我正在开发一个使用 SQLite 数据库的 iOS 应用程序,并且我使用 SQLite.swift 库(https://github.com/stephencelis/SQLite.swift)。
我正在尝试在 Swift 3 中迁移我的应用程序,所以我更改了我的库以使用分支swift3-mariotaku
(https://github.com/stephencelis/SQLite.swift/tree/swift3-mariotaku)
当我尝试使用 a 时,我仍然遇到一个问题join
:Ambiguous reference to member ==
这是我的代码:
class ArticlesDAO {
static let articles = Table("Article")
static let id = Expression<Int?>("id")
}
class FiltresVehiculesDAO {
let vfiltres = Table("VFiltre")
let idVehi = Expression<Int?>("vehicule")
func get(_ idVehicule: Int) throws -> [FiltreVehicule] {
let sqlQuery = vfiltres.join(
ArticlesDAO.articles,
// Next Line : "Ambiguous reference to member ==" error
on: vfiltres[idArticle] == ArticlesDAO.articles[ArticlesDAO.id]
)
//[...]
}
}
经过几次搜索,我发现这个线程Swift 3 URLSession.shared() Ambiguous reference to member 'dataTask(with:completionHandler:) error (bug)。所以我尝试应用在on
参数中指定返回类型的解决方案,如下所示:
on: (vfiltres[idArticle] == ArticlesDAO.articles[ArticlesDAO.id]) as Expression<Bool?>
我也尝试精确每个元素:
on: ((vfiltres[idArticle] as Expression<Int?>) == (ArticlesDAO.articles[ArticlesDAO.id] as Expression<Int?>)) as Expression<Bool?>
错误仍然相同。
我检查了库代码,但我不知道如何解决这个问题,所以这是使用的库代码,也许它应该有助于理解:
join
方法:
public func join(_ table: QueryType, on condition: Expression<Bool>) -> Self {
return join(table, on: Expression<Bool?>(condition))
}
==
过载:
public func ==<V : Value>(lhs: Expression<V>, rhs: Expression<V>) -> Expression<Bool> where V.Datatype : Equatable {
return "=".infix(lhs, rhs)
}
String
扩展名(用于方法infix
):
extension String {
func infix<T>(_ lhs: Expressible, _ rhs: Expressible, wrap: Bool = true) -> Expression<T> {
let expression = Expression<T>(" \(self) ".join([lhs, rhs]).expression)
guard wrap else {
return expression
}
return "".wrap(expression)
}
func wrap<T>(_ expression: Expressible) -> Expression<T> {
return Expression("\(self)(\(expression.expression.template))", expression.expression.bindings)
}
func wrap<T>(_ expressions: [Expressible]) -> Expression<T> {
return wrap(", ".join(expressions))
}
}
谢谢您的帮助