Mongodb query MongoDB：如何从查询中删除重复记录？

示例集合：

员工

{"FNAME" : "John", "LNAME" : "Smith", "SSN" : "123456789", "SALARY" : 30000, "SUPERSSN" : "333445555"}
{"FNAME" : "Franklin", "LNAME" : "Wong", "SSN" : "333445555", "SALARY" : 40000, "SUPERSSN" : "888665555"}
{"FNAME" : "Joyce", "LNAME" : "English", "SSN" : "453453453", "SALARY" : 25000, "SUPERSSN" : "333445555"}
{"FNAME" : "Ramesh", "LNAME" : "Narayan", "SSN" : "666884444", "SALARY" : 38000, "SUPERSSN" : "333445555"}
{"FNAME" : "James", "LNAME" : "Borg", "SSN" : "888665555", "SALARY" : 55000, "SUPERSSN" : "", "DNO" : 1 }
{"FNAME" : "Jennifer", "LNAME" : "Wallace", "SSN" : "987654321", "SALARY" : 43000, "SUPERSSN" : "888665555"}
{"FNAME" : "Ahmad", "LNAME" : "Jabbar", "SSN" : "987987987", "SALARY" : 25000, "SUPERSSN" : "987654321"}
{"FNAME" : "Alicia", "LNAME" : "Zelaya", "SSN" : "999887777", "SALARY" : 25000, "SUPERSSN" : "987654321"}
{"FNAME" : "John", "LNAME" : "Smith", "SSN" : "123456789", "SALARY" : 30000, "SUPERSSN" : "333445555"}
{"FNAME" : "Franklin", "LNAME" : "Wong", "SSN" : "333445555", "SALARY" : 40000, "SUPERSSN" : "888665555"}
{"FNAME" : "Joyce", "LNAME" : "English", "SSN" : "453453453", "SALARY" : 25000, "SUPERSSN" : "333445555"}
{"FNAME" : "Ramesh", "LNAME" : "Narayan", "SSN" : "666884444", "SALARY" : 38000, "SUPERSSN" : "333445555"}
{"FNAME" : "James", "LNAME" : "Borg", "SSN" : "888665555",  "SALARY" : 55000, "SUPERSSN" : "", "DNO" : 1 }
{"FNAME" : "Jennifer", "LNAME" : "Wallace", "SSN" : "987654321",  "SALARY" : 43000, "SUPERSSN" : "888665555"}
{"FNAME" : "Ahmad", "LNAME" : "Jabbar", "SSN" : "987987987",  "SALARY" : 25000, "SUPERSSN" : "987654321"}
{"FNAME" : "Alicia", "LNAME" : "Zelaya", "SSN" : "999887777", "SALARY" : 25000, "SUPERSSN" : "987654321"}

工作于

{ "ESSN" : "123456789", "PNO" : 1, "HOURS" : 32.5 }
{ "ESSN" : "123456789", "PNO" : 2, "HOURS" : 7.5 }
{ "ESSN" : "333445555", "PNO" : 2, "HOURS" : 10 }
{ "ESSN" : "333445555", "PNO" : 3, "HOURS" : 10 }
{ "ESSN" : "333445555", "PNO" : 10, "HOURS" : 10 }
{ "ESSN" : "333445555", "PNO" : 20, "HOURS" : 10 }
{ "ESSN" : "453453453", "PNO" : 1, "HOURS" : 20 }
{ "ESSN" : "453453453", "PNO" : 2, "HOURS" : 20 }
{ "ESSN" : "666884444", "PNO" : 3, "HOURS" : 40 }
{ "ESSN" : "888665555", "PNO" : 20, "HOURS" : 0 }
{ "ESSN" : "987654321", "PNO" : 20, "HOURS" : 15 }
{ "ESSN" : "987654321", "PNO" : 30, "HOURS" : 20 }
{ "ESSN" : "987987987", "PNO" : 10, "HOURS" : 35.5 }
{ "ESSN" : "987987987", "PNO" : 30, "HOURS" : 5.5 }
{ "ESSN" : "999887777", "PNO" : 10, "HOURS" : 10 }
{ "ESSN" : "999887777", "PNO" : 30, "HOURS" : 30 }

我想从以下MongoDB查询的“连接”中删除重复记录：

db.employee.aggregate([
    {
        $lookup:{
            from: "works_on",    
            localField: "SSN",   
            foreignField: "ESSN",
            as: "works_on_here"  
        }
    },
    {   $unwind:"$works_on_here" },
    {
        $group:{
          _id:"$_id",
          nodes:{
            $addToSet:"$works_on_here"
        }
    },
    {   
        $project:{
            _id : 1,
            FNAME : 1,
            LNAME : 1,
            HOURS : "$works_on_here.HOURS",
        } 
    }
]);

预期结果是：

{ "FNAME" : "John", "LNAME" : "Smith", "HOURS" : 32.5 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }
{ "FNAME" : "Joyce", "LNAME" : "English", "HOURS" : 20 }
{ "FNAME" : "Ramesh", "LNAME" : "Narayan", "HOURS" : 40 }
{ "FNAME" : "James", "LNAME" : "Borg", "HOURS" : 0 }
{ "FNAME" : "Jennifer", "LNAME" : "Wallace", "HOURS" : 20 }
{ "FNAME" : "Ahmad", "LNAME" : "Jabbar", "HOURS" : 5.5 }
{ "FNAME" : "Alicia", "LNAME" : "Zelaya", "HOURS" : 30 }
{ "FNAME" : "John", "LNAME" : "Smith", "HOURS" : 7.5 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }

实际输出没有“$group”部分，如下所示：

{ "FNAME" : "John", "LNAME" : "Smith", "HOURS" : 32.5 }
{ "FNAME" : "John", "LNAME" : "Smith", "HOURS" : 7.5 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }
{ "FNAME" : "Joyce", "LNAME" : "English", "HOURS" : 20 }
{ "FNAME" : "Joyce", "LNAME" : "English", "HOURS" : 20 }
{ "FNAME" : "Ramesh", "LNAME" : "Narayan", "HOURS" : 40 }
{ "FNAME" : "James", "LNAME" : "Borg", "HOURS" : 0 }
{ "FNAME" : "Jennifer", "LNAME" : "Wallace", "HOURS" : 15 }
{ "FNAME" : "Jennifer", "LNAME" : "Wallace", "HOURS" : 20 }
{ "FNAME" : "Ahmad", "LNAME" : "Jabbar", "HOURS" : 35.5 }
{ "FNAME" : "Ahmad", "LNAME" : "Jabbar", "HOURS" : 5.5 }
{ "FNAME" : "Alicia", "LNAME" : "Zelaya", "HOURS" : 10 }
{ "FNAME" : "Alicia", "LNAME" : "Zelaya", "HOURS" : 30 }
{ "FNAME" : "John", "LNAME" : "Smith", "HOURS" : 32.5 }
{ "FNAME" : "John", "LNAME" : "Smith", "HOURS" : 7.5 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }
{ "FNAME" : "Franklin", "LNAME" : "Wong", "HOURS" : 10 }

我有两个集合“employee”和“works_on”，我试着做一些像这样的事情”

带有$group部分的代码不返回任何内容。这应作为复制过滤器或？

• $group

  by

  SSN

  并首先获取

  FNAME

  和

  LNAME

  字段，如果需要其他字段，可以添加与

  FNAME

  和

  LNANE

• $lookup

  与

  一起工作

• $project

  显示必填字段，并使用

  $sum

---

。请参阅此更新的我的问题-$group部分不起作用。我正在做smth。错误？你能在问题中发布收集和预期结果的样本文档吗？当然，问题已更新！员工集合中的唯一字段是什么，是否为

SSN

？好的，此解决方案工作正常。投影也可以吗？我试图写一些类似于一个小转换器的东西，所以我在寻找一个通用的解决方案。如果我不想做基本的计算，那么投影是可行的，但它是否也适用于计算？关键字$first是什么？是的，您可以使用$project而不是$ADDFIERS stage关键字$first是什么？$group stage将按SSN对您的根文档进行分组，因此实际上它在group stage内循环，因此$first表示我们正在从分组的根元素中选择第一个元素。您能否提供带有$project和$sum的解决方案？不适合我。。。

db.employee.aggregate([
  {
    $group: {
      _id: "$SSN",
      FNAME: { $first: "$FNAME" },
      LNAME: { $first: "LNAME" }
    }
  },
  {
    $lookup: {
      from: "works_on",
      localField: "_id",
      foreignField: "ESSN",
      as: "HOURS"
    }
  },
  {
    $project: {
      _id: 0,
      FNAME: 1,
      LNAME: 1,
      HOURS: {
        $sum: "$HOURS.HOURS"
      }
    }
  }
])