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
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
其他回答
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"
semver
npm使用的语义版本解析器。
$ npm install semver
var semver = require('semver');
semver.diff('3.4.5', '4.3.7') //'major'
semver.diff('3.4.5', '3.3.7') //'minor'
semver.gte('3.4.8', '3.4.7') //true
semver.ltr('3.4.8', '3.4.7') //false
semver.valid('1.2.3') // '1.2.3'
semver.valid('a.b.c') // null
semver.clean(' =v1.2.3 ') // '1.2.3'
semver.satisfies('1.2.3', '1.x || >=2.5.0 || 5.0.0 - 7.2.3') // true
semver.gt('1.2.3', '9.8.7') // false
semver.lt('1.2.3', '9.8.7') // true
var versions = [ '1.2.3', '3.4.5', '1.0.2' ]
var max = versions.sort(semver.rcompare)[0]
var min = versions.sort(semver.compare)[0]
var max = semver.maxSatisfying(versions, '*')
语义版本控制链接:https://www.npmjs.com/package/semver#prerelease-identifiers
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"]
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?
}
