Question

我创建了一个函数，它将返回一个没有重复分支的对象和每个分支的总销售额。

    var salesPersons = [
        {"name": {"first": "Aleen", "last" :"Atkins"}, "sales" : "$2314", "branch": "Austin, TX"},
        {"name": {"first": "Alvaro", "last" :"Angelos"}, "sales" : "$1668", "branch": "New York City, NY"},
        {"name": {"first": "Denese", "last" :"Dossett"}, "sales" : "$9248", "branch": "Austin, TX"},
        {"name": {"first": "Douglas", "last" :"Denney"}, "sales" : "$5058", "branch": "New York City, NY"},
        {"name": {"first": "Earline", "last" :"Erickson"}, "sales" : "$18876", "branch": "New York City, NY"},
        {"name": {"first": "Herman", "last" :"Hazell"}, "sales" : "$2746", "branch": "New York City, NY"},
        {"name": {"first": "Homer", "last" :"Hirth"}, "sales" : "$474", "branch": "Orlando, FL"},
        {"name": {"first": "Hwa", "last" :"Heidt"}, "sales" : "$9607", "branch": "San Francisco, CA"},
        {"name": {"first": "Hyon", "last" :"Hampshire"}, "sales" : "$13598", "branch": "New York City, NY"},
        {"name": {"first": "Issac", "last" :"Ingerson"}, "sales" : "$5225", "branch": "Austin, TX"},
        {"name": {"first": "Jeraldine", "last" :"Joplin"}, "sales" : "$2891", "branch": "New York City, NY"},
        {"name": {"first": "Jin", "last" :"Jeffrey"}, "sales" : "$14402", "branch": "Austin, TX"},
        {"name": {"first": "Joleen", "last" :"Jolin"}, "sales" : "$15736", "branch": "Austin, TX"},
        {"name": {"first": "Jude", "last" :"Jarrett"}, "sales" : "$7557", "branch": "San Francisco, CA"},
        {"name": {"first": "Magda", "last" :"Mireles"}, "sales" : "$1498", "branch": "Austin, TX"},
        {"name": {"first": "Mistie", "last" :"Montealegre"}, "sales" : "$6920", "branch": "New York City, NY"},
        {"name": {"first": "Nancy", "last" :"Napoli"}, "sales" : "$5255", "branch": "New York City, NY"},
        {"name": {"first": "Regine", "last" :"Rohrbaugh"}, "sales" : "$7881", "branch": "Orlando, FL"},
        {"name": {"first": "Rolando", "last" :"Riebel"}, "sales" : "$8573", "branch": "San Francisco, CA"},
        {"name": {"first": "Scarlett", "last" :"Stagg"}, "sales" : "$7126", "branch": "Orlando, FL"},
        {"name": {"first": "Sherron", "last" :"Strawn"}, "sales" : "$8848", "branch": "New York City, NY"},
        {"name": {"first": "Susan", "last" :"Shilling"}, "sales" : "$8542", "branch": "Seattle, WA"},
        {"name": {"first": "Tama", "last" :"Tworek"}, "sales" : "$9200", "branch": "Seattle, WA"},
        {"name": {"first": "Tonisha", "last" :"Taunton"}, "sales" : "$5219", "branch": "Orlando, FL"},
        {"name": {"first": "Vergie", "last" :"Villescas"}, "sales" : "$8712", "branch": "New York City, NY"}
    ];

    function convertSalesToInt(person){
      var salesInt = person.sales.slice(1);
      return parseInt(salesInt, 10);
    }

    function salesPerBranch(salesPersons){
    var myObj = {};
    var uniq = [];
    var total = 0;

    for(var i = 0; i < salesPersons.length; i++){
      var splitTeam = salesPersons[i].branch;
      if(uniq.indexOf(salesPersons) < 0){
        myObj[salesPersons] = total+= convertSalesToInt(salesPersons[i]);
      }
    }

    return myObj;


    }

    console.log(salesPerBranch(salesPersons));
    /* { 'Austin, TX': 110898,
  'New York City, NY': 187174,
  'Orlando, FL': 178462,
  'San Francisco, CA': 139527,
  'Seattle, WA': 173243 } */

答案应如下所示：

{
    'Austin, TX': 48423,
    'New York City, NY': 74572,
    'Orlando, FL': 20700,
    'San Francisco, CA': 25737,
    'Seattle, WA': 17742
}

我现在被困住的唯一部分是我为每个分支添加总销售额的部分。现在它没有提高适量。这部分负责：

if(uniq.indexOf(salesPersons) < 0){
        myObj[salesPersons] = total+= convertSalesToInt(salesPersons[i]);
      }

