如何将类型应用于NSFetchRequest实例？

Question

在Swift 2中,以下代码正在运行:

let request = NSFetchRequest(entityName: String)

但是在Swift 3中它给出了错误:

无法推断通用参数"ResultType"

因为NSFetchRequest现在是通用类型.他们在文件中写道:

let request: NSFetchRequest<Animal> = Animal.fetchRequest

所以如果我的结果类是例如Level我应该如何正确请求？

因为这不起作用:

let request: NSFetchRequest<Level> = Level.fetchRequest

Answer 1

let request: NSFetchRequest<NSFetchRequestResult> = Level.fetchRequest()

要么

let request: NSFetchRequest<Level> = Level.fetchRequest()

取决于您想要的版本.

您必须指定泛型类型,否则方法调用是不明确的.

定义第一个版本NSManagedObject,使用扩展名为每个对象自动生成第二个版本,例如:

extension Level {
    @nonobjc class func fetchRequest() -> NSFetchRequest<Level> {
        return NSFetchRequest<Level>(entityName: "Level");
    }

    @NSManaged var timeStamp: NSDate?
}

重点是删除String常量的用法.

Answer 2

我想通过这样做我得到了它:

let request:NSFetchRequest<NSFetchRequestResult> = NSFetchRequest(entityName: "Level")

至少它保存并加载DataBase中的数据.

但感觉它不是一个合适的解决方案,但它现在有效.

Answer 3

我发现在3.0中工作的最简单的结构如下:

let request = NSFetchRequest<Country>(entityName: "Country")

数据实体类型为国家/地区的位置.

但是,在尝试创建Core Data BatchDeleteRequest时,我发现此定义不起作用,您似乎需要使用表单:

let request: NSFetchRequest<NSFetchRequestResult> = Country.fetchRequest()

即使ManagedObject和FetchRequestResult格式应该是等效的.

Answer 4

以下是一些可能回答您问题的通用CoreData方法:

import Foundation
import Cocoa

func addRecord<T: NSManagedObject>(_ type : T.Type) -> T
{
    let entityName = T.description()
    let context = app.managedObjectContext
    let entity = NSEntityDescription.entity(forEntityName: entityName, in: context)
    let record = T(entity: entity!, insertInto: context)
    return record
}

func recordsInTable<T: NSManagedObject>(_ type : T.Type) -> Int
{
    let recs = allRecords(T.self)
    return recs.count
}


func allRecords<T: NSManagedObject>(_ type : T.Type, sort: NSSortDescriptor? = nil) -> [T]
{
    let context = app.managedObjectContext
    let request = T.fetchRequest()
    do
    {
        let results = try context.fetch(request)
        return results as! [T]
    }
    catch
    {
        print("Error with request: \(error)")
        return []
    }
}

func query<T: NSManagedObject>(_ type : T.Type, search: NSPredicate?, sort: NSSortDescriptor? = nil, multiSort: [NSSortDescriptor]? = nil) -> [T]
{
    let context = app.managedObjectContext
    let request = T.fetchRequest()
    if let predicate = search
    {
        request.predicate = predicate
    }
    if let sortDescriptors = multiSort
    {
        request.sortDescriptors = sortDescriptors
    }
    else if let sortDescriptor = sort
    {
        request.sortDescriptors = [sortDescriptor]
    }

    do
    {
        let results = try context.fetch(request)
        return results as! [T]
    }
    catch
    {
        print("Error with request: \(error)")
        return []
    }
}


func deleteRecord(_ object: NSManagedObject)
{
    let context = app.managedObjectContext
    context.delete(object)
}

func deleteRecords<T: NSManagedObject>(_ type : T.Type, search: NSPredicate? = nil)
{
    let context = app.managedObjectContext

    let results = query(T.self, search: search)
    for record in results
    {
        context.delete(record)
    }
}

func saveDatabase()
{
    let context = app.managedObjectContext

    do
    {
        try context.save()
    }
    catch
    {
        print("Error saving database: \(error)")
    }
}

假设有一个NSManagedObject设置为Contact,如下所示:

class Contact: NSManagedObject
{
    @NSManaged var contactNo: Int
    @NSManaged var contactName: String
}

可以通过以下方式使用这些方法:

let name = "John Appleseed"

let newContact = addRecord(Contact.self)
newContact.contactNo = 1
newContact.contactName = name

let contacts = query(Contact.self, search: