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
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)
这是另一种递归算法。
这段代码只使用了Array。shift和递归，这意味着它可以在Internet Explorer 6+中运行。如果你有任何疑问，你可以访问我的GitHub页面。
(function(root, factory) {
if (typeof exports === 'object') {
return module.exports = factory();
} else if (typeof define === 'function' && define.amd) {
return define(factory);
} else {
return root.compareVer = factory();
}
})(this, function() {
'use strict';
var _compareVer;
_compareVer = function(newVer, oldVer) {
var VER_RE, compareNum, isTrue, maxLen, newArr, newLen, newMatch, oldArr, oldLen, oldMatch, zerofill;
VER_RE = /(\d+\.){1,9}\d+/;
if (arguments.length !== 2) {
return -100;
}
if (typeof newVer !== 'string') {
return -2;
}
if (typeof oldVer !== 'string') {
return -3;
}
newMatch = newVer.match(VER_RE);
if (!newMatch || newMatch[0] !== newVer) {
return -4;
}
oldMatch = oldVer.match(VER_RE);
if (!oldMatch || oldMatch[0] !== oldVer) {
return -5;
}
newVer = newVer.replace(/^0/, '');
oldVer = oldVer.replace(/^0/, '');
if (newVer === oldVer) {
return 0;
} else {
newArr = newVer.split('.');
oldArr = oldVer.split('.');
newLen = newArr.length;
oldLen = oldArr.length;
maxLen = Math.max(newLen, oldLen);
zerofill = function() {
newArr.length < maxLen && newArr.push('0');
oldArr.length < maxLen && oldArr.push('0');
return newArr.length !== oldArr.length && zerofill();
};
newLen !== oldLen && zerofill();
if (newArr.toString() === oldArr.toString()) {
if (newLen > oldLen) {
return 1;
} else {
return -1;
}
} else {
isTrue = -1;
compareNum = function() {
var _new, _old;
_new = ~~newArr.shift();
_old = ~~oldArr.shift();
_new > _old && (isTrue = 1);
return _new === _old && newArr.length > 0 && compareNum();
};
compareNum();
return isTrue;
}
}
};
return _compareVer;
});
好吧，我希望这段代码能帮助到一些人。
下面是测试。
console.log(compareVer("0.0.2","0.0.1"));//1
console.log(compareVer("0.0.10","0.0.1")); //1
console.log(compareVer("0.0.10","0.0.2")); //1
console.log(compareVer("0.9.0","0.9")); //1
console.log(compareVer("0.10.0","0.9.0")); //1
console.log(compareVer("1.7", "1.07")); //1
console.log(compareVer("1.0.07", "1.0.007")); //1
console.log(compareVer("0.3","0.3")); //0
console.log(compareVer("0.0.3","0.0.3")); //0
console.log(compareVer("0.0.3.0","0.0.3.0")); //0
console.log(compareVer("00.3","0.3")); //0
console.log(compareVer("00.3","00.3")); //0
console.log(compareVer("01.0.3","1.0.3")); //0
console.log(compareVer("1.0.3","01.0.3")); //0
console.log(compareVer("0.2.0","1.0.0")); //-1
console.log(compareVer('0.0.2.2.0',"0.0.2.3")); //-1
console.log(compareVer('0.0.2.0',"0.0.2")); //-1
console.log(compareVer('0.0.2',"0.0.2.0")); //-1
console.log(compareVer("1.07", "1.7")); //-1
console.log(compareVer("1.0.007", "1.0.07")); //-1
console.log(compareVer()); //-100
console.log(compareVer("0.0.2")); //-100
console.log(compareVer("0.0.2","0.0.2","0.0.2")); //-100
console.log(compareVer(1212,"0.0.2")); //-2
console.log(compareVer("0.0.2",1212)); //-3
console.log(compareVer('1.abc.2',"1.0.2")); //-4
console.log(compareVer('1.0.2',"1.abc.2")); //-5
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();
把它转换成一个数字，然后比较。 假设每个主要/次要/补丁使用不超过3个数字，并且没有标签 (像这个1.12.042)
const versionNumber = +versionString
.split('.')
.map(v => '000' + v)
.map(v => v.slice(-3))
.join('');
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?
}
