4","4.8.5","4.8.4","2.10.4"],
sorted = arr.sort((a,b) => {var aa = a.split("."),
ba = b.split(".");
return +aa[0] < +ba[0] ? -1
: aa[0] === ba[0] ? +aa[1] < +ba[1] ? -1
: aa[1] === ba[1] ? +aa[2] < +ba[2] ? -1
: 1
: 1
: 1;
});
console.log(sorted);这是我基于 @trincot 开发的一个解决方案，即使字符串不完全是“1.2.3”，它也会按 semver 排序 - 它们可能是“v1.2.3”或“2.4”
function sortSemVer(arr, reverse = false) {
let semVerArr = arr.map(i => i.replace(/(\d+)/g, m => +m + 100000)).sort(); // +m is just a short way of converting the match to int
if (reverse)
semVerArr = semVerArr.reverse();
return semVerArr.map(i => i.replace(/(\d+)/g, m => +m - 100000))
}
console.log(sortSemVer(["1.0.1", "1.0.9", "1.0.10"]))
console.log(sortSemVer(["v2.1", "v2.0.9", "v2.0.12", "v2.2"], true))如果点之间只有数字，这似乎可行：
var a = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0']
a = a.map(function (x) {
return x.split('.').map(function (x) {
return parseInt(x)
})
}).sort(function (a, b) {
var i = 0, m = a.length, n = b.length, o, d
o = m < n ? n : m
for (; i < o; ++i) {
d = (a[i] || 0) - (b[i] || 0)
if (d) return d
}
return 0
}).map(function (x) {
return x.join('.')
})
'use strict';
var arr = ['5.1.2', '5.1.1', '5.1.1', '5.1.0', '5.7.2.2'];
Array.prototype.versionSort = function () {
var arr = this;
function isNexVersionBigger (v1, v2) {
var a1 = v1.split('.');
var b2 = v2.split('.');
var len = a1.length > b2.length ? a1.length : b2.length;
for (var k = 0; k < len; k++) {
var a = a1[k] || 0;
var b = b2[k] || 0;
if (a === b) {
continue;
} else
return b < a;
}
}
for (var i = 0; i < arr.length; i++) {
var min_i = i;
for (var j = i + 1; j < arr.length; j++) {
if (isNexVersionBigger(arr[i], arr[j])) {
min_i = j;
}
}
var temp = arr[i];
arr[i] = arr[min_i];
arr[min_i] = temp;
}
return arr;
}
console.log(arr.versionSort());
此解决方案考虑了可能不是完整的三部分格式的版本号（例如，如果版本号之一仅为 2 或 2.0 或 0.1 等）。
我编写的自定义排序函数可能主要是您正在寻找的，它只需要一个格式为
{"major":X, "minor":X, "revision":X}var versionArr = ['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'];
var versionObjectArr = [];
var finalVersionArr = [];
/*
split each version number string by the '.' and separate them in an
object by part (major, minor, & revision). If version number is not
already in full, 3-part format, -1 will represent that part of the
version number that didn't exist. Push the object into an array that
can be sorted.
*/
for(var i = 0; i < versionArr.length; i++){
var splitVersionNum = versionArr[i].split('.');
var versionObj = {};
switch(splitVersionNum.length){
case 1:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":-1,
"revision":-1
};
break;
case 2:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":parseInt(splitVersionNum[1]),
"revision":-1
};
break;
case 3:
versionObj = {
"major":parseInt(splitVersionNum[0]),
"minor":parseInt(splitVersionNum[1]),
"revision":parseInt(splitVersionNum[2])
};
}
versionObjectArr.push(versionObj);
}
//sort objects by parts, going from major to minor to revision number.
versionObjectArr.sort(function(a, b){
if(a.major < b.major) return -1;
else if(a.major > b.major) return 1;
else {
if(a.minor < b.minor) return -1;
else if(a.minor > b.minor) return 1;
else {
if(a.revision < b.revision) return -1;
else if(a.revision > b.revision) return 1;
}
}
});
/*
loops through sorted object array to recombine it's version keys to match the original string's value. If any trailing parts of the version
number are less than 0 (i.e. they didn't exist so we replaced them with
-1) then leave that part of the version number string blank.
*/
for(var i = 0; i < versionObjectArr.length; i++){
var versionStr = "";
for(var key in versionObjectArr[i]){
versionStr = versionObjectArr[i].major;
versionStr += (versionObjectArr[i].minor < 0 ? '' : "." + versionObjectArr[i].minor);
versionStr += (versionObjectArr[i].revision < 0 ? '' : "." + versionObjectArr[i].revision);
}
finalVersionArr.push(versionStr);
}
console.log('Original Array: ',versionArr);
console.log('Expected Output: ',['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0']);
console.log('Actual Output: ', finalVersionArr);受到已接受答案的启发，但与 ECMA5 兼容，并且具有常规字符串填充（请参阅我对答案的评论）：
function sortCallback(a, b) {
function padParts(version) {
return version
.split('.')
