MongoDB的 - 如何投影从加入集合两个或多个字段

Question

我有以下问题，尽管我的搜索，我不出来...

我有两个集合具有以下值：

DOCTOROFFICES COLLECTION

{ _id: 1, name: "doctoroffice1", appointment_hours:["09:00","10:00","11:00"]} 
{ _id: 2, name: "doctoroffice2", appointment_hours:["09:30","11:30"]} 

医师COLLECTION

{_id: 1, name: "Kostas", lastname: "Pap", tel: "1234567890", doctoroffice: "doctoroffice1", rating: { totalRate: 300, totalVotes: 100 } } 
{_id: 2, name: "George", lastname: "Geo", tel: "7890246899", doctoroffice: "doctoroffice1", rating: { totalRate: 400, totalVotes: 200 } } 
{_id: 3, name: "Mark", lastname: "Ma", tel: "24689001122", doctoroffice: "doctoroffice2", rating: { totalRate: 450, totalVotes: 310 } } 
{_id: 4, name: "Paul", lastname: "Pa", tel: "2244668800", doctoroffice: "doctoroffice2", rating: { totalRate: 40, totalVotes: 10 } } 

如上所述，我有（2）医生办公室收集“医生办公室”。每个医生办公室包括（2）医生，保存在“医生”集合中。 我的目标是从我的MongoDB产生以下JSON答案：

{ 
    doctorOfficeLabel: "doctoroffice1", 
    ratingData: [ 
     { doctorName: "Kostas", 
     doctorLastname: "Pap", 
     totalRate: 300, 
     totalVotes: 100 }, 
     { doctorName: "George", 
     doctorLastname: "Geo", 
     totalRate: 400, 
     totalVotes: 200 } 
    ] 
}, 
{ 
    doctorOfficeLabel: "doctoroffice2", 
    ratingData: [ 
     { doctorName: "Mark", 
     doctorLastname: "Ma", 
     totalRate: 450, 
     totalVotes: 310 }, 
     { doctorName: "Paul", 
     doctorLastname: "Pa", 
     totalRate: 40, 
     totalVotes: 10 } 
    ] 
} 

我的解决方案迄今：

db.doctoroffices.aggregate([ 
    {$lookup: { 
       from: "doctors", 
       localField: "name", 
       foreignField: "doctoroffice", 
       as: "doctor_docs" 
       }  
    }, 
    {$unwind: "$doctor_docs"}, 
    {$project: { 
       _id: 0, 
       doctorOfficeLabel: "$name", 
       ratingData: { 
           doctorName: "$doctor_docs.name", 
           doctorLastname: "$doctor_docs.lastname", 
           totalRate: "$doctor_docs.rating.totalRate", 
           totalVotes: "$doctor_docs.rating.totalVotes", 
          } 
       } 
    } 
] 

这里的问题是，用相同“doctorOfficeLabel”不止一个对象创建，但我想要的医生办公室数量尽可能多的对象。 我也尝试了上面的代码，没有{$ unwind：“$ doctor_docs”}，但结果也不像预期的那样。

任何人都可以帮我解决我的问题，或者如果没有解决我的问题，建议我一个解决方法？

非常感谢您的时间！

Answer 1

你应该尝试在客户端做格式化，但如果你真的想在服务器端做，你可以使用$map。

服务器端

喜欢的东西

db.doctoroffices.aggregate([{ 
     $lookup: { 
      from: "doctors", 
      localField: "name", 
      foreignField: "doctoroffice", 
      as: "doctor_docs" 
     } 
    }, 
    { 
     $project: { 
      _id: 0, 
      doctorOfficeLabel: "$name", 
      ratingData: { 
       $map: { 
        input: "$doctor_docs", 
        as: "result", 
        in: { 
         doctorName: "$$result.name", 
         doctorLastname: "$$result.lastname", 
         totalRate: "$$result.rating.totalRate", 
         totalVotes: "$$result.rating.totalVotes", 
        } 
       } 
      } 
     } 
    } 
]) 

客户端

db.doctoroffices.aggregate({ 
    $lookup: { 
     from: "doctors", 
     localField: "name", 
     foreignField: "doctoroffice", 
     as: "doctor_docs" 
    } 
}).map(a => { 
    var doctoroffice = { 
     doctorOfficeLabel: a.name 
    }; 
    doctoroffice.ratingData = a.doctor_docs.map(b => { 
     var doctordoc = { 
      doctorName: b.name, 
      doctorLastname: b.lastname, 
      totalRate: b.rating.totalRate, 
      totalVotes: b.rating.totalVotes 
     } 
     return doctordoc; 
    }); 
    return doctoroffice; 
});