对 Javascript 中的版本点数字符串进行排序？

Question

I have an array of following strings:

['5.5.1', '4.21.0', '4.22.0', '6.1.0', '5.1.0', '4.5.0'] 

...etc.

I need a solution that will give me following ordered result

['4.5.0', '4.21.0', '4.22.0', '5.1.0', '5.5.1', '6.1.0'].

I tried to implement a sort so it first sorts by the numbers in the first position, than in case of equality, sort by the numbers in the second position (after the first dot), and so on...

I tried using sort() and localeCompare() , but if I have elements '4.5.0' and '4.11.0' , I get them sorted as ['4.11.0','4.5.0'] , but I need to get ['4.5.0','4.11.0'] .

How can I achieve this?

Answer 1

You could prepend all parts to fixed size strings, then sort that, and finally remove the padding again.

 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)

Obviously you have to choose the size of the number 100000 wisely: it should have at least one more digit than your largest number part will ever have.

With regular expression

The same manipulation can be achieved without having to split & join, when you use the callback argument to the replace method:

 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.replace(/\\d+/g, n => +n+100000 ) ).sort() .map( a => a.replace(/\\d+/g, n => +n-100000 ) ); console.log(arr)

Defining the padding function once only

As both the padding and its reverse functions are so similar, it seemed a nice exercise to use one function f for both, with an extra argument defining the "direction" (1=padding, -1=unpadding). This resulted in this quite obscure, and extreme code. Consider this just for fun, not for real use:

 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);

Use the sort compare callback function

You could use the compare function argument of sort to achieve the same:

arr.sort( (a, b) => a.replace(/\d+/g, n => +n+100000 )
                     .localeCompare(b.replace(/\d+/g, n => +n+100000 )) );

But for larger arrays this will lead to slower performance. This is because the sorting algorithm will often need to compare a certain value several times, each time with a different value from the array. This means that the padding will have to be executed multiple times for the same number. For this reason, it will be faster for larger arrays to first apply the padding in the whole array, then use the standard sort, and then remove the padding again.

But for shorter arrays, this approach might still be the fastest. In that case, the so-called natural sort option -- that can be achieved with the extra arguments of localeCompare -- will be more efficient than the padding method:

 var 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);

More about the padding and unary plus

To see how the padding works, look at the intermediate result it generates:

[ "100005.100005.100001", "100004.100021.100000", "100004.100022.100000", 
  "100006.100001.100000", "100005.100001.100000" ]

Concerning the expression +n+100000 , note that the first + is the unary plus and is the most efficient way to convert a string-encoded decimal number to its numerical equivalent. The 100000 is added to make the number have a fixed number of digits. Of course, it could just as well be 200000 or 300000. Note that this addition does not change the order the numbers will have when they would be sorted numerically.

The above is just one way to pad a string. See this Q&A for some other alternatives.

Answer 2

If you are looking for a npm package to compare two semver version, https://www.npmjs.com/package/compare-versions is the one.

Then you can sort version like this:

// 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);

Answer 3

You could split the strings and compare the parts.

Answer 4

You can check in loop if values are different, return difference, else continue

 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)

Answer 5

Though slightly late this would be my solution;

Answer 6

Here's a solution I developed based on @trincot's that will sort by semver even if the strings aren't exactly "1.2.3" - they could be ie "v1.2.3" or "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))

Answer 7

This seems to work provided there are only digits between the dots:

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('.')
})

Answer 8

'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());

Answer 9

This solution accounts for version numbers that might not be in the full, 3-part format (for example, if one of the version numbers is just 2 or 2.0 or 0.1, etc).

The custom sort function I wrote is probably mostly what you're looking for, it just needs an array of objects in the format {"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 (ie 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);

Answer 10

Inspired from the accepted answer, but ECMA5-compatible, and with regular string padding (see my comments on the answer):

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);
}

Usage:

['1.1', '1.0'].sort(sortCallback);

Answer 11

 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);<\/code><\/pre>

"

Answer 12

• sort 1.0a notation correct
• use native localeCompare to sort 1.090 notation

 function log(label,val){ document.body.append(label,String(val).replace(/,/g," - "),document.createElement("BR")); } const sortVersions = ( x, v = s => s.match(/[az]|\\d+/g).map(c => c==~~c ? String.fromCharCode(97 + c) : c) ) => x.sort((a, b) => (a + b).match(/[az]/) ? 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))]);

Answer 13

This can be in an easier way using the sort method without hardcoding any numbers and in a more generic way.

 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"]

Answer 14

const arr = ["5.2.0.1","5.2.1SW","5.2.1","6.1.0SW","6.1.0","6.2.0.1TSH","6.2.0.1SW","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);

Answer 15

In ES6 you can go without regex.

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)]

Answer 16

If ES6 I do this:

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);
});

Answer 17

**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!!!