通过公共键对对象数组进行分组，将嵌套属性合并到数组中[重复]

Question

假设我有这个数组：

[
    {"type": "A",   "status": "CREATED",        "name": "Jack"}, 
    {"type": "A",   "status": "CREATED",        "name": "John"}, 
    {"type": "A",   "status": "UPDATED",        "name": "Alex"}, 
    {"type": "B",   "status": "UPDATED",        "name": "Jane"}
]

我希望有一个按“type”和“status”分组的新数组，并将名称放在一个列表中。 这是所需的输出：

[
    {"type": "A",   "status": "CREATED",        "name-list": ["Jack", "John"]}, 
    {"type": "A",   "status": "UPDATED",        "name-list": ["Alex"]}, 
    {"type": "B",   "status": "UPDATED",        "name-list": ["Jane"]}
]

如您所见，数组的第一个对象包含 2 个名称，因为它们属于相同的“type”和“status”。 （name-list只是一个例子，也可以保留name，重要的是它应该是一个字符串/对象数组）

我必须在 html 页面中表示这些数据，因此需要能够通过 JavaScript 循环它们。

Answer 1

您可以使用Array.prototype.reduce() 遍历您的源数组，构建Map，将type 和status 组合为键和各自的对象，并合并name 作为值。然后从Map中提取Map.prototype.values()：

const src = [{"type":"A","status":"CREATED","name":"Jack"},{"type":"A","status":"CREATED","name":"John"},{"type":"A","status":"UPDATED","name":"Alex"},{"type":"B","status":"UPDATED","name":"Jane"}],

     result = [...src
        .reduce((r, o) => {
          const key = o.type+'\ud8ff'+o.status,
                match = r.get(key)
          match ? match.name.push(o.name) : r.set(key, {...o, name: [o.name]})
          return r
        }, new Map())
        .values()
      ]
      
console.log(result)

.as-console-wrapper{min-height:100%;}

Answer 2

我会尝试使用.reduce() 和.find()：

const data = [
    {"type": "A",   "status": "CREATED",        "name": "Jack"}, 
    {"type": "A",   "status": "CREATED",        "name": "John"}, 
    {"type": "A",   "status": "UPDATED",        "name": "Alex"}, 
    {"type": "B",   "status": "UPDATED",        "name": "Jane"}
];

const result = data.reduce((a, { type, status, name}) => {
  const found = a.find(e => e.type === type && e.status === status);
  
  if (found) {
    found['name-list'].push(name);
  } else {
    a.push({
      type,
      status,
      ['name-list']: [name]
    });
  }
  
  return a;
}, []);

console.log(result);

Answer 3

const arr = [
    {"type": "A",   "status": "CREATED",        "name": "Jack"}, 
    {"type": "A",   "status": "CREATED",        "name": "John"}, 
    {"type": "A",   "status": "UPDATED",        "name": "Alex"}, 
    {"type": "B",   "status": "UPDATED",        "name": "Jane"}
]

const res = arr.reduce((acc, cur) => {
  
  acc.forEach((obj, idx) => {
      if(obj.type == cur.type && obj.status == cur.status){
          acc[idx]["name-list"].push(cur.name)
      }
  })
  
  if(!acc.some(obj => obj["name-list"].includes(cur.name))){
      acc.push({type: cur.type, status: cur.status, "name-list": [cur.name]})
  }
  
  return acc
},[])

console.log(res)

Answer 4

这是你想要的：

const ar = [
    { "type": "A", "status": "CREATED", "name": "Jack" },
    { "type": "A", "status": "CREATED", "name": "John" },
    { "type": "A", "status": "UPDATED", "name": "Alex" },
    { "type": "B", "status": "UPDATED", "name": "Jane" }
];
const res = Array.from(ar.reduce((a, c) => {
    if (a.has(c.type + c.status)) {
        a.get(c.type + c.status)['name-list'].push(c.name);
    } else {
        c['name-list'] = [c.name];
        delete c.name;
        a.set(c.type + c.status, c)
    }
    return a;
}, new Map()).values());

console.log(res);

Answer 5

使用reduce 和Object.values

let arr = [
    {"type": "A",   "status": "CREATED",        "name": "Jack"}, 
    {"type": "A",   "status": "CREATED",        "name": "John"}, 
    {"type": "A",   "status": "UPDATED",        "name": "Alex"}, 
    {"type": "B",   "status": "UPDATED",        "name": "Jane"}
];

arr = Object.values(
    arr.reduce((acc, cur) => {
        if(acc[`${cur.type}-${cur.status}`]) acc[`${cur.type}-${cur.status}`]["name-list"].push(cur.name)

        else {
            acc[`${cur.type}-${cur.status}`] = {type:cur.type, status:cur.status, "name-list":[cur.name]}
        }    
        return acc;
    }, {})
);

console.log(arr)