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我需要对不同的表进行大量查询，并在一个响应中返回结果。我想知道我可以展平极其嵌套的函数吗？

final class CodeController {
    internal func indexCodes(_ req: Request) throws -> EventLoopFuture<Response> {
        CodeCountry.query(on: req).sort(\.name).all().flatMap { countries -> EventLoopFuture<Response> in
            CodeFloor.query(on: req).sort(\.name).all().flatMap { floors -> EventLoopFuture<Response> in
                CodeRegion.query(on: req).sort(\.name).all().flatMap { regions -> EventLoopFuture<Response> in
                    CodeObjectType.query(on: req).sort(\.name).all().flatMap { objectTypes -> EventLoopFuture<Response> in
                        CodePropertyType.query(on: req).sort(\.name).all().flatMap { propertyTypes -> EventLoopFuture<Response> in
                            CodeRooms.query(on: req).sort(\.name).all().flatMap { rooms -> EventLoopFuture<Response> in
                                let codes = CodesContent(
                                    countries: countries,
                                    floors: floors,
                                    regions: regions,
                                    objectTypes: objectTypes,
                                    propertyTypes: propertyTypes,
                                    rooms: rooms
                                )

                                return codes.encode(status: .ok, for: req)
                            }
                        }
                    }
                }
            }
        }
    }
}
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1 回答 1

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这可能不是最好看的代码，但我认为可以解决问题。我喜欢@Nick 将新数据模型作为集合引入的方法，所以也会这样做。您可以合并两个期货并获得一个。这来自 SwiftNIO

let futureThingOne: Future<ThingOne> = ...
let futureThingTwo: Future<ThingTwo> = ...
let futureBothThings: Future<(ThingOne, ThingTwo)> = futureThingOne.and(futureThingTwo)

因此，通过合并来自查询的期货，您可以构建一个扁平的结构。

struct Codes: Content {
    let countries: [CodeCountry]
    let floors: [CodeFloor]
    let regions: [CodeRegion]
    let objectTypes: [CodeObjectType]
    let propertyTypes: [CodePropertyType]
    let rooms: [CodeRooms]

    static func codes(req: Request) -> Future<Codes> {
        let futureCountries = CodeCountry.query(on: req).sort(\.name).all()
        let futureFloors = CodeFloor.query(on: req).sort(\.name).all()
        let futureRegions = CodeRegion.query(on: req).sort(\.name).all()
        let futureObjectTypes = CodeObjectType.query(on: req).sort(\.name).all()
        let futurePropertyTypes = CodePropertyType.query(on: req).sort(\.name).all()
        let futureRooms = CodeRooms.query(on: req).sort(\.name).all()

        let combined = futureCountries.and(futureFloors).and(futureRegions).and(futureObjectTypes).and(futurePropertyTypes).and(futureRooms)
       return combined.map {
            Codes.init(
                countries: $0.0.0.0.0.0,
                floors: $0.0.0.0.0.1,
                regions: $0.0.0.0.1,
                objectTypes: $0.0.0.1,
                propertyTypes: $0.0.1,
                rooms: $0.1
            )
        }

    }
}
于 2020-03-30T07:38:08.890 回答