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

你可以遍历每个以句点分隔的字符并将其转换为int类型:

var parts = versionString.split('.');

for (var i = 0; i < parts.length; i++) {
  var value = parseInt(parts[i]);
  // do stuffs here.. perhaps build a numeric version variable?
}

其他回答

我找到了最简单的方法来比较它们,但我不确定这是否是你想要的。

当我在控制台中运行下面的代码时,它是有意义的,并且使用sort()方法,我可以获得版本字符串的排序数组。它是根据字母顺序排列的。

"1.0" < "1.0.1" //true
var arr = ["1.0.1", "1.0", "3.2.0", "1.3"]
arr.sort();     //["1.0", "1.0.1", "1.3", "3.2.0"]

replace()函数只替换字符串中的第一个出现项。我们来替换。与,。然后全部删除。然后做,to。再次将其解析为float。

for(i=0; i<versions.length; i++) {
    v = versions[i].replace('.', ',');
    v = v.replace(/\./g, '');
    versions[i] = parseFloat(v.replace(',', '.'));
}

最后,排序:

versions.sort();

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

摘自http://java.com/js/deployJava.js:

    // return true if 'installed' (considered as a JRE version string) is
    // greater than or equal to 'required' (again, a JRE version string).
    compareVersions: function (installed, required) {

        var a = installed.split('.');
        var b = required.split('.');

        for (var i = 0; i < a.length; ++i) {
            a[i] = Number(a[i]);
        }
        for (var i = 0; i < b.length; ++i) {
            b[i] = Number(b[i]);
        }
        if (a.length == 2) {
            a[2] = 0;
        }

        if (a[0] > b[0]) return true;
        if (a[0] < b[0]) return false;

        if (a[1] > b[1]) return true;
        if (a[1] < b[1]) return false;

        if (a[2] > b[2]) return true;
        if (a[2] < b[2]) return false;

        return true;
    }

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);