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

你可以使用带有选项的String#localeCompare

sensitivity Which differences in the strings should lead to non-zero result values. Possible values are: "base": Only strings that differ in base letters compare as unequal. Examples: a ≠ b, a = á, a = A. "accent": Only strings that differ in base letters or accents and other diacritic marks compare as unequal. Examples: a ≠ b, a ≠ á, a = A. "case": Only strings that differ in base letters or case compare as unequal. Examples: a ≠ b, a = á, a ≠ A. "variant": Strings that differ in base letters, accents and other diacritic marks, or case compare as unequal. Other differences may also be taken into consideration. Examples: a ≠ b, a ≠ á, a ≠ A. The default is "variant" for usage "sort"; it's locale dependent for usage "search". numeric Whether numeric collation should be used, such that "1" < "2" < "10". Possible values are true and false; the default is false. This option can be set through an options property or through a Unicode extension key; if both are provided, the options property takes precedence. Implementations are not required to support this property.

var版本=[" 2.0.1”、“2.0”、“1.0”、“1.0.1”,“2.0.0.1”); 版本。sort((a, b) => a.localeCompare(b, undefined, {numeric: true,灵敏度:'base'})); console.log(版本);

其他回答

// 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
    }

我不喜欢任何一个解决方案,所以我根据自己的编码偏好重新编写了它。请注意,最后四个检查结果与接受的答案略有不同。对我有用。

function v_check(version_a, version_b) {
    // compares version_a as it relates to version_b
    // a = b => "same"
    // a > b => "larger"
    // a < b => "smaller"
    // NaN   => "invalid"

    const arr_a = version_a.split('.');
    const arr_b = version_b.split('.');

    let result = "same"; // initialize to same // loop tries to disprove

    // loop through a and check each number against the same position in b
    for (let i = 0; i < arr_a.length; i++) {
        let a = arr_a[i];
        let b = arr_b[i];

        // same up to this point so if a is not there, a is smaller
        if (typeof a === 'undefined') {
            result = "smaller";
            break;

        // same up to this point so if b is not there, a is larger
        } else if (typeof b === 'undefined') {
            result = "larger";
            break;

        // otherwise, compare the two numbers
        } else {

            // non-positive numbers are invalid
            if (a >= 0 && b >= 0) {

                if (a < b) {
                    result = "smaller";
                    break;
                }
                else if (a > b) {
                    result = "larger";
                    break;
                }

            } else {
                result = "invalid";
                break;
            }
        }
    }

    // account for the case where the loop ended but there was still a position in b to evaluate
    if (result == "same" && arr_b.length > arr_a.length) result = "smaller";

    return result;
}


console.log(v_check("1.7.1", "1.7.10"));  // smaller
console.log(v_check("1.6.1", "1.7.10"));  // smaller
console.log(v_check("1.6.20", "1.7.10")); // smaller
console.log(v_check("1.7.1", "1.7.10"));  // smaller
console.log(v_check("1.7", "1.7.0"));     // smaller
console.log(v_check("1.7", "1.8.0"));     // smaller

console.log(v_check("1.7.10", "1.7.1"));  // larger
console.log(v_check("1.7.10", "1.6.1"));  // larger
console.log(v_check("1.7.10", "1.6.20")); // larger
console.log(v_check("1.7.0", "1.7"));     // larger
console.log(v_check("1.8.0", "1.7"));     // larger

console.log(v_check("1.7.10", "1.7.10")); // same
console.log(v_check("1.7", "1.7"));       // same

console.log(v_check("1.7", "1..7")); // larger
console.log(v_check("1.7", "Bad"));  // invalid
console.log(v_check("1..7", "1.7")); // smaller
console.log(v_check("Bad", "1.7"));  // invalid

在排序之前对版本进行预处理意味着parseInt不会被不必要地多次调用。 使用array# map类似于Michael Deal的建议,下面是我用来查找标准3部分semver的最新版本的排序:

var semvers = ["0.1.0", "1.0.0", "1.1.0", "1.0.5"]; var versions = semvers.map(function(semver) { return semver.split(".").map(function(part) { return parseInt(part); }); }); versions.sort(function(a, b) { if (a[0] < b[0]) return 1; else if (a[0] > b[0]) return -1; else if (a[1] < b[1]) return 1; else if (a[1] > b[1]) return -1; else if (a[2] < b[2]) return 1; else if (a[2] > b[2]) return -1; return 0; }); var newest = versions[0].join("."); console.log(newest); // "1.1.0"

如果不允许使用字母或符号,那么代码行数就很少。如果您控制了版本控制方案,并且不是第三方提供的,那么这种方法是有效的。

// we presume all versions are of this format "1.4" or "1.10.2.3", without letters // returns: 1 (bigger), 0 (same), -1 (smaller) function versionCompare (v1, v2) { const v1Parts = v1.split('.') const v2Parts = v2.split('.') const length = Math.max(v1Parts.length, v2Parts.length) for (let i = 0; i < length; i++) { const value = (parseInt(v1Parts[i]) || 0) - (parseInt(v2Parts[i]) || 0) if (value < 0) return -1 if (value > 0) return 1 } return 0 } console.log(versionCompare('1.2.0', '1.2.4') === -1) console.log(versionCompare('1.2', '1.2.0') === 0) console.log(versionCompare('1.2', '1') === 1) console.log(versionCompare('1.2.10', '1.2.1') === 1) console.log(versionCompare('1.2.134230', '1.2.2') === 1) console.log(versionCompare('1.2.134230', '1.3.0.1.2.3.1') === -1)

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));