任何知道我的错误以及如何解决错误的人？

谢谢！

Answer 1

＆＃13;

var salesPersons = [
    {"name": {"first": "Aleen", "last" :"Atkins"}, "sales" : "$2314", "branch": "Austin, TX"},
    {"name": {"first": "Alvaro", "last" :"Angelos"}, "sales" : "$1668", "branch": "New York City, NY"},
    {"name": {"first": "Denese", "last" :"Dossett"}, "sales" : "$9248", "branch": "Austin, TX"},
    {"name": {"first": "Douglas", "last" :"Denney"}, "sales" : "$5058", "branch": "New York City, NY"},
    {"name": {"first": "Earline", "last" :"Erickson"}, "sales" : "$18876", "branch": "New York City, NY"},
    {"name": {"first": "Herman", "last" :"Hazell"}, "sales" : "$2746", "branch": "New York City, NY"},
    {"name": {"first": "Homer", "last" :"Hirth"}, "sales" : "$474", "branch": "Orlando, FL"},
    {"name": {"first": "Hwa", "last" :"Heidt"}, "sales" : "$9607", "branch": "San Francisco, CA"},
    {"name": {"first": "Hyon", "last" :"Hampshire"}, "sales" : "$13598", "branch": "New York City, NY"},
    {"name": {"first": "Issac", "last" :"Ingerson"}, "sales" : "$5225", "branch": "Austin, TX"},
    {"name": {"first": "Jeraldine", "last" :"Joplin"}, "sales" : "$2891", "branch": "New York City, NY"},
    {"name": {"first": "Jin", "last" :"Jeffrey"}, "sales" : "$14402", "branch": "Austin, TX"},
    {"name": {"first": "Joleen", "last" :"Jolin"}, "sales" : "$15736", "branch": "Austin, TX"},
    {"name": {"first": "Jude", "last" :"Jarrett"}, "sales" : "$7557", "branch": "San Francisco, CA"},
    {"name": {"first": "Magda", "last" :"Mireles"}, "sales" : "$1498", "branch": "Austin, TX"},
    {"name": {"first": "Mistie", "last" :"Montealegre"}, "sales" : "$6920", "branch": "New York City, NY"},
    {"name": {"first": "Nancy", "last" :"Napoli"}, "sales" : "$5255", "branch": "New York City, NY"},
    {"name": {"first": "Regine", "last" :"Rohrbaugh"}, "sales" : "$7881", "branch": "Orlando, FL"},
    {"name": {"first": "Rolando", "last" :"Riebel"}, "sales" : "$8573", "branch": "San Francisco, CA"},
    {"name": {"first": "Scarlett", "last" :"Stagg"}, "sales" : "$7126", "branch": "Orlando, FL"},
    {"name": {"first": "Sherron", "last" :"Strawn"}, "sales" : "$8848", "branch": "New York City, NY"},
    {"name": {"first": "Susan", "last" :"Shilling"}, "sales" : "$8542", "branch": "Seattle, WA"},
    {"name": {"first": "Tama", "last" :"Tworek"}, "sales" : "$9200", "branch": "Seattle, WA"},
    {"name": {"first": "Tonisha", "last" :"Taunton"}, "sales" : "$5219", "branch": "Orlando, FL"},
    {"name": {"first": "Vergie", "last" :"Villescas"}, "sales" : "$8712", "branch": "New York City, NY"}
];

function convertSalesToInt(person){
  var salesInt = person.sales.slice(1);
  return parseInt(salesInt, 10);
}

function salesPerBranch(salesPersons){
  var myObj = {};

  for(var i = 0; i < salesPersons.length; i++){
    var splitTeam = salesPersons[i].branch;
    myObj[splitTeam] = myObj[splitTeam] ? myObj[splitTeam]: 0;
    myObj[splitTeam] += convertSalesToInt(salesPersons[i]);
  }

  return myObj;
}

console.log(salesPerBranch(salesPersons));

＆＃13;

如果您已经添加了销售计数数据，则需要检查是否有销售计数数据。

Answer 2

这是另一个受到@ Durga答案启发的略有不同的方式。我使用for ... of循环和简化的if语句来理解他的答案。

   function salesPerBranch(salesPersons) {
      var myObj = {};

      //loop through sales people
      for (let person of salesPersons) {
        let branch = person.branch;

        //test for branches
        if (!myObj[branch]) {
          //create branches in the obj
          myObj[branch] = myObj[branch]
            //set default value to 0
          myObj[branch] = 0;

        }
        //Add sales for each branch
        myObj[branch] += convertSalesToInt(person);

      }
      console.log(myObj)

    }

     salesPerBranch(salesPersons);

Answer 3

这应该可以解决问题

salesPersons.map(function(x){return x.branch}).filter(function(value, index, self){return self.indexOf(value) === index;}).forEach(function(x){
        console.log(x + ' : $' + salesPersons.filter(function(y){if(y.branch === x) return y}).map(function(z){return z.sales.slice(1)}).reduce(function(a,b){return +a + +b;},0));
    });

获得总销售额

3 个答案: