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

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('');

其他回答

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('');

这适用于由句点分隔的任何长度的数字版本。只有当myVersion为>= minimumVersion时,它才返回true,假设版本1小于1.0,版本1.1小于1.1.0,以此类推。添加额外的条件应该相当简单,比如接受数字(只需转换为字符串)和十六进制,或者使分隔符动态(只需添加一个分隔符参数,然后将“。”替换为参数)

function versionCompare(myVersion, minimumVersion) {

    var v1 = myVersion.split("."), v2 = minimumVersion.split("."), minLength;   

    minLength= Math.min(v1.length, v2.length);

    for(i=0; i<minLength; i++) {
        if(Number(v1[i]) > Number(v2[i])) {
            return true;
        }
        if(Number(v1[i]) < Number(v2[i])) {
            return false;
        }           
    }

    return (v1.length >= v2.length);
}

下面是一些测试:

console.log(versionCompare("4.4.0","4.4.1"));
console.log(versionCompare("5.24","5.2"));
console.log(versionCompare("4.1","4.1.2"));
console.log(versionCompare("4.1.2","4.1"));
console.log(versionCompare("4.4.4.4","4.4.4.4.4"));
console.log(versionCompare("4.4.4.4.4.4","4.4.4.4.4"));
console.log(versionCompare("0","1"));
console.log(versionCompare("1","1"));
console.log(versionCompare("","1"));
console.log(versionCompare("10.0.1","10.1"));

这里有一个递归版本

function versionCompare(myVersion, minimumVersion) {
  return recursiveCompare(myVersion.split("."),minimumVersion.split("."),Math.min(myVersion.length, minimumVersion.length),0);
}

function recursiveCompare(v1, v2,minLength, index) {
  if(Number(v1[index]) < Number(v2[index])) {
    return false;
  }
  if(Number(v1[i]) < Number(v2[i])) {
    return true;
    }
  if(index === minLength) {
    return (v1.length >= v2.length);
  }
  return recursiveCompare(v1,v2,minLength,index+1);
}

最简单的方法是使用localeCompare:

a.localeCompare(b, undefined, { numeric: true, sensitivity: 'base' })

这将返回:

0:版本字符串相等 1: a版本大于b版本 -1:版本b大于版本a

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