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

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/

其他回答

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/

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"]

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。

这里有一个面向对象的有趣方法:

    function versionString(str) {
    var parts = str.split('.');
    this.product = parts.length > 0 ? parts[0] * 1 : 0;
    this.major = parts.length > 1 ? parts[1] * 1 : 0;
    this.minor = parts.length > 2 ? parts[2] * 1 : 0;
    this.build = parts.length > 3 ? parts[3] * 1 : 0;

    this.compareTo = function(vStr){
        vStr = this._isVersionString(vStr) ? vStr : new versionString(vStr);
        return this.compare(this, vStr);
    };

    this.toString = function(){
        return this.product + "." + this.major + "." + this.minor + "." + this.build;
    }

    this.compare = function (str1, str2) {
        var vs1 = this._isVersionString(str1) ? str1 : new versionString(str1);
        var vs2 = this._isVersionString(str2) ? str2 : new versionString(str2);

        if (this._compareNumbers(vs1.product, vs2.product) == 0) {
            if (this._compareNumbers(vs1.major, vs2.major) == 0) {
                if (this._compareNumbers(vs1.minor, vs2.minor) == 0) {
                    return this._compareNumbers(vs1.build, vs2.build);
                } else {
                    return this._compareNumbers(vs1.minor, vs2.minor);
                }
            } else {
                return this._compareNumbers(vs1.major, vs2.major);
            }
        } else {
            return this._compareNumbers(vs1.product, vs2.product);
        }
    };

    this._isVersionString = function (str) {
        return str !== undefined && str.build !== undefined;
    };

    this._compareNumbers = function (n1, n2) {
        if (n1 > n2) {
            return 1;
        } else if (n1 < n2) {
            return -1;
        } else {
            return 0;
        }
    };
}

还有一些测试:

var v1 = new versionString("1.0");
var v2 = new versionString("1.0.1");
var v3 = new versionString("2.0");
var v4 = new versionString("2.0.0.1");
var v5 = new versionString("2.0.1");


alert(v1.compareTo("1.4.2"));
alert(v3.compareTo(v1));
alert(v5.compareTo(v4));
alert(v4.compareTo(v5));
alert(v5.compareTo(v5));

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/