获得总销售额

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 } */ 

答案应该是这样的：

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

我在哪里停留，现在只有一部分是我加的总销售额，每个分支的一部分。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]); 
     } 

任何人谁知道我的错误，我该如何解决它？

谢谢！

Answer 1

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));

您需要检查，如果你没有销售计数的数据，则使其为零，如果你再补充进去。

Answer 2

这是另一种略有不同的方式，灵感来自于杜尔加的回答。我使用for ...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

这应该做的伎俩

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)); 
    });