按公共键对对象数组进行分组，将嵌套属性合并到数组中[duplicate]

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"}
]

我想让一个新数组按“类型”和“状态”分组，并将名称列在一个列表中。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两个名称，因为它们属于相同的“类型”和“状态”。（名称列表只是一个示例，也可以保留名称，重要的是它应该是字符串/对象的数组）

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

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 2

我会尝试使用。减少（）和。查找（）：

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

您可以使用数组遍历源数组。原型reduce（）。然后提取Map。原型值（）来自该映射：

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)

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