Question

我的db集合就像

 {  "_id" : "57f4c1323eb5c694041ghea3",
 "example" : [ 
    {
        "exId" : "57f4c0d43eb5c694041fdebd",
        "_id" : ObjectId("57f4c1323eb5c694041fded2"),
        "projects" : [ 
            "57f4c1303eb5c694041fdec6"
        ]
    }, 
    {
        "exId" : "57f24ee56da92f9c19b0efd4",
        "_id" : ObjectId("57f4cfa93eb5c694041fdf4c"),
        "projects" : [ 
            "57f4cfa83eb5c694041fdf45"
        ]
    }, 
    {
        "exId" : "57f24ec16da92f9c19b0efcd",
        "_id" : ObjectId("580864867a36f5bc1058c2c8"),
        "projects" : [ 
            "580864837a36f5bc1058c2a9", 
            "57f4c1303eb5c694041fdec6"
        ]
    }
],

我需要输出

"example": [
{
  "exId": "57f24ee56da92f9c19b0efd4",
  "_id": "57f4cfa93eb5c694041fdf4c",
  "projects": [
    "57f4cfa83eb5c694041fdf45"
  ]
},
{
      "exId" : "57f24ec16da92f9c19b0efcd",
      "_id" : ObjectId("580864867a36f5bc1058c2c8"),
      "projects" : [ 
          "580864837a36f5bc1058c2a9", 
          "57f4c1303eb5c694041fdec6"
      ]
  }
 ]
 }

我的查询是

     db.userDb.findOne({ '_id': req.params.userId }, { example: { $elemMatch: { exId: { $in: accIds } } } })

和accIds是一个包含值“57f24ec16da92f9c19b0efcd”的数组， “57f24ee56da92f9c19b0efd4” “57f24fa26da92f9c19b0f005”

但是我的输出是

"example": [
{
  "exId": "57f24ee56da92f9c19b0efd4",
  "_id": "57f4cfa93eb5c694041fdf4c",
  "projects": [
    "57f4cfa83eb5c694041fdf45"
  ]
}

]，

感谢任何帮助

Answer 1

db.collection.aggregate([
            { $unwind: '$example' },
            { $unwind: '$example.projects' },
            { $match: { 'example.exId': { $in: accIds} } },
            { $group: { _id: null, projectList: { $addToSet: '$example.projects'  } } },
        ])

希望这会奏效。

如何从mongodb中的对象数组中获取项目

1 个答案: