在對象數組中合並對象數組

Question

我有以下對象

Person {
   name: string
   birthday: Date
   lifeEvents: LifeEvent[]
}

LifeEvent {
   eventId: number
   message: string
   comments: string
}

由於數據以塊的形式出現，我將有一個Person數組，其中一個對象具有name和birthday以及值，但lifeEvents是空的（將此視為父對象。）

所有其他對象都不會填充生日和姓名，並且只有一個LifeEvent帶有eventId和帶有值的message或eventId和comments

在該數組中，我需要獲取具有名稱和生日的父對象，然后從其余對象中獲取所有 lifeEvents，將包含相同 eventId 的所有項目合並到LifeEvent中，然后將其推送到父對象的 lifeEvents。

我已經嘗試過 array.reduce、array.map 但無法找到將這些對象組合成一個的方法。

我的輸出應該只有一個Person ，所有lifeEvents都由eventId合並

樣本數據：

let results = [
{ 
   name: 'Test1',
   birthday: '2022-06-14',
   lifeEvents: null
},
{ 
   name: null,
   birthday: null,
   lifeEvents: [
        {
           eventId: 1,
           message: 'First event',
           comments: null
        }
   ]
},
{ 
   name: null,
   birthday: null,
   lifeEvents: [
        {
           eventId: 2,
           message: 'Second event',
           comments: null
        }
   ]
},
{ 
   name: null
   birthday: null
   lifeEvents: [
        {
           eventId: 1,
           message: null,
           comments: 'First event comment'
        }
   ]
},
{ 
   name: null
   birthday: null
   lifeEvents: [
        {
           eventId: 2,
           message: null,
           comments: 'Second event comment'
        }
   ]
},
]

感謝任何幫助。

Answer 1

以下根據提供的示例生成請求的結果：

 let results = [ { name: 'Test1', birthday: '2022-06-14', lifeEvents: null }, { name: null, birthday: null, lifeEvents: [ { eventId: 1, message: 'First event', comments: null } ] }, { name: null, birthday: null, lifeEvents: [ { eventId: 2, message: 'Second event', comments: null } ] }, { name: null, birthday: null, lifeEvents: [ { eventId: 1, message: null, comments: 'First event comment' } ] }, { name: null, birthday: null, lifeEvents: [ { eventId: 2, message: null, comments: 'Second event comment' } ] }, ]; // extract parent const parentResult = results.find((result) => result.name); // generate unique events from sibling entities const uniqueEvents = new Map(); results.forEach((result) => result.lifeEvents?.forEach( (lifeEvent) => { if (uniqueEvents.has(lifeEvent.eventId)) { updateEvent(lifeEvent); } else { uniqueEvents.set(lifeEvent.eventId, { eventId: lifeEvent.eventId, message: lifeEvent.message, comments: lifeEvent.comments}); } }) ); // function to update event that is already stored in uniqueEvents function updateEvent(lifeEvent) { const existingLifeEvent = uniqueEvents.get(lifeEvent.eventId); if (lifeEvent.message) existingLifeEvent.message = lifeEvent.message; if (lifeEvent.comments) { if (existingLifeEvent.comments) { existingLifeEvent.comments.concat(lifeEvent.comments) } else { existingLifeEvent.comments = lifeEvent.comments; } } } // populate lifeEvents inside the parentResult parentResult.lifeEvents = []; uniqueEvents.forEach((uniqueId) => { parentResult.lifeEvents.push(uniqueId); }); console.log(parentResult);

Answer 2

前提：您使用的數據結構是錯誤的，您應該嘗試使用具有同質模型的數組。

也就是說，我使用了一種 reduce 方法，條件是以與其他元素不同的方式處理第一個元素。

最后一件事，你說合並lifeEvents，我假設你的意思是覆蓋具有相同 id 的事件的無效值，如果你想覆蓋所有值，那么你可以省略我寫的合並實用程序函數。

 let results = [{ name: 'Test1', birthday: '2022-06-14', lifeEvents: null, }, { name: null, birthday: null, lifeEvents: [{ eventId: 1, message: 'First event', comments: null, }, ], }, { name: null, birthday: null, lifeEvents: [{ eventId: 2, message: 'Second event', comments: null, }, ], }, { name: null, birthday: null, lifeEvents: [{ eventId: 1, message: null, comments: 'First event comment', }, ], }, { name: null, birthday: null, lifeEvents: [{ eventId: 2, message: null, comments: 'Second event comment', }, ], }, ]; const merge = (o1, o2) => { const r = {...o1} Object.keys(o1).forEach(k => r[k] = o2[k] || o1[k]) return r } const r = results.reduce((o, curr, i) => { if (i === 0) { return { ...o, lifeEvents: [] }; } else { const currentEvent = curr.lifeEvents[0] const idx = o.lifeEvents.findIndex((_o) => _o.eventId === currentEvent.eventId); if (idx !== -1) return { ...o, lifeEvents: o.lifeEvents.map((_o, i) => i === idx ? merge(_o, currentEvent) : _o) } else return { ...o, lifeEvents: [...o.lifeEvents, currentEvent] } } }, results[0]); console.log("RESULT:", r);