无法创建用于查询嵌套父对象的流畅ORM查询

Question

我有3张桌子，LastName，MiddleName和FirstName。。。FirstName以MiddleNameID作为父项，MiddleName具有LastName。作为父项的id

模型看起来像这样

final class LastName: Model, Content {
    static let schema = "lastnames"
    
    @ID(key: .id)
    var id: UUID?

    @Field(key: "name")
    var name: String

    
    @Children(for: \.$lastname)
    var middle_names: [MiddleName]
    
    init() { }

    init(id: UUID? = nil, name: String) {
        self.id = id
        self.name = name
    }
}

final class MiddleName: Model, Content {
    static let schema = "middlenames"
    
    @ID(key: .id)
    var id: UUID?

    @Parent(key: "last_name_id")
    var lastname: LastName
    
    @Field(key: "name")
    var name: String

    @Children(for: \.$middleNameId)
    var firstNames: [FirstName]
    
    init() { }

    init(id: UUID? = nil, lastname: LastName, name: String ) {
        self.id = id
        self.lastname = lastname
        self.name = name
    }
}


final class FirstName: Model, Content {
    static let schema = "firstnames"
    
    @ID(key: .id)
    var id: UUID?
    
    @Parent(key: "middle_name_id")
    var middleNameId: MiddleName
    
    
    @Field(key: "name")
    var name: String
   
    
    init() { }

    init(id: UUID? = nil, middleNameId: MiddleName, name: String) {
        self.id = id
        self.middleNameId = testBundleId
        self.name = name
    }
}

我想要这样的查询，比如给我所有的名字记录，其中lastname==Smith和/或middle==James。。。不确定如何使用Fluent DSL进行多重连接？或者给我所有与中间名和/或姓氏匹配的名字

如果我必须编写一个SQL查询，可能会像这样

                  SELECT DISTINCT ON (ln.id, mn.id)
                  ln.name, mn.name, fn.*
                  FROM last_name ln
                  LEFT JOIN middle_name mn ON ln.id = mn.last_name_id
                  LEFT JOIN first_name fn ON fn.middle_name_id = mn.id
                  WHERE ( mn.name <> '' AND fn.name <> '' ) AND
                  (true AND ln.name ~ 'smith')
                  ORDER BY ln.id, mn.id, fn.id

Answer 1

您可以在流畅的查询中执行此操作，例如：

return Lastname.query(on: req)
    .filter(\.$name == "Smith")
    .with(\.$middle_names) { $0.with(\.$firstnames) }
    .all()
    .flatMap { wholeNames in
    let result = wholeNames.filter { $0.middle_names.contains("James") }
    // continue
}

这将给你所有中间名为詹姆斯的史密斯一家。您可以将其用作OR变体的基础。它首先过滤史密斯的姓氏，然后用连接中间名，但闭包确保名字也连接在一起。

我认为没有办法将中间名过滤器构建到查询中，所以我认为您必须按照我所展示的那样做。