Question

我有一个简单的问题。我有两个数组A和B，如果B具有相同的ID，我想保留A对象。例如：

const A = [{id: "price", value: "1"}]

const B = [{id: "price", value: "0"}, {id: "number", value: "0"}]

预期结果：

[{id: "price", value: "1"}, {id: "number", value: "0"}}]

我该怎么做？

我试图将A和A映射到A内，但没有用。

Answer 1

const result = A.concat(B.filter(bo => A.every(ao => ao.id != bo.id)));

将A中的所有对象与B中不在的A中的对象连接起来（这是通过仅过滤B中不存在的对象来完成的） A中具有相同ID的对象）。

示例：

const A = [{id: "price", value: "1"}];

const B = [{id: "price", value: "0"}, {id: "number", value: "0"}];

const result = A.concat(B.filter(bo => A.every(ao => ao.id != bo.id)));

console.log(result);

Answer 2

您将在合并数组上使用reduce-还将value转换为数字：

const A = [{id: "price", value: "1"}];
const B = [{id: "price", value: "0"}, {id: "number", value: "0"}];
const res = Object.values([...A, ...B].reduce((acc, { id, value }) => {
  if (acc[id]) acc[id].value += parseInt(value);
  else acc[id] = { id, value: parseInt(value) };
  return acc;
}, {}));
console.log(res);

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Answer 3

您可以尝试的另一种选择（我相信它将是O（n））是将数组转换为以id为键的对象，然后扩展（jquery.extend或纯js实现），然后将合并的对象转换回数组。

const A = [{id: "price", value: "1"}];

const B = [{id: "price", value: "0"}, {id: "number", value: "0"}];

//convert arrays to objects
var Bx = {};
B.forEach(i => Bx[i.id] = i);

var Ax = {};
A.forEach(i => Ax[i.id] = i);

//copy all matching id properties from A to B
A.forEach(i => Bx[i.id] = Ax[i.id]);

//convert the merged object to array
var C = [];
Object.getOwnPropertyNames(Bx).forEach(i => C.push(Bx[i]));

console.log(C);

如何在JavaScript中合并两个对象数组

3 个答案: