Question

您好，我是mongoDB的新手，我希望将一些数据汇总到一个将使用 $ out 创建的数据中。基本上，我有一个数据集合，其中的信息存储在数组中，如果相同的数据至少已插入到数组中一次，我想将数据放入新的集合中。最好的功能是执行此操作。我正在查看 $ filter 和 $ elemMatch 。

基本上，如果卡车的注册号中的任何一个至少出现在任何用户类型的机械师 maintenances 数组中，那么我想将该信息保存到新集合中类似于{“ registration”：“ truck-registration”}，没有重复项。

什么是最好的方法？

这些只是许多数据库插入中的一部分（用户很多，卡车也很多）。

卡车

db.transport.insert ( {"_id":"GFT008", "TRUCK": { "registration":"GFT008", "capacity":40000, "weight":15000, "status":"AVAILABLE"} });
db.transport.insert ( {"_id":"PKR008", "TRUCK": { "registration":"PKR008", "capacity":22000, "weight":8800, "status":"AVAILABLE"} });  
db.transport.insert ( {"_id":"QRT834", "TRUCK": { "registration":"QRT834", "capacity":5550, "weight":400, "status":"USED"} } );

员工

db.transport.insert ( {"_id":"6", "EMPLOYEE": { "e#":"6", "name":"Michael Jones", "dob":"05-OCT-65", "address":"23 Waterloo Ave. Surry Hills, NSW 2502", "hiredate":"12-JAN-93","position":"mechanic","licence":"7773","status":"ON_LEAVE","experience":"STANDARD","maintenances":[{"registration":"QRT834","time":"40","maintenance date":"12-JUN-99"},{"registration":"QRT834","time":"40","maintenance date":"15-JUN-98"},{"registration":"SYF777","time":"30","maintenance date":"01-AUG-98"},{"registration":"SYF777","time":"30","maintenance date":"05-AUG-95"},{"registration":"SYF777","time":"30","maintenance date":"06-AUG-00"},{"registration":"LUCY01","time":"200","maintenance date":"12-MAR-97"},{"registration":"SYF777","time":"30","maintenance date":"02-AUG-00"},{"registration":"PKR768","time":"200","maintenance date":"12-AUG-00"},{"registration":"QRT834","time":"200","maintenance date":"30-JUN-00"},{"registration":"SYF777","time":"300","maintenance date":"02-AUG-02"},{"registration":"PKR768","time":"460","maintenance date":"12-AUG-02"},{"registration":"LUCY01","time":"40","maintenance date":"29-JUL-02"},{"registration":"QRT834","time":"40","maintenance date":"30-JUN-02"}]} } );

Answer 1

看看：

MongoPlayground上的实时演示

db.employee.aggregate([
  {
    $unwind: "$EMPLOYEE.maintenances"
  },
  {
    $lookup: {
      from: "transport",
      localField: "EMPLOYEE.maintenances.registration",
      foreignField: "TRUCK.registration",
      as: "matches"
    }
  },
  {
    $match: {
      "matches": {
        $size: 1
      }
    }
  },
  {
    $group: {
      _id: "$EMPLOYEE.maintenances.registration",

    }
  }
])

这是一个多阶段聚合： -展开数组 -加入运输记录的放松阶段 -筛选完全匹配（？）的记录 -最后-通过分组删除重复项

结果：

[{"_id": "QRT834"}]

来自阵列的MongoDB聚合

1 个答案: