如何将父 object 的 id 推送到子对象？

Question

我有一个 object 数组，它有一个 object 的内部数组，我想将父 object 的 id 推送给每个子 ZA8CFDE6331BD5B6666

   a = [
        {id: 'abc', stage: [{name: 'car' , value: '123'},{name: 'bus' , value: '345'},{name: 'truck' , value: '567'}],
        {id: 'def', stage: [{name: 'bike' , value: '890'},{name: 'cycle' , value: '123'},{name: 'car' , value: '456'}]}
    ]

expected output = [
    {name: 'car' , value: '123', id: 'abc'},{name: 'bus' , value: '345', 'abc'},{name: 'truck' , value: '567', 'abc'}, {name: 'bike' , value: '890', id: 'def'},{name: 'cycle' , value: '123',id: 'def',id: 'def'},{name: 'car' , value: '456', id: 'def'}
]

我能够获得唯一的舞台，但无法将 id 推送到每个 object。 请帮忙

const getAllStages = [].concat(...map(a, el => el.stage));
console.log(getAllStages )

Answer 1

再次使用.map()将el.stage添加到el.id的每个元素。

您可以在外部映射中使用.flatMap()将所有结果连接到一个数组中。

 const a = [ {id: 'abc', stage: [{name: 'car', value: '123'},{name: 'bus', value: '345'},{name: 'truck', value: '567'}]}, {id: 'def', stage: [{name: 'bike', value: '890'},{name: 'cycle', value: '123'},{name: 'car', value: '456'}]} ] result = a.flatMap(({ id, stage }) => stage.map(s => ({ id: id, ...s }))); console.log(result);

Answer 2

您可以在不使用库的情况下这样编写

let a = [
    {
        id: 'abc', stage:
            [
                { name: 'car', value: '123' },
                { name: 'bus', value: '345' },
                { name: 'truck', value: '567' }
            ]
    },
    {
        id: 'def', stage:
            [
                { name: 'bike', value: '890' },
                { name: 'cycle', value: '123' },
                { name: 'car', value: '456' }
            ]
    }
]    
let output = [];
    for (const element of a) {
        for (const stageElement of element.stage) {
            let newElement = {
                name: stageElement.name,
                value: stageElement.value,
                id: element.id
            };
            output.push(newElement);
        }
    }

Answer 3

如果您尝试合并id ：

let remapped = a.map(e => ({ id: a.id, ...e }));

其中将每个条目转换为 object 继承a.id值并添加 object 中的其他任何内容。

Answer 4

这是一个例子和结果:)

a.forEach(function(row) {
    row.stage.map(function (child) {child.id = row.id})
})

结果：

[{"id":"abc","stage":[{"name":"car","value":"123","id":"abc"},{"name":"bus","value":"345","id":"abc"},{"name":"truck","value":"567","id":"abc"}]},{"id":"def","stage":[{"name":"bike","value":"890","id":"def"},{"name":"cycle","value":"123","id":"def"},{"name":"car","value":"456","id":"def"}]}]

Answer 5

const a = [
    {id: 'abc', stage: [{name: 'car' , value: '123'},{name: 'bus' , value: '345'},{name: 'truck' , value: '567'}]},
    {id: 'def', stage: [{name: 'bike' , value: '890'},{name: 'cycle' , value: '123'},{name: 'car' , value: '456'}]}
]

a.forEach(item => {
    const itemId = item.id;
    const stages = item.stage;

    stages.forEach(stage => {
        stage['id'] = itemId;
    })
});

console.log(a);