在对象数组中合并 object 数组

Question

我有两个 arrays，所以我想将数组中的值分配给对象数组，如果有相同的 id 这是我的代码，但 foreach 重复第一个数组，

let cars =[
  {
    sap_code:"audi",
    products:[
      {id:23,name:"a4"},
      {id:55,name:"a5"},
      
    ]
  },
  {
    sap_code:"chevi",
    products:[
      {id:23,name:"x4"},
      {id:55,name:"x5"},
       
    ]
  }
]

let cars2= [{id:55, value:25000}]
let joinCar = []
cars.forEach((value)=> {
  value.products.forEach((value2)=>{  
    const existing = cars2.filter((v, i)=> {
      return (v.id == value2.id);
    });
    if (existing.length) {
      value2.value = existing[0].value;
      joinCar.push(value)
    } else {
      value2.value = null;
      joinCar.push(value);
    }
  });
})

我想要这样的东西

[
      {
        sap_code:"audi",
        products:[
          {id:23,name:"a4"},
          {id:55,name:"a5", value:25000},
          
        ]
      },
      {
        sap_code:"chevi",
        products:[
          {id:23,name:"x4"},
          {id:55,name:"x5", value:25000},
          {id:29,name:"x6"},
        ]
      }
]

Answer 1

在外部forEach中push ，而不是在内部推动。

 let cars =[ { sap_code:"audi", products:[ {id:23,name:"a4"}, {id:55,name:"a5"}, ] }, { sap_code:"chevi", products:[ {id:23,name:"x4"}, {id:55,name:"x5"}, ] } ] let cars2= [{id:55, value:25000}] let joinCar = [] cars.forEach((value)=> { value.products.forEach((value2)=>{ const existing = cars2.filter((v, i)=> { return (v.id == value2.id); }); if (existing.length) { value2.value = existing[0].value; } else { value2.value = null; } }); joinCar.push(value); }) console.log(joinCar);

Answer 2

使用for-of和 A rray.prototype.find()

 let cars =[ { sap_code:"audi", products:[ {id:23,name:"a4"}, {id:55,name:"a5"}, ] }, { sap_code:"chevi", products:[ {id:23,name:"x4"}, {id:55,name:"x5"}, ] } ] let cars2 = [{id:55, value:25000}] const joinCar = [...cars] // you originally created a new array for(const {products} of joinCar) for(const product of products) { const car2 = cars2.find(({id})=>id===product.id) if(car2) product.value = car2.value } console.log(joinCar)

Answer 3

我认为如果您使用hashtable并reduce ，可以显着降低复杂性。 因此，最好的方法是创建一个带有cars2的 id 和值的哈希表，以便找到匹配项并在cars上使用reduce ，如下所示：

 let cars =[ { sap_code:"audi", products:[ {id:23,name:"a4"}, {id:55,name:"a5"}, ] }, { sap_code:"chevi", products:[ {id:23,name:"x4"}, {id:55,name:"x5"}, ] } ] let cars2= [{id:55, value:25000}]; var hasTable = {}; for(let i of cars2){ hasTable[i.id] = i.value //create a hashtable with id and values } cars.reduce((a,v)=>{ let val = v.products.find(k=> hasTable[k.id]); if(val){ val.value = hasTable[val.id]; } return a }) console.log(cars)

Answer 4

 let cars =[ { sap_code:"audi", products:[ {id:23,name:"a4"}, {id:55,name:"a5",value:4500}, ] }, { sap_code:"chevi", products:[ {id:23,name:"x4"}, {id:55,name:"x5"}, ] } ]; let cars2 = [{id:55, value:25000}]; cars.forEach(car=>{ cars2.forEach(newCar=>{ let fc = car.products.findIndex(c=>c.id===newCar.id); if (fc.==-1) car.products[fc].value=newCar;value; }); }). console.log(JSON.stringify(cars))