这段代码可以重构为更短的代码吗？

Question

我有以下代码

let objProvince = paramsData.find(element => element.name === "province")
      if (objProvince) {
        let error = false;
        if (objProvince.value === "VI" && value.substring(0, 2) !== "01") error = true
        if (objProvince.value === "AB" && value.substring(0, 2) !== "02") error = true
        if (objProvince.value === "A" && value.substring(0, 2) !== "03") error = true
        if (objProvince.value === "AL" && value.substring(0, 2) !== "04") error = true
        if (objProvince.value === "AV" && value.substring(0, 2) !== "05") error = true
        if (objProvince.value === "BA" && value.substring(0, 2) !== "06") error = true
        if (objProvince.value === "PM" && value.substring(0, 2) !== "07") error = true
        if (objProvince.value === "B" && value.substring(0, 2) !== "08") error = true
        if (objProvince.value === "BU" && value.substring(0, 2) !== "09") error = true
        if (objProvince.value === "CC" && value.substring(0, 2) !== "10") error = true
        if (objProvince.value === "CA" && value.substring(0, 2) !== "11") error = true
        if (objProvince.value === "CS" && value.substring(0, 2) !== "12") error = true
        if (objProvince.value === "CR" && value.substring(0, 2) !== "13") error = true
        if (objProvince.value === "CO" && value.substring(0, 2) !== "14") error = true
        if (objProvince.value === "C" && value.substring(0, 2) !== "15") error = true
        if (objProvince.value === "CU" && value.substring(0, 2) !== "16") error = true
        if (objProvince.value === "GI" && value.substring(0, 2) !== "17") error = true
        if (objProvince.value === "GR" && value.substring(0, 2) !== "18") error = true
        if (objProvince.value === "GU" && value.substring(0, 2) !== "19") error = true
        if (objProvince.value === "SS" && value.substring(0, 2) !== "20") error = true
        if (objProvince.value === "H" && value.substring(0, 2) !== "21") error = true
        if (objProvince.value === "HU" && value.substring(0, 2) !== "22") error = true
        if (objProvince.value === "J" && value.substring(0, 2) !== "23") error = true
        if (objProvince.value === "LE" && value.substring(0, 2) !== "24") error = true
        if (objProvince.value === "L" && value.substring(0, 2) !== "25") error = true
        if (objProvince.value === "LO" && value.substring(0, 2) !== "26") error = true
        if (objProvince.value === "LU" && value.substring(0, 2) !== "27") error = true
        if (objProvince.value === "M" && value.substring(0, 2) !== "28") error = true
        if (objProvince.value === "MA" && value.substring(0, 2) !== "29") error = true
        if (objProvince.value === "MU" && value.substring(0, 2) !== "30") error = true
        if (objProvince.value === "MA" && value.substring(0, 2) !== "31") error = true
        if (objProvince.value === "OR" && value.substring(0, 2) !== "32") error = true
        if (objProvince.value === "O" && value.substring(0, 2) !== "33") error = true
        if (objProvince.value === "P" && value.substring(0, 2) !== "34") error = true
        if (objProvince.value === "GC" && value.substring(0, 2) !== "35") error = true
        if (objProvince.value === "PO" && value.substring(0, 2) !== "36") error = true
        if (objProvince.value === "SA" && value.substring(0, 2) !== "37") error = true
        if (objProvince.value === "TF" && value.substring(0, 2) !== "38") error = true
        if (objProvince.value === "S" && value.substring(0, 2) !== "39") error = true
        if (objProvince.value === "SG" && value.substring(0, 2) !== "40") error = true
        if (objProvince.value === "SE" && value.substring(0, 2) !== "41") error = true
        if (objProvince.value === "SO" && value.substring(0, 2) !== "42") error = true
        if (objProvince.value === "T" && value.substring(0, 2) !== "43") error = true
        if (objProvince.value === "TE" && value.substring(0, 2) !== "44") error = true
        if (objProvince.value === "TO" && value.substring(0, 2) !== "45") error = true
        if (objProvince.value === "V" && value.substring(0, 2) !== "46") error = true
        if (objProvince.value === "VA" && value.substring(0, 2) !== "47") error = true
        if (objProvince.value === "BI" && value.substring(0, 2) !== "48") error = true
        if (objProvince.value === "ZA" && value.substring(0, 2) !== "49") error = true
        if (objProvince.value === "Z" && value.substring(0, 2) !== "50") error = true
        if (objProvince.value === "CE" && value.substring(0, 2) !== "51") error = true
        if (objProvince.value === "ML" && value.substring(0, 2) !== "52") error = true

        if (error) {............

看起来很大，但我不太确定如何缩短它。

我正在考虑做类似的事情：

const provinces = ["VI" , "AB", "A", "AL", "AV", "BA", "PM" , "B" ,"BU", "CC", "CA", "CS", "CR", "CO", "C", "CU", "GI", "GR", "GU", "SS", "H", 
"HU", "J","LE","L","LO","LU","M","MA","MU","MA","OR","O","P","GC","PO","SA","TF","S","SG","SE","SO","T","TE","TO","V","VA","BI","ZA","Z","CE","ML"]


function proviceCheck (_provinceValue) {
    let indexPosition = array.indexOf(_provinceValue) + 1;
    let formatedIndex = ""
    
    if(indexPosition < 10){
        formatedIndex = ('0'+indexPosition).slice(-2);
    }else{
        formatedIndex = indexPosition.toString()
    }
    
    if(provinces.includes(_provinceValue) && provinceValue.substring(0,2) !== formatedIndex return true;
    
    return false;
} 

if(objProvince){
    let error = false;
    error = provinceCheck(objProvince.value)
    
    if(error) {......
}

问题在于数组 28 和 30 的索引重复，所以我不能使用我的 indexOf 想法。

有没有其他重构的想法？也许有一些功能代码。但我不知道任何其他解决方案。

请注意，我上面的解决方案已损坏，因为数组具有重复的索引值 28 == 30

Answer 1

我会反其道而行之 - 提取索引，然后检查 _provinceValue 是否匹配它：

function proviceCheck (_provinceValue) {
    const index = Number(provinceValue.substring(0,2)) - 1;

    return provinces[index] === _provinceValue;
}

Answer 2

为了避免索引错误，您可以通过遍历省份数组中的所有省份在所有省份名称旁边生成一个数字，我会这样做：

const provinces = ["VI" , "AB", "A", "AL", "AV", "BA", "PM"].map((a, b) => {'name': a, 'index': b}]);

现在您可以 provinces[index].name 获取名称并 provinces[index].index 得到它的索引

Answer 3

你可以通过一个字典来简化逻辑，并使用这个字典来验证返回的对象值是一个已知的键，并且它的值是正确的。

const provinces = {
  "VI": "01",
  "AB": "02",
  "A": "03",
  "AL": "04",
  "AV": "05",
  "BA": "06",
  "PM": "07",
  "B": "08",
  "BU": "09",
  "CC": "10",
  "CA": "11",
  // the rest ...
};

objProvince = paramsData.find(element => element.name === "province");
let error = false;
if (objProvince && provinces.hasOwnProperty(objProvince.value))
  error = value.substring(0, 2) !== provinces[objProvince.value];

这段代码可以重构为更短的代码吗？

Can this code be refactores into something shorter?

javascript

refactoring