mongoDB：當對象中存在引用時，無法獲得完整的對象響應

Question

我正在嘗試使用find查詢對象。 整個對象就像這樣

{
"_id" : ObjectId("5aaf4133b72ed853c09d60d9"),
"invoiceItems" : [ 
    {
        "$ref" : "JobsPostings",
        "$id" : ObjectId("5aad26b6f78a341ef831b4f7"),
        "price" : 15,
        "publishDate" : ISODate("2018-03-17T00:00:00.000Z"),
        "_id" : ObjectId("5aaf4133b72ed853c09d60db")
    }, 
    {
        "$ref" : "JobsPostings",
        "$id" : ObjectId("5aad27e4f78a341ef831b513"),
        "price" : 20,
        "publishDate" : ISODate("2018-03-17T00:00:00.000Z"),
        "_id" : ObjectId("5aaf4133b72ed853c09d60da")
    }
  ], 
  {...}
}

我正在嘗試獲取invoiceItems對象，該對象是：

"invoiceItems" : [ 
    {
        "$ref" : "JobsPostings",
        "$id" : ObjectId("5aad26b6f78a341ef831b4f7"),
        "price" : 15,
        "publishDate" : ISODate("2018-03-17T00:00:00.000Z"),
        "_id" : ObjectId("5aaf4133b72ed853c09d60db")
    }, 
    {
        "$ref" : "JobsPostings",
        "$id" : ObjectId("5aad27e4f78a341ef831b513"),
        "price" : 20,
        "publishDate" : ISODate("2018-03-17T00:00:00.000Z"),
        "_id" : ObjectId("5aaf4133b72ed853c09d60da")
    }
]

這是我的獲取代碼：

PostPaidInvoice.find({_id: { $in : [ObjectId("5aaf4133b72ed853c09d60d9")]}}).then(result => {
      console.log(JSON.stringify(result[0].invoiceItems));
    }).catch(err => {
      console.log(err);
    });

我收到的回復不包含price和publishDate結果：

[{"_bsontype":"DBRef","namespace":"JobsPostings","oid":"5aad26b6f78a341ef831b4f7"},{"_bsontype":"DBRef","namespace":"JobsPostings","oid":"5aad27e4f78a341ef831b513"}]

但是，如果我從引用ref刪除美元符號（$），則它將返回整個對象。

Answer 1

您只需要在找到條件后添加一個關鍵的invoiceItems: 1 ：

PostPaidInvoice.find({
  _id: {
      $in: [ObjectId("5aaf4133b72ed853c09d60d9")]
  }
 }, { invoiceItems: 1, _id: 0 }).then(result => {
  console.log(JSON.stringify(result[0].invoiceItems));
 }).catch(err => {
  console.log(err);
});