Question

我正在接收下一个数据

    [
      { id: "1", name: "test1", rName: "the1" },
      { id: "1", name: "test1", rName: "the2" },
      { id: "1", name: "test1", rName: "the3" },
      { id: "2", name: "test2", rName: "the1" },
      { id: "2", name: "test2", rName: "the2" },
      { id: "3", name: "test3", rName: "the1" }
    ]

我想通过id合并它，并将rName推送到具有此结构的数组中

    [
      { id: "1", name: "test1", rName: ["the1", "the2","the3"] },
      { id: "2", name: "test2", rName: ["the1", "the2"] },
      { id: "3", name: "test3", rName: ["the1"] }
    ]

我以为减少操作会成功，但没有成功，如果有人能指出正确的方向，那将不胜感激。

Answer 1

这是重新格式化数据的非常简单的情况。代码在下面。

var data = [
  { id: "1", name: "test1", rName: "the1" },
  { id: "1", name: "test1", rName: "the2" },
  { id: "1", name: "test1", rName: "the3" },
  { id: "2", name: "test2", rName: "the1" },
  { id: "2", name: "test2", rName: "the2" },
  { id: "3", name: "test3", rName: "the1" }
];
    
var reduced = Object.values(data.reduce(function(accumulator, element) {
  if (!accumulator[element.id]) {
    accumulator[element.id] = { id: element.id, name: element.name, rName: [] };
  }
  
  accumulator[element.id].rName.push(element.rName);
  return accumulator;
}, {}));
console.log(reduced);

累加器检查累加器中是否存在element.id键。如果没有，它将创建它。然后，它将新的rName推送到现有堆栈中。然后使用Object.values()将转换返回到数组。

Answer 2

您可以像这样使用reduce和find：

在化简accumulator中，检查是否存在与要迭代的当前项目相同的id的项目。如果是，请将当前项目的rName推入rName数组。否则，将一个新项目推送到accumulator

var data = [
      { id: "1", name: "test1", rName: "the1" },
      { id: "1", name: "test1", rName: "the2" },
      { id: "1", name: "test1", rName: "the3" },
      { id: "2", name: "test2", rName: "the1" },
      { id: "2", name: "test2", rName: "the2" },
      { id: "3", name: "test3", rName: "the1" }
      ];

const newArray = data.reduce((acc, {id,name,rName}) => {
    const existing = acc.find(a => a.id == id);
    if (existing)
        existing["rName"].push(rName);
    else
        acc.push({id,name,rName: [rName]})
    return acc
}, []);

console.log(newArray)

这是一个单行代码高尔夫球的答案。（从@Sébastien的答案中得到了想法）：

var data = [
  { id: "1", name: "test1", rName: "the1" },
  { id: "1", name: "test2", rName: "the2" },
  { id: "1", name: "test2", rName: "the3" },
  { id: "2", name: "test2", rName: "the1" },
  { id: "2", name: "test1", rName: "the2" },
  { id: "3", name: "test3", rName: "the1" }
]

const anotherArray = Object.values(data.reduce((acc, {id,name,rName}) =>
      ((acc[id] = acc[id] || {id,name,rName:[]})["rName"].push(rName), acc), {}));

console.log(anotherArray)

Answer 3

纯ES6

let result = obj.reduce((acc, item) => {
  let found = acc.find(i => i.id === item.id);
  found ? (found.rName = [...found.rName, item.rName]) : (acc = [...acc, { ...item, rName: [item.rName] }]);
  return acc;
}, []);

Answer 4

如果每个id都可以确定一个唯一的name，那么它将起作用：

let data = [
  { id: "1", name: "test1", rName: "the1" },
  { id: "1", name: "test1", rName: "the2" },
  { id: "1", name: "test1", rName: "the3" },
  { id: "2", name: "test2", rName: "the1" },
  { id: "2", name: "test2", rName: "the2" },
  { id: "3", name: "test3", rName: "the1" }
];

let result = [];

for (let item of data) 
{
    const index = result.findIndex(i => i.id === item.id);
    if (index < 0) {
        result.push({ id: item.id, name: item.name, rName: [item.rName] });
    }
    else {
        result[index].rName.push(item.rName);
    }
}

请注意，当以下数据应被接受时，这将不起作用：

[
  { id: "1", name: "test1", rName: "the1" },
  { id: "1", name: "test2", rName: "the2" },
  { id: "1", name: "test2", rName: "the3" },
  { id: "2", name: "test2", rName: "the1" },
  { id: "2", name: "test1", rName: "the2" },
  { id: "3", name: "test3", rName: "the1" }
]

Answer 5

如果您坚持使用'reduce'方法，您可以这样做：

var arr = [
        { id: "1", name: "test1", rName: "the1" },
        { id: "1", name: "test1", rName: "the2" },
        { id: "1", name: "test1", rName: "the3" },
        { id: "2", name: "test2", rName: "the1" },
        { id: "2", name: "test2", rName: "the2" },
        { id: "3", name: "test3", rName: "the1" }
    ]
    var arr1 = arr.reduce(function (a1, a2) {
        if (a1 instanceof Array) {
            let lastItem = a1[a1.length - 1]
            if (lastItem.id == a2.id) {
                lastItem.rName.push(a2.rName)
            } else {
                a1.push({ ...a2, rName: [a2.rName] })
            }
            return a1
        } else {
            let result = []
            if (a1.id == a2.id) {
                result.push({ ...a1, rName: [a1.rName, a2.rName] })
            } else {
                result.push({ ...a1, rName: [a1, rName] })
                result.push({ ...a2, rName: [a2, rName] })
            }
            return result

        }

    })
    console.log(JSON.stringify(arr1))

但是我认为您应该这样做，因为代码更清晰：

var arr = [
    { id: "1", name: "test1", rName: "the1" },
    { id: "1", name: "test1", rName: "the2" },
    { id: "1", name: "test1", rName: "the3" },
    { id: "2", name: "test2", rName: "the1" },
    { id: "2", name: "test2", rName: "the2" },
    { id: "3", name: "test3", rName: "the1" }
]
var map = new Map()
arr.forEach(function(item){
    if(map.has(item.id)){
        map.get(item.id).rName.push(item.rName)
    }else{
        map.set(item.id, {...item, rName: [item.rName]})
    }
})
var arr1 = Array.from(map.values())
console.log(JSON.stringify(arr1))

如何将具有相同属性的对象合并到单个数组中？

5 个答案: