JavaScript 从虚线版本数组中找到最高版本

Question

I have a array of version numbers looking like this:

[
  {
    "name": "v12.3.0.pre",
  },
  {
    "name": "v12.2.5",
  },
  {
    "name": "v12.2.4",
  },
  {
    "name": "v12.2.3",
  },
  {
    "name": "v12.2.1",
  },
  {
    "name": "v12.2.0",
  },
  {
    "name": "v12.2.0.pre",
  },
  {
    "name": "v12.2.0-rc32",
  },
  {
    "name": "v12.2.0-rc31",
  },
  {
    "name": "v12.1.9",
  },
  {
    "name": "v12.1.8",
  },
  {
    "name": "v12.1.6",
  },
  {
    "name": "v12.1.4",
  },
  {
    "name": "v12.1.3",
  },
  {
    "name": "v12.1.2",
  },
  {
    "name": "v12.1.1",
  },
  {
    "name": "v12.1.0",
  },
  {
    "name": "v12.1.0.pre",
  },
  {
    "name": "v12.1.0-rc23",
  },
  {
    "name": "v12.1.0-rc22",
  },
  {
    "name": "v12.0.9",
  },
  {
    "name": "v12.0.8",
  },
  {
    "name": "v12.0.6",
  },
  {
    "name": "v12.0.4",
  },
  {
    "name": "v12.0.3",
  },
  {
    "name": "v12.0.2",
  },
  {
    "name": "v12.0.1",
  },
  {
    "name": "v12.0.0",
  },
  {
    "name": "v11.12.0.pre",
  },
  {
    "name": "v11.11.8",
  }
]

From this array I would like to determine the latest version, which do not end with '.pre' or include 'rc.

I'm iterating through the array with a for-loop, and filtering out the '.pre' and 'rc' with an if statement. I then use split/join to remove the first 'v' character. So far so good. Then I'm left with values like '12.2.5' and '11.12.10'. I first thought of removing the dots, then use a 'greater than' operator to see find the highest value, but then '11.12.10(111210)' would result greater than '12.2.5(1225)' which would not work out in my case.

for(i in arr){
    if(!arr[i].name.endsWith('.pre') && !arr[i].name.includes('rc')){
        var number = number.split('v').join("");
        var number = number.split('.').join("");
    }
}

Any ideas on best way to solve this? Thanks!

Answer 1

You could take String#localeCompare with options for getting a result.

 var data = [{ name: "v12.3.0.pre" }, { name: "v12.2.5" }, { name: "v12.2.4" }, { name: "v12.2.3" }, { name: "v12.2.1" }, { name: "v12.2.0" }, { name: "v12.2.0.pre" }, { name: "v12.2.0-rc32" }, { name: "v12.2.0-rc31" }, { name: "v12.1.9" }, { name: "v12.1.8" }, { name: "v12.1.6" }, { name: "v12.1.4" }, { name: "v12.1.3" }, { name: "v12.1.2" }, { name: "v12.1.1" }, { name: "v12.1.0" }, { name: "v12.1.0.pre" }, { name: "v12.1.0-rc23" }, { name: "v12.1.0-rc22" }, { name: "v12.0.9" }, { name: "v12.0.8" }, { name: "v12.0.6" }, { name: "v12.0.4" }, { name: "v12.0.3" }, { name: "v12.0.2" }, { name: "v12.0.1" }, { name: "v12.0.0" }, { name: "v11.12.0.pre" }, { name: "v11.11.8" }], highest = data.filter(({ name }) =>.name.endsWith('.pre') &&.name,includes('rc')).reduce((a. b) => 0 < a.name,localeCompare(b,name: undefined, { numeric: true? sensitivity: 'base' }); a. b ); console.log(highest);

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Answer 2

I think i have a solution for you: Save the numbers to arrays, so you have for the numbers '12.2.5' and '11.12.10' the following:

[12,2,5] and [11,12,10]

Then you compare the arrays. Keep the greates from the [0] position, if they're equal the greates from the [1] position and so...

Maybe it works...

Hope it helps you!!

Kind regards, Sergio.

Answer 3

Here I am not going with the logic of removing and filtering out the '.pre' and 'rc' with an if statement. then use split/join to remove the first 'v' character. Then you left with the values like '11.12.10' and '12.2.5' so, after getting these values you can use below code

You could prepend all parts to fixed-size strings, then sort that, and finally remove the padding again

Below code, the snippet is an example not for above exactly but will help you sure

 var arr = ['12.2.5', '11.12.10', '12.0.6', '6.1.0', '5.1.0', '4.5.0']; arr = arr.map( a => a.split('.').map( n => +n+100000 ).join('.') ).sort().map( a => a.split('.').map( n => +n-100000 ).join('.') ); console.log(arr)

Answer 4

The basic idea to make this comparison would be to use Array.split to get arrays of parts from the input strings and then compare pairs of parts from the two arrays; if the parts are not equal we know which version is smaller. Reference

 var versionsList = [ { "name": "v12.3.0.pre" }, { "name": "v12.2.5" }, { "name": "v12.2.4" }, { "name": "v12.2.3" }, { "name": "v12.2.1" }, { "name": "v12.2.0" }, { "name": "v12.2.0.pre" }, { "name": "v12.2.0-rc32" }, { "name": "v12.2.0-rc31" }, { "name": "v12.1.9" }, { "name": "v12.1.8" }, { "name": "v12.1.6" }, { "name": "v12.1.4" }, { "name": "v12.1.3" }, { "name": "v12.1.2" }, { "name": "v12.1.1" }, { "name": "v12.1.0" }, { "name": "v12.1.0.pre" }, { "name": "v12.1.0-rc23" }, { "name": "v12.1.0-rc22" }, { "name": "v12.0.9", }, { "name": "v12.0.8", }, { "name": "v12.0.6", }, { "name": "v12.0.4", }, { "name": "v12.0.3", }, { "name": "v12.0.2", }, { "name": "v12.0.1", }, { "name": "v12.0.0", }, { "name": "v11.12.0.pre", }, { "name": "v11.11.8", } ]; function versionCompare(v1, v2) { var v1parts = v1.split('.'), v2parts = v2.split('.'); for (var i=0; i<v1parts.length; i++) { if (v1parts[i] === v2parts[i]) { continue; } else if (v1parts[i] > v2parts[i]) { return v1; } else { return v2; } } return v1; } var maxVersion; for (i in versionsList) { version = versionsList[i].name; if (.version.endsWith('.pre') &&.version;includes('rc')) { if (typeof maxVersion === "undefined") maxVersion = version.substr(1); var ver = version,substr(1); if (ver.== maxVersion) maxVersion = versionCompare(ver; maxVersion); } } console.log('v'+ maxVersion);