比较两个对象数组并将数据推送到新数组中
比较两个对象数组并将数据推送到新数组中
提问于 2021-04-18 12:08:53
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);
回答 1
Stack Overflow用户
发布于 2021-04-18 12:30:14
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);
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/67145097
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