Mongoose聚合回傳整個物件而不是1個物件及其屬性-有解無憂

我的用戶有一個“通知”欄位，即一個陣列

我正在嘗試獲取尚未閱讀的通知，因此我撰寫了以下查詢：

await User.aggregate([
        { $unwind: '$notifications' },
        { $match: { 'notifications.read': false } },
        { $project: { "notifications": 1 } }
    ])

我預計它會回傳如下內容：

[
    {
        "_id": "6201a79305489c2ae98b2753",
        "notifications": [
            {
                "_id": "62473e2f76b38858303b41d8",
                "text": "A tarefa tarefa 2 foi atualizada pela sua equipe!",
                "read": false,
                "date": "2022-04-01T18:02:23.813Z"
            },
            {
                "_id": "62473c4776b38858303b41a6",
                "text": "A tarefa finalizei foi atualizada pela sua equipe!",
                "read": false,
                "date": "2022-04-01T17:54:15.142Z"
            },
            {
                "_id": "62473e2f76b38858303b41d9",
                "text": "A tarefa tarefa 2 foi atualizada pela sua equipe!",
                "read": false,
                "date": "2022-04-01T18:02:23.814Z"
            }
        }
    ]
]

但它回傳：[

    {
        "_id": "6201a79305489c2ae98b2753",
        "notifications": {
            "_id": "62473e2f76b38858303b41d8",
            "text": "A tarefa tarefa 2 foi atualizada pela sua equipe!",
            "read": false,
            "date": "2022-04-01T18:02:23.813Z"
        }
    },
    {
        "_id": "6227b07b04cd1196302dba06",
        "notifications": {
            "_id": "62473c4776b38858303b41a6",
            "text": "A tarefa finalizei foi atualizada pela sua equipe!",
            "read": false,
            "date": "2022-04-01T17:54:15.142Z"
        }
    },
    {
        "_id": "6227b07b04cd1196302dba06",
        "notifications": {
            "_id": "62473e2f76b38858303b41d9",
            "text": "A tarefa tarefa 2 foi atualizada pela sua equipe!",
            "read": false,
            "date": "2022-04-01T18:02:23.814Z"
        }
    }
]

如何通過 read:false 過濾我的用戶通知并將其回傳到同一個陣列中？

uj5u.com熱心網友回復：

我會像這樣使用 $filter 來完成這個任務：

db.collection.aggregate([
 { $match: { 'notifications.read': false } },
 {
   "$addFields": {
    "notifications": {
      "$filter": {
        "input": "$notifications",
        "as": "n",
        "cond": {
         $eq: [
            "$$n.read",
             false
         ]
      }
     }
    }
   }
  }
])

解釋：

匹配所有具有通知的檔案。讀取：假
僅過濾具有 read:false 的通知

操場

轉載請註明出處，本文鏈接：https://www.uj5u.com/gongcheng/455113.html

標籤：mongodb 猫鼬

上一篇：我想找到一個四舍五入的方程的值

下一篇：使用$in查詢時，如何獲取mongoDB中每個唯一ID的n個檔案？