我有这个要求:
router.get("/fetchOngoingReleases") { (request) -> Future<[ReleaseOut]> in
return Release.query(on: request).filter(\.inprogress == true).all().map { releases in
var result: [ReleaseOut] = []
for r in releases {
var pageEvents: [Event] = []
let num = r.releaseUsers.query(on: request).filter(\.user.fbId ~~ "something").count()
var needAuthentication: Bool
if num == 0 {
needAuthentication = true
} else {
needAuthentication = false
}
let rOut = ReleaseOut(fbId: r.fbId, name: r.name, purpose: r.purpose, needAuthentication: needAuthentication)
result.append(rOut)
}
return result
}
}
}
说我无法访问查询中的(???) releaseUser.user.fbId
?
这里的数据模型:
和代码中
final class Release: Content {
var id: Int?
var fbId: String
var inprogress: Bool?
var name: String
var purpose: String
/// Creates a new `Release`.
init(id: Int? = nil, fbId: String, name: String, purpose: String = "normal selling") {
self.id = id
self.fbId = fbId
self.name = name
self.purpose = purpose
}
}
extension Release {
var releaseUsers: Children<Release, ReleaseUser> {
return children(\.releaseId)
}
}
final class ReleaseUser: Content {
var id: Int?
var releaseId: Release.ID
var userId: User.ID
init(id: Int? = nil, releaseId: Release.ID, userId: User.ID) {
self.id = id
self.releaseId = releaseId
self.userId = userId
}
}
extension ReleaseUser {
var user: Parent<ReleaseUser, User> {
return parent(\.userId)
}
}
final class User: Content {
var id: Int?
var fbId: String
var name: String
init(id: Int? = nil, fbId: String, name: String) {
self.id = id
self.fbId = fbId
self.name = name
}
}
答案 0 :(得分:2)
好吧,这里发生了几件事,但是主要的概念是您不能像这样跨不同的表跳转-您需要使用JOIN
来加入ReleaseUser
表到User
表中,以便随后可以在fbId
答案 1 :(得分:2)
尝试将查询更改为:
Release.query(on: request).filter(\.inprogress == true).join(\ReleaseUser.releaseId, to:\Release.Id).join(\ReleaseUser.userId, to:\User.Id).alsoDecode(User.self).all()
alsoDecode
将为您提供一个元组,第一个位置包含原始Release
实例,第二个位置包含相应的User
实例。因此,fbId
应该可用:
r.1.fbId
以您为例。