比较不同条件下的功能:

const compareVer = (ver1, middle, ver2) => {
  const res = new Intl.Collator("en").compare(ver1, ver2)
  let comp

  switch (middle) {
    case "=":
      comp = 0 === res
      break

    case ">":
      comp = 1 === res
      break

    case ">=":
      comp = 1 === res || 0 === res
      break

    case "<":
      comp = -1 === res
      break

    case "<=":
      comp = -1 === res || 0 === res
      break
  }

  return comp
}

console.log(compareVer("1.0.2", "=", "1.0.2")) // true
console.log(compareVer("1.0.3", ">", "1.0.2")) // true
console.log(compareVer("1.0.1", ">=", "1.0.2")) // false
console.log(compareVer("1.0.3", ">=", "1.0.2")) // true
console.log(compareVer("1.0.1", "<", "1.0.2")) // true
console.log(compareVer("1.0.1", "<=", "1.0.2")) // true

// Return 1 if a > b
// Return -1 if a < b
// Return 0 if a == b
function compare(a, b) {
    if (a === b) {
       return 0;
    }

    var a_components = a.split(".");
    var b_components = b.split(".");

    var len = Math.min(a_components.length, b_components.length);

    // loop while the components are equal
    for (var i = 0; i < len; i++) {
        // A bigger than B
        if (parseInt(a_components[i]) > parseInt(b_components[i])) {
            return 1;
        }

        // B bigger than A
        if (parseInt(a_components[i]) < parseInt(b_components[i])) {
            return -1;
        }
    }

    // If one's a prefix of the other, the longer one is greater.
    if (a_components.length > b_components.length) {
        return 1;
    }

    if (a_components.length < b_components.length) {
        return -1;
    }

    // Otherwise they are the same.
    return 0;
}

console.log(compare("1", "2"));
console.log(compare("2", "1"));

console.log(compare("1.0", "1.0"));
console.log(compare("2.0", "1.0"));
console.log(compare("1.0", "2.0"));
console.log(compare("1.0.1", "1.0"));

// Returns true if v1 is bigger than v2, and false if otherwise.
function isNewerThan(v1, v2) {
      v1=v1.split('.');
      v2=v2.split('.');
      for(var i = 0; i<Math.max(v1.length,v2.length); i++){
        if(v1[i] == undefined) return false; // If there is no digit, v2 is automatically bigger
        if(v2[i] == undefined) return true; // if there is no digit, v1 is automatically bigger
        if(v1[i] > v2[i]) return true;
        if(v1[i] < v2[i]) return false;
      }
      return false; // Returns false if they are equal
    }

你可以使用JavaScript的localeCompare方法:

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

这里有一个例子:

"1.1".localeCompare("2.1.1"， undefined， {numeric: true}) => -1

"1.0.0".localeCompare("1.0"， undefined， {numeric: true}) =>

"1.0.0".localeCompare("1.0.0"， undefined， {numeric: true}) => 0

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