swift 3 - 将实体作为参数传递给函数

Question

在早期版本的swift中，我有一个核心数据函数

func retrieveItemsForRelatedEntity(entity: String, relatedEntity: String, identifier: String, sortDescriptors: [NSSortDescriptor]?) -> Array<AnyObject>? {

    let appDel: AppDelegate = UIApplication.sharedApplication().delegate as! AppDelegate
    let context: NSManagedObjectContext = appDel.managedObjectContext!
    let frequest = NSFetchRequest(entityName: entity)
    frequest.returnsObjectsAsFaults = false

    if sortDescriptors != nil {
        frequest.sortDescriptors = sortDescriptors
    } 

    switch relatedEntity {
    case "CostCentre":
        frequest.predicate = NSPredicate(format: "costCentre.identifier == '\(identifier)'")
        return try! context.executeFetchRequest(frequest)
    case "CostCentreDay":
        frequest.predicate = NSPredicate(format: "costCentreDay.identifier == '\(identifier)'")
        return try! context.executeFetchRequest(frequest)
    case "Resource":
        frequest.predicate = NSPredicate(format: "resource.identifier == '\(identifier)'")
        return try! context.executeFetchRequest(frequest)
    case "ResourceDay":
        frequest.predicate = NSPredicate(format: "resourceDay.identifier == '\(identifier)'")
        return try! context.executeFetchRequest(frequest)
    case "CostedDay":
        frequest.predicate = NSPredicate(format: "costedDay.identifier == '\(identifier)'")
        return try! context.executeFetchRequest(frequest)
    default:
        print("wrong entity for this function")
        return nil
    }
}

在 Swift 3.0 中，“无法推断通用结果类型”，所以我可以传入实体类型，而不是传入实体标题字符串，然后为我的获取请求打开它吗？

Answer 1

回答你的问题

所以我可以不传入实体标题字符串，而是传入实体类型，然后为我的获取请求打开它吗？

是的，那将是最好的方法。类似的东西

 retrieveItemsForRelatedEntity(entity: NSManagedObject, relatedEntity: String, identifier: String, sortDescriptors: [NSSortDescriptor]?) -> Array<AnyObject>? {

From another question

您必须指定泛型类型，否则方法调用不明确。

第一个版本是为 NSManagedObject 定义的，第二个版本是为每个使用扩展名的对象自动生成的，例如：

这个扩展是自动生成的（如下图Animal 类型。

来自 Apple 的关于从 Swift 2 更改为 Swift 3 的文档

NSFetchRequest 现在是基于新 NSFetchRequestResult 协议的参数化类型。

executeFetchRequest(...) 的函数名称在 Swift 3 中也发生了变化

public func fetch<T : NSFetchRequestResult>(_ request: NSFetchRequest<T>) throws -> [T]

NSFetchRequest 更改的 Apple 文档示例

斯威夫特 2

func findAnimals() {
    let request = NSFetchRequest(entityName:”Animal")
    do {
        guard let searchResults = try context.executeFetchRequest(request) as? [Animal] else {
            print("Results were not of the expected structure")
        }
        ... use(searchResults) ...
    } catch {
    print("Error ocurred during execution: \(error)")
}

}

斯威夫特 3

func findAnimals() {
    let request: NSFetchRequest<Animal> = Animal.fetchRequest
    do {
        let searchResults = try context.fetch(request)
        ... use(searchResults) ...
    } catch {
        print("Error with request: \(error)")
    }
}

因此在 Swift 2 中转换您的代码以创建 NSFetchRequest

let frequest = NSFetchRequest(entityName: entity)

到 Swift 3

let frequest: NSFetchRequest<YourEntityType> = YourEntityType.fetchRequest()

和

context.executeFetchRequest(frequest)

将会

context.fetch(frequest)

注意：从网络上的其他示例来看，Animal.fetchRequest 的 Apple 文档代码示例可能是错误的，而应该是 Animal.fetchRequest()

Answer 2

请试试这个。它帮助了我。

func retrieveItemsForRelatedEntity(entity: String, relatedEntity: NSEntityDescription, identifier: String, sortDescriptors: [NSSortDescriptor]?) -> Array<AnyObject>? {

    let appDelegate = UIApplication.shared.delegate as! AppDelegate
    let context: NSManagedObjectContext = appDelegate.persistentContainer.viewContext

    let frequest = NSFetchRequest<NSFetchRequestResult>(entityName: entity) 

和

switch relatedEntity {
    case NSEntityDescription.entity(forEntityName: "CostCentre", in: context)!:

Answer 3

试试这个：

let frequest = NSFetchRequest<NSFetchRequestResult>(entityName: entity)