Question

我有2个JSON数组和1个日期来表示关键，如何在每个日期的基础上将它排成一行并总结钢和混凝土？

var a = ['Nov 1 2016','Nov 2 2016','Nov 3 2016','Nov 4 2016'];
var b = [{Date:'Nov 1 2016', steel:10.98},{Date:'Nov 3 2016', steel:5.67},{Date:'Nov 4 2016', steel:3.14}]
var c = [{Date:'Nov 1 2016', concrete:9.10},{Date:'Nov 2 2016', concrete:16.8},{Date:'Nov 4 2016', concrete:7.20}]

//output sum of steel and concrete
var x  =  [{Date:'Nov 1 2016', val:20.08},{Date:'Nov 2 2016', val:16.8},{Date:'Nov 3 2016', val:5.67},{Date:'Nov 4 2016', val:10.34}]

Answer 1

你可以做的是遍历日期数组，并且对于每个日期，找到b和c中的相应项目（如果找不到则提供具有0值的默认项目），并总结他们各自的steel和concrete属性。

换句话说，将每个日期字符串转换（map）到包含该日期的对象以及该日期的steel和concrete之和：

// dates is the `a` array, steelDays is the `b` array, concreteDays is the `c` array
const result = dates.map(date =>
  toDateAndValueObj(
    date,
    sumSteelAndConcrete(
      findByDate(steelDays, date),   
      findByDate(concreteDays, date) 
    )
  )
);

功能将在哪里

// converts a date and a value into an object containing that date and that value
const toDateAndValueObj = (date, val) => ({ date, val });

// takes two objects (or the objects with default values if passed undefined) and sums up their respective `steel` and `concrete` properties
const sumSteelAndConcrete = (s = { steel: 0 }, c = { concrete: 0 }) => s.steel + c.concrete;

// finds an item in the array who's Date property lexicographically equals the given date string
const findByDate = (arr, date) => arr.find(item => item.Date === date);

请注意，为了简单起见，我在这里使用字符串比较。如果您使用的是实际日期，则可能需要构建Date个对象并使用getTime进行比较：

const findByDate = (arr, date) => arr.find(item => new Date(item.Date).getTime() === new Date(date).getTime());

完整演示：

＆＃13;

const dates = ['Nov 1 2016','Nov 2 2016','Nov 3 2016','Nov 4 2016'];
const steelDays = [{Date:'Nov 1 2016', steel:10.98},{Date:'Nov 3 2016', steel:5.67},{Date:'Nov 4 2016', steel:3.14}];
const concreteDays = [{Date:'Nov 1 2016', concrete:9.10},{Date:'Nov 2 2016', concrete:16.8},{Date:'Nov 4 2016', concrete:7.20}];

const toDateAndValueObj = (date, val) => ({ date, val });
const sumSteelAndConcrete = (s = { steel: 0 }, c = { concrete: 0 }) => s.steel + c.concrete;
const findByDate = (arr, date) => arr.find(item => item.Date === date);

const result = dates.map(date =>
  toDateAndValueObj(
    date,
    sumSteelAndConcrete(
      findByDate(steelDays, date),
      findByDate(concreteDays, date)
    )
  )
);

console.log(result);

＆＃13;

Answer 2

var b = [{Date:'Nov 1 2016', steel:'10.98'},{Date:'Nov 3 2016', steel:'5.67'},{Date:'Nov 4 2016', steel:'3.14'}];
var c = [{Date:'Nov 1 2016', concrete:'9.10'},{Date:'Nov 2 2016', concrete:'16.8'},{Date:'Nov 4 2016', concrete:'7.20'}];
var obj = {};
for(var index in b){
  obj[b[index]['Date']] = b[index]['steel'];
}
for(index in c){
  if(obj.hasOwnProperty(c[index]['Date'])){
    obj[c[index]['Date']] += c[index]['concrete'];
  }else{
    obj[c[index]['Date']] = c[index]['concrete'];
  }
}
var result = [];
for(var key in obj){
  result.push({Date:key, val:obj[key]});
}
console.log(result);

Javascript - 使用与密钥相同的日期的JSON数组的总和

2 个答案: