將具有不同屬性名稱的物件陣列拆分為一個物件，并用給定名稱分隔它們

我有一個包含物件的陣列。現在我想將陣列切片為一個新物件，該物件僅包含與某個屬性名稱匹配并按此屬性名稱分組的那些物件。問題是我也有它們之間不同的屬性名稱。名稱和 ID 在物件陣列中重復，但在新物件內部，它應該只包含一次 ID 和名稱。

原始陣列如下所示：

let personArray = [
    {
        id_dentist: 1,
        dentist_name: 'John',
        id_secretary: 6,
        secretary_name: 'Paul',
        id_security: 3,
        security_name: 'Carl'
    },
    {
        id_dentist: 2,
        dentist_name: 'Lisa',
        id_secretary: 9,
        secretary_name: 'Beth',
        id_security: 5,
        security_name: 'Monica'
    },
    {
        id_dentist: 1,
        dentist_name: 'John',
        id_secretary: 6,
        secretary_name: 'Paul',
        id_security: 3,
        security_name: 'Carl'
    }
];

新物件應如下所示：

let personObject = {
    dentist: [
        { id_dentist: 1, dentist_name: 'John' },
        { id_dentist: 2, dentist_name: 'Lisa' },
    ],
    secretary: [
        { id_secretary: 6, secretary_name: 'Paul' },
        { id_secretary: 9, secreatary_name: 'Beth' },
    ],
    security: [
        { id_security: 3, security_name: 'Carl' },
        { id_security: 5, security_name: 'Monica' }
    ]
};

我很感激幫助。

根據要求，我嘗試使用reduce()and filter()，但我無法將它們拆分。這是代碼：

const obj = personArray.reduce((acc, cur) => {
    const key = Object.keys(cur).filter(f => /^id_/.test(f))[0].split('_')[1];
    if (!acc[key]) acc[key] = [];
    acc[key].push(cur);
    return acc;
}, {});

console.log(obj);

關于奇怪的資料結構，我使用SELECTSQL 語法從資料庫中獲取這些資料。

uj5u.com熱心網友回復：

給你，這可以進一步增強

let personArray = [{"id_dentist":1,"dentist_name":"John","id_secretary":6,"secretary_name":"Paul","id_security":3,"security_name":"Carl"},{"id_dentist":2,"dentist_name":"Lisa","id_secretary":9,"secretary_name":"Beth","id_security":5,"security_name":"Monica"},{"id_dentist":1,"dentist_name":"John","id_secretary":6,"secretary_name":"Paul","id_security":3,"security_name":"Carl"}];

const personObject = { dentist: [], secretary: [], security: [] };
const isExist = (arr, id, key) => arr.find(x => x[key] === id);


personArray.reduce((personObj, person) => {
  
  const isDentistExists = isExist(personObj.dentist, person.id_dentist, 'id_dentist');
  
  if (!isDentistExists) {
    personObj.dentist.push({
      id_dentist: person.id_dentist,
      dentist_name: person.dentist_name
    });
  }
  
  const isSecretaryExists = isExist(personObj.secretary, person.id_secretary, 'id_secretary');
  
  if (!isSecretaryExists) {
    personObj.secretary.push({
      id_secretary: person.id_secretary,
      secretary_name: person.secretary_name
    });
  }
  
  const isSecurityExists = isExist(personObj.security, person.id_security, 'id_security');
  
  if (!isSecurityExists) {
    personObj.security.push({
      id_security: person.id_security,
      security_name: person.security_name
    });
  }
  
  return personObj;
}, personObject);

console.log(personObject);

uj5u.com熱心網友回復：

這不是一個簡單的演算法。這是一個 [大部分] 功能實作，可以處理任意數量的 id 和名稱

let personArray = [
  {
    id_dentist: 1,
    dentist_name: 'John',
    id_secretary: 6,
    secretary_name: 'Paul',
    id_security: 3,
    security_name: 'Carl',
  },
  {
    id_dentist: 2,
    dentist_name: 'Lisa',
    id_secretary: 9,
    secretary_name: 'Beth',
    id_security: 5,
    security_name: 'Monica',
  },
  {
    id_dentist: 1,
    dentist_name: 'John',
    id_secretary: 6,
    secretary_name: 'Paul',
    id_security: 3,
    security_name: 'Carl',
  },
]

const parsed = Object.fromEntries(
  Object.keys(personArray[0])
    .filter(key => key.startsWith('id_'))
    .map(id => {
      const uniqIds = [...new Set(personArray.map(person => person[id]))]
      const [, name] = id.split('_')
      const matchingPeople = uniqIds.map(uniqId => {
        return personArray.find(person => uniqId === person[id])
      })
      return matchingPeople.map(person => ({
        [id]: person[id],
        [`${name}_name`]: person[`${name}_name`],
      }))
    })
    .filter(entry => entry.length > 0)
    .map(groupedPeople => {
      const [name] = Object.keys(groupedPeople[0])
        .find(key => key.includes('_name'))
        .split('_')
      return [name, groupedPeople]
    })
)
console.log(parsed)

uj5u.com熱心網友回復：

是的，我也承認，這并不是微不足道的。

更改您的 SQL 選擇可能是一個更好的主意，以避免接收到需要進行這些冗長轉換的冗余資料。

const arr = [
{
    id_dentist: 1,
    dentist_name: 'John',
    id_secretary: 6,
    secretary_name: 'Paul',
    id_security: 3,
    security_name: 'Carl'
},
{
    id_dentist: 2,
    dentist_name: 'Lisa',
    id_secretary: 9,
    secretary_name: 'Beth',
    id_security: 5,
    security_name: 'Monica'
},
{
    id_dentist: 1,
    dentist_name: 'John',
    id_secretary: 6,
    secretary_name: 'Paul',
    id_security: 3,
    security_name: 'Carl'
}
], types=Object.keys(arr[0]).reduce((a,c,k)=>{
 k=c.match(/id_(.*)/);
 if(k) a.push(k[1]);
 return a;
},[]);

const res=Object.entries(arr.reduce((a,c)=>{
 types.forEach((t,id)=>{
  id="id_" t;
  a[t ":" c[id]]={[id]:c[id],[t "_name"]:c[t "_name"]}
 });
 return a;
},{})).reduce((a,[k,o],n)=>{
 [n]=k.split(":");
 (a[n]=a[n]||[]).push(o)
 return a;
},{});

console.log(res);

標籤：javascript

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