快速3-將實體作為函數的參數傳遞

Question

在Swift的早期版本中，我具有核心數據功能

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
    }
}

使用Swift 3.0，“無法推斷出通用結果類型”，因此我可以傳入實體類型，然后為我的提取請求打開該實體，而不是傳入實體標題字符串。

Answer 1

回答你的問題

因此，除了傳遞實體標題字符串之外，我還可以傳遞實體類型，然后為我的提取請求啟用它嗎？

是的，那將是最好的方法。 就像是

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

從另一個問題

您必須指定泛型類型，因為否則方法調用是不明確的。

第一個版本是為NSManagedObject定義的，第二個版本是使用擴展名自動為每個對象生成的，例如：

此擴展名是自動生成的（如下面的Animal類型所示。

從Apple文檔中了解從Swift 2到Swift 3更改

NSFetchRequest現在是基於新NSFetchRequestResult協議的參數化類型。

Swift 3中的executeFetchRequest(...)函數名稱也已更改

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

Apple文檔對NSFetchRequest更改的NSFetchRequest

迅捷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)")
}

}

迅捷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中轉換代碼以創建NSFetchRequest

let frequest = NSFetchRequest(entityName: entity)

到Swift 3

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

和

context.executeFetchRequest(frequest)

將會

context.fetch(frequest)

注意：從網絡上的其他示例來看，Apple docs Animal.fetchRequest示例代碼可能是錯誤的，而應該是Animal.fetchRequest()

Answer 2

請嘗試這個。 它幫助了我。

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

嘗試這個 ：

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