.map(function (part) {
return '00000000'.substr(0, 8 - part.length) + part;
})
.join('.');
}
a = padParts(a);
b = padParts(b);
return a.localeCompare(b);
}
用途：
['1.1', '1.0'].sort(sortCallback);
const arr = ["5.1.1","5.1.12","5.1.2","3.7.6","2.11.4","4.8.5","4.8.4","2.10.4"];
const sorted = arr.sort((a,b) => {
const ba = b.split('.');
const d = a.split('.').map((a1,i)=>a1-ba[i]);
return d[0] ? d[0] : d[1] ? d[1] : d[2]
});
console.log(sorted);这可以通过使用排序方法以更简单的方式实现，无需对任何数字进行硬编码，并且采用更通用的方式。
enter code here
var arr = ['5.1.2', '5.1.1', '5.1.1', '5.1.0', '5.7.2.2'];
splitArray = arr.map(elements => elements.split('.'))
//now lets sort based on the elements on the corresponding index of each array
//mapped.sort(function(a, b) {
// if (a.value > b.value) {
// return 1;
// }
// if (a.value < b.value) {
// return -1;
// }
// return 0;
//});
//here we compare the first element with the first element of the next version number and that is [5.1.2,5.7.2] 5,5 and 1,7 and 2,2 are compared to identify the smaller version...In the end use the join() to get back the version numbers in the proper format.
sortedArray = splitArray.sort((a, b) => {
for (i in a) {
if (parseInt(a[i]) < parseInt(b[i])) {
return -1;
break
}
if (parseInt(a[i]) > parseInt(b[i])) {
return +1;
break
} else {
continue
}
}
}).map(p => p.join('.'))
sortedArray = ["5.1.0", "5.1.1", "5.1.1", "5.1.2", "5.7.2.2"]1.0alocaleCompare1.090function log(label,val){
document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR"));
}
const sortVersions = (
x,
v = s => s.match(/[a-z]|\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c)
) => x.sort((a, b) => (a + b).match(/[a-z]/)
? v(b) < v(a) ? 1 : -1
: a.localeCompare(b, 0, {numeric: true}))
let v=["1.90.1","1.090","1.0a","1.0.1","1.0.0a","1.0.0b","1.0.0.1","1.0a"];
log(' input : ',v);
log('sorted: ',sortVersions(v));
log('no dups:',[...new Set(sortVersions(v))]);在 ES6 中你可以不用正则表达式。
const versions = ["0.4", "0.11", "0.4.1", "0.4", "0.4.2", "2.0.1","2", "0.0.1", "0.2.3"];
const splitted = versions.map(version =>
version
.split('.')
.map(i => +i))
.map(i => {
let items;
if (i.length === 1) {
items = [0, 0]
i.push(...items)
}
if (i.length === 2) {
items = [0]
i.push(...items)
}
return i
})
.sort((a, b) => {
for(i in a) {
if (a[i] < b[i]) {
return -1;
}
if (a[i] > b[i]) {
return +1;
}
}
})
.map(item => item.join('.'))
const sorted = [...new Set(splitted)]
如果 ES6 我这样做：
versions.sort((v1, v2) => {
let [, major1, minor1, revision1 = 0] = v1.match(/([0-9]+)\.([0-9]+)(?:\.([0-9]+))?/);
let [, major2, minor2, revision2 = 0] = v2.match(/([0-9]+)\.([0-9]+)(?:\.([0-9]+))?/);
if (major1 != major2) return parseInt(major1) - parseInt(major2);
if (minor1 != minor2) return parseInt(minor1) - parseInt(major2);
return parseInt(revision1) - parseInt(revision2);
});
**Sorted Array Object by dotted version value**
var sampleData = [
{ name: 'Edward', value: '2.1.2' },
{ name: 'Sharpe', value: '2.1.3' },
{ name: 'And', value: '2.2.1' },
{ name: 'The', value: '2.1' },
{ name: 'Magnetic', value: '2.2' },
{ name: 'Zeros', value: '0' },
{ name: 'Zeros', value: '1' }
];
arr = sampleData.map( a => a.value).sort();
var requireData = [];
arr.forEach(function(record, index){
var findRecord = sampleData.find(arr => arr.value === record);
if(findRecord){
requireData.push(findRecord);
}
});
console.log(requireData);
[check on jsfiddle.net][1]
[1]: https://jsfiddle.net/jx3buswq/2/
It is corrected now!!!