JavaScript - 从对象数组中计算和删除重复项 ES6

2022-01-19 00:00:00 frontend javascript ecmascript-6

我基本上是在寻找一个简洁的解决方案(可能使用 ES6)来减少重复对象的数组并计算重复的数量.

I am basically looking for a neat and succinct solution (possibly using ES6) for reducing an array of duplicate objects and counting the quantity of duplicates.

例如

const items = [
   {
     id: 1,
     type: 'PEPSI',
   },
  {
     id: 1,
     type: 'PEPSI',
   },
  {
     id: 1,
     type: '7UP',
   },
  {
     id: 1,
     type: 'FANTA',
   },
  {
     id: 2,
     type: 'FANTA',
   },
  {
     id: 2,
     type: '7UP',
   },
  {
     id: 2,
     type: '7UP',
   }
];

function getItems(data, id) {
      return data.filter(x => x.id === id).map((item) => {
        const obj = {};
        obj.quantity = **I want the quantity to go here**
        obj.type = item.type;
        return obj;
      });
    }

所以运行函数 getItems(items, 1) 会给我:

So running the function getItems(items, 1) would give me:

[
 { quantity: 2, type: 'PEPSI' },
 { quantity: 1, type: '7UP' },
 { quantity: 1, type: 'FANTA' }
]

提前致谢！

推荐答案

你可以使用这个代码:

const items = [
   {
     id: 1,
     type: 'PEPSI',
   },
  {
     id: 1,
     type: 'PEPSI',
   },
  {
     id: 1,
     type: '7UP',
   },
  {
     id: 1,
     type: 'FANTA',
   },
  {
     id: 2,
     type: 'FANTA',
   },
  {
     id: 2,
     type: '7UP',
   },
  {
     id: 2,
     type: '7UP',
   }
];

const getItems = (data) => {
    return data.reduce((a,b) => {
        const item = a.find(i => i.type === b.type);
        if(item){
            item.quantity = item.quantity + 1;
        }
        else {
            a.push({
                quantity: 0,
                type: b.type
            });
        }
        return a;
    }, []);
};
console.log(getItems(items));

const getItemsWithId = (data, id) => {
return data.reduce((a,b) => {
    if(b.id === id) {
        const item = a.find(i => i.type === b.type);
        if(item){
            item.quantity = item.quantity + 1;
        }
        else {
            a.push({
                quantity: 0,
                type: b.type
            });
        }   
    }
    return a;
}, []);
};

console.log(getItemsWithId(items, 1));

相关文章