Question

我正在尝试使用find查询对象。整个对象就是这样

{
"_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")
    }
  ], 
  {...}
}

我正在尝试获取invoiceItems对象：

"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")
    }
]

这是我的提取代码：

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

我收到的回复不包括price和publishDate的结果：

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

但如果我从引用ref中删除美元符号（$），则返回整个对象。

Answer 1

在查找条件后，您只需添加一个键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);
});

mongoDB：当对象中存在引用时，无法获得完整对象响应

1 个答案: