下面是一个适合Array使用的coffeescript实现。从其他答案中得到启发:

# Returns > 0 if v1 > v2 and < 0 if v1 < v2 and 0 if v1 == v2
compareVersions = (v1, v2) ->
  v1Parts = v1.split('.')
  v2Parts = v2.split('.')
  minLength = Math.min(v1Parts.length, v2Parts.length)
  if minLength > 0
    for idx in [0..minLength - 1]
      diff = Number(v1Parts[idx]) - Number(v2Parts[idx])
      return diff unless diff is 0
  return v1Parts.length - v2Parts.length

我的答案比这里的大多数答案都要简洁

/**
 * Compare two semver versions. Returns true if version A is greater than
 * version B
 * @param {string} versionA
 * @param {string} versionB
 * @returns {boolean}
 */
export const semverGreaterThan = function(versionA, versionB){
  var versionsA = versionA.split(/\./g),
    versionsB = versionB.split(/\./g)
  while (versionsA.length || versionsB.length) {
    var a = Number(versionsA.shift()), b = Number(versionsB.shift())
    if (a == b)
      continue
    return (a > b || isNaN(b))
  }
  return false
}

我必须比较我的扩展版本，但我没有 在这里找到一个可行的解决方案。在比较1.89 > 1.9或1.24.1 == 1.240.1时，几乎所有提议的期权都被打破了

这里，我从仅在最后的记录1.1 == 1.10和1.10.1 > 1.1.1中0下降的事实开始

compare_version = (new_version, old_version) => {
    new_version = new_version.split('.');
    old_version = old_version.split('.');
    for(let i = 0, m = Math.max(new_version.length, old_version.length); i<m; i++){
        //compare text
        let new_part = (i<m-1?'':'.') + (new_version[i] || 0)
        ,   old_part = (i<m-1?'':'.') + (old_version[i] || 0);
        //compare number (I don’t know what better)
      //let new_part = +((i<m-1?0:'.') + new_version[i]) || 0
      //,   old_part = +((i<m-1?0:'.') + old_version[i]) || 0;
        //console.log(new_part, old_part);
        if(old_part > new_part)return 0;    //change to -1 for sort the array
        if(new_part > old_part)return 1
    }
    return 0
};
compare_version('1.0.240.1','1.0.240.1');   //0
compare_version('1.0.24.1','1.0.240.1');    //0
compare_version('1.0.240.89','1.0.240.9');  //0
compare_version('1.0.24.1','1.0.24');       //1

我不是一个大专家，但我构建了简单的代码来比较两个版本，将第一个返回值更改为-1以对版本数组进行排序

['1.0.240', '1.0.24', '1.0.240.9', '1.0.240.89'].sort(compare_version)
//results ["1.0.24", "1.0.240", "1.0.240.89", "1.0.240.9"]

和短版本的比较全字符串

c=e=>e.split('.').map((e,i,a)=>e[i<a.length-1?'padStart':'padEnd'](5)).join('');

//results "    1    0  2409    " > "    1    0  24089   "

c('1.0.240.9')>c('1.0.240.89')              //true

如果您有意见或改进，请不要犹豫提出建议。

你不能把它们转换成数字，然后按大小排序吗?在长度< 4的数的1后面加上0

在主机上玩:

$(["1.0.0.0", "1.0.1.0", "2.0.0.0", "2.0.0.1", "2.0.1", "3.0"]).each(function(i,e) {
    var n =   e.replace(/\./g,"");
    while(n.length < 4) n+="0" ; 
    num.push(  +n  )
});

版本越大，数字越大。 编辑:可能需要调整，以考虑更大的版本系列

如果数组包含数字，则可以对数组进行比较。 转换为没有点的数组并比较它们。

函数VersionCompare(version1, version2){ var regExStrip0 = /(\.0+)+$/; var segmentsA = version1。替换(regExStrip0”).split (' . '); var segmentsB = version2。替换(regExStrip0”).split (' . '); if(segmentsA > segmentsB){ 返回1 }else if (segmentsA< segmentsB){ 返回1 其他}{ 返回0 ｝ ｝

一个非常简单的方法:

function compareVer(previousVersion, currentVersion) {
 try {
    const [prevMajor, prevMinor = 0, prevPatch = 0] = previousVersion.split('.').map(Number);
    const [curMajor, curMinor = 0, curPatch = 0] = currentVersion.split('.').map(Number);

    if (curMajor > prevMajor) {
      return 'major update';
    }
    if (curMajor < prevMajor) {
      return 'major downgrade';
    }
    if (curMinor > prevMinor) {
      return 'minor update';
    }
    if (curMinor < prevMinor) {
      return 'minor downgrade';
    }
    if (curPatch > prevPatch) {
      return 'patch update';
    }
    if (curPatch < prevPatch) {
      return 'patch downgrade';
    }
    return 'same version';
  } catch (e) {
    return 'invalid format';
  }
}

输出:

compareVer("3.1", "3.1.1") // patch update
compareVer("3.1.1", "3.2") // minor update
compareVer("2.1.1", "1.1.1") // major downgrade
compareVer("1.1.1", "1.1.1") // same version