我有两个这样的查询:
db.employee.aggregate({$match: {lname:"Smith"}}, {$project: {"SSN": 1, "_id": 0}})
db.works_on.aggregate({$match: {essn: 123456789}}, {$project: {"pno": 1, "_id": 0}})
有没有办法把它变成一个单一的查询,用第一个聚合输出的SSN替换essn编号?
是的,您可以使用$lookup来完成。请尝试以下查询。
db.employee.aggregate([
{
$lookup:{
from: "works_on",
localField: "SSN",
foreignField: "essn",
as: "employee_works"
}
},
{ $unwind:"$employee_works" },
{
$project:{
lname: 1,
SSN: 1,
pno : "$employee_works.pno",
}
}
]);