如何组合具有相同键和值的多个JSON对象

Question

如何在相同的响应中组合JSON对象，具有相同的键和值与javascript?这是我的数据，例如:

{
"data": [
{
"name": "A",
"description": {
"location": "location1",
"floor": "floor1",
},
},
{
"name": "A",
"description": {
"location": "location2",
"floor": "floor1",
},
},
{
"name": "B",
"description": {
"location": "location3",
"floor": "floor3",
},
},
]
}

然后变成这样:

{
"data": [
{
"name": "A",
"description": {
"location": ["location1","location2"],
"floor": "floor1",
},
},
{
"name": "B",
"description": {
"location": "location3",
"floor": "floor3",
},
},
]
}

基本上我是一个新学习javascript的人。任何帮助都将非常有帮助，谢谢。

Answer 1

你可以这样做:

const data = {data: [{name: 'A',description: {location: 'location1',floor: 'floor1',},},{name: 'A',description: {location: 'location2',floor: 'floor1',},},{name: 'B',description: {location: 'location3',floor: 'floor3',},},],}
const result = {
data: data.data.reduce((a, { name, description }) => {
const index = a.findIndex((d) => d.name === name)
if (index >= 0) {
let location = a[index].description.location
location = Array.isArray(location) ? location : [location]
a[index].description.location = [...location, description.location]
} else {
a.push({ name, description })
}
return a
}, []),
}
console.log(result)

Answer 2

const list = {
"data": [
{
"name": "A",
"description": {
"location": "location1",
"floor": "floor1",
},
},
{
"name": "A",
"description": {
"location": "location2",
"floor": "floor1",
},
},
{
"name": "B",
"description": {
"location": "location3",
"floor": "floor3",
},
},
]
};

const consolidatedData = [];
for (const ele of list.data) {
const isExist = consolidatedData.find(x => x.name === ele.name);
if (!isExist) {
consolidatedData.push({
...ele
})
} else {
const objectKey = consolidatedData.findIndex(x => x.name === ele.name);
if (objectKey > -1) {
const description = consolidatedData[objectKey].description;
const newDes = ele.description;
if (newDes.location !== description.location) {
const data = consolidatedData[objectKey].description;
const added = [data.location, ele.description.location];
delete consolidatedData[objectKey].description.location
consolidatedData[objectKey].description["location"] = added
}
if (newDes.floor !== description.floor){
const data = consolidatedData[objectKey].floor;
const added = [data.floor, ele.description.floor];
delete consolidatedData[objectKey].description.floor
consolidatedData[objectKey].description["floor"] = added
}
}
}
}
console.log(JSON.stringify(consolidatedData, null, 2));

Answer 3

这里是一个使用中间桶对象的解决方案。然后从桶对象构造所需的结果对象:

const input = { "data": [ { "name": "A", "description": { "location": "location1", "floor": "floor1", }, }, { "name": "A", "description": { "location": "location2", "floor": "floor1", }, }, { "name": "B", "description": { "location": "location3", "floor": "floor3", }, }, ] };
let buckets = input.data.reduce((acc, obj) => {
if(!acc[obj.name]) {
acc[obj.name] = {
locations: {},
floors: {}
};
}
acc[obj.name].locations[obj.description.location] = true; 
acc[obj.name].floors[obj.description.floor] = true; 
return acc;
}, {});
console.log('buckets: ', buckets);
let result = {
data: Object.keys(buckets).map(name => {
let locations = Object.keys(buckets[name].locations);
let floors = Object.keys(buckets[name].floors);
return {
name: name,
description: {
location: locations.length == 1 ? locations[0] : locations,
floor: floors.length == 1 ? floors[0] : floors
}
}
})
};
console.log('result:', result);

指出:

• 桶对象:
  • 是使用数组.reduce()
  创建的。
  • array.reduce()docs: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/reduce
  • 位置和楼层使用对象而不是数组收集，这是为了避免重复的名称
• 结果对象:
  • 使用Object.keys(buckets)获取名称数组
  • .map()将每个名称转换为所需对象
  • 您的位置和地板的不寻常数组或字符串值是用条件
  构造的