问比较两个对象数组并将数据推送到新数组中
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Stack Overflow用户

提问于 2021-04-18 04:08:53

回答 1查看 30关注 0票数 0

这里我比较了对象的b1和b2数组，并将数据添加到新的数组b3中，一切都很好，但是在数组b3中，只想获取一次有id和没有id的数组，在这种情况下，在新的数组b3中获得重复的数组元素。

let b3 = [];
let idss;
let b1 = [
  { batchReferenceNo: '118', receivedQty: 1 },
  { batchReferenceNo: '120', receivedQty: 1 },
  { batchReferenceNo: '100', receivedQty: 1 },
];
let b2 = [
  { id: 1, batchReferenceNo: '118', receivedQty: 1 },
  { id: 3, batchReferenceNo: '120', receivedQty: 1 },
];
console.log(b2);
b1.forEach((bno1) => {
  b2.forEach((bno2) => {
    if (bno1.batchReferenceNo == bno2.batchReferenceNo) {
      idss = bno2.id;
      let c1 = {
        id: idss,batchReferenceNo: bno1.batchReferenceNo,qty: bno1.receivedQty
      };
      b3.push(c1);
    }
    if (bno1.batchReferenceNo != bno2.batchReferenceNo) {
      idss = '';
      let c1 = {
        id: idss,batchReferenceNo: bno1.batchReferenceNo,qty: bno1.receivedQty
      };
      b3.push(c1);
    }
  });
});
console.log(b3);

请帮我解决这个问题，提前谢谢。

let b3 = [];
let idss;
let b1 = [
  { batchReferenceNo: '118', receivedQty: 1 },
  { batchReferenceNo: '120', receivedQty: 1 },
  { batchReferenceNo: '100', receivedQty: 1 },
];
let b2 = [
  { id: 1, batchReferenceNo: '118', receivedQty: 1 },
  { id: 3, batchReferenceNo: '120', receivedQty: 1 },
];
console.log(b2);
b1.forEach((bno1) => {
  b2.forEach((bno2) => {
    if (bno1.batchReferenceNo == bno2.batchReferenceNo) {
      idss = bno2.id;
      let c1 = {
        id: idss,batchReferenceNo: bno1.batchReferenceNo,qty: bno1.receivedQty
      };
      b3.push(c1);
    }
    if (bno1.batchReferenceNo != bno2.batchReferenceNo) {
      idss = '';
      let c1 = {
        id: idss,batchReferenceNo: bno1.batchReferenceNo,qty: bno1.receivedQty
      };
      b3.push(c1);
    }
  });
});
console.log(b3);

javascript

arrays

回答 1

Stack Overflow用户

发布于 2021-04-18 04:30:14

如果b2是b1的子集

从b2数组中构造一个Map，其中映射键是batchReferenceNo，映射值是整个对象。现在，循环遍历b1数组并查看当前对象是否在映射中，如果是，则添加相应的id，如果否，则添加一个空字符串。

const 
  b1 = [{ batchReferenceNo: "118", receivedQty: 1 }, { batchReferenceNo: "120", receivedQty: 1 }, { batchReferenceNo: "100", receivedQty: 1 }],
  b2 = [{ id: 1, batchReferenceNo: "118", receivedQty: 1 }, { id: 3, batchReferenceNo: "120", receivedQty: 1 }],
      
  b2Map = new Map(b2.map((b) => [b.batchReferenceNo, b])),
  res = b1.map((b) =>
    b2Map.has(b.batchReferenceNo)
      ? { ...b, id: b2Map.get(b.batchReferenceNo).id }
      : { ...b, id: "" }
  );

console.log(res);

如果b2不是b1的子集

保留b2的所有元素，并添加b1中不在b2中的元素。

const 
  b1 = [{ batchReferenceNo: "118", receivedQty: 1 }, { batchReferenceNo: "120", receivedQty: 1 }, { batchReferenceNo: "100", receivedQty: 1 }],
  b2 = [{ id: 1, batchReferenceNo: "118", receivedQty: 1 }, { id: 3, batchReferenceNo: "120", receivedQty: 1 }],
  b2Map = new Map(b2.map((b) => [b.batchReferenceNo, b])),
  res = b1
    .reduce(
      (r, b) => (!b2Map.has(b.batchReferenceNo) && r.push({ ...b, id: "" }), r),
      []
    )
    .concat(b2);

console.log(res);