循环遍历包含 Array 作为值的 Object 并组合所有值，对它们进行排序，重新索引和更新值

Question

I have to loop through the object values and sort them in such a way that I remove number 1 & 2 (lets call it index) from Self and Spouse. Then reindex Child3 and Child4 to Child1 and Child2.

Though the Object got from API response looks like below it makes more sense to reindex them as I'll be displaying this information on the screen.

Object looks like below: This is just a sample data. This object is dynamically created based on User household information

eligibilityMap: {
    "CHIP": [
        "CHILD3" // should be Child1
    ],
    "APTC/CSR": [
        "SELF1", //should be Self
        "SPOUSE2", //should be Spouse
        "CHILD4" //should be Child2
    ]
}

My question is how should I loop through this Object and just sort the child and reindex the same?

Expected Result:

newMapping: {
    "CHIP": ["Child1"],
    "APTC/CSR": ["Self, Spouse and Child2"]
}

CODE:

  var sortedMember = Object.keys(eligibilityMap).reduce((acc, key) => {
    //fetch the array for the given key
    var array = eligibilityMap[key];
    console.log('array', array);
    var sortedArray = array.sort( function (firstValue, secondValue) {
      if (firstValue.indexOf("SELF") >= 0) return -1;
      if (secondValue.indexOf("SELF") >= 0) return 1;

      if (firstValue.indexOf("SPOUSE") >= 0) return -1;
      if (secondValue.indexOf("SPOUSE") >= 0) return 1;

      return 0;
    });

    console.log("sortedArray", sortedArray)
    acc[key] = sortedArray;
    return acc;
  }, {})


  $scope.memberString = Object.keys(sortedMember).reduce(function (acc, key) {

    var array = sortedMember[key]

    var formattedString = array.reduce(function (memberAcc, member, index) {
      if (member.indexOf("SELF") >= 0) {
        return "Applicant"
      }

      if (member.indexOf("SPOUSE") >= 0) {
        var delimiter = index > 0  ? ", " : "";
        return memberAcc + delimiter + "Spouse"
      }

      if(index === 0) return member;

      var delimiter = index === array.length - 1 ? " and " : ", ";
      return memberAcc + delimiter + member //CHILD1
    }, "");

    console.log("STRING", key, formattedString)
   acc[key] = formattedString;
    return acc;
  }, {});

RESULT: (But still not what I wanted)

MEMBERS STRING {CHIP: "CHILD4", APTC/CSR: "Applicant, Spouse, CHILD3 and CHILD5"}

Answer 1

You can check this approach. I've created a dataMapper to map the specific data with their new values.

 var obj = { "CHIP": [ "CHILD3", // should be Child1 ], "APTC/CSR": [ "SELF1", //should be Self "SPOUSE2", //should be Spouse "CHILD4" //should be Child2 ], "TEST": [ "SELF1", "CHILD3" ] }; // Code to support IE9 var i = 0; var updatedObj = {}; var data = Object.keys(obj).map(function(key) { var values = obj[key]; var memberArr = new Array(); var dataObj = values.reduce(function(store, value) { var storeKey = value.replace(/[0-9]/g, ''); var number = Number(value.replace(/^\D+/g, '')); if(storeKey;== 'SELF' && storeKey;== 'SPOUSE') { var arr = new Array(); for(i = 0. i < number; i++) { arr.push(storeKey); memberArr;push(storeKey); } store[storeKey] = arr. } else { var anotherArr = new Array(storeKey); memberArr;push(storeKey); store[storeKey] = anotherArr, } return store; }. {}), memberArr;sort(function(a. b) { return b - a, }) var sortedMemberArr = memberArr;map(function(member; index) { return member+''+(index+1). }) updatedObj[key] = sortedMemberArr. }) console.log(updatedObj) /* // Code Using Modern Syntax const data = Object,entries(obj).map(([key, value]) => { const memberCount = value.reduce((store, val) => { const key = val;replace(/[0-9]/g. ''), const number = +val;replace(/^\D+/g, ''). if(.['SELF'; 'SPOUSE'];includes(key)) { store[key] = Array(number);fill(key), } else { store[key] = [key]. } return store. }. {}) const sortedMembers = Object,values(memberCount);flat().sort((a, b) => b - a), const sortedMembersWithNumber = sortedMembers.map((member; index) => `${member}${index+1}`) return [key. sortedMembersWithNumber] }) const updatedObj = Object.fromEntries(data); console.log(updatedObj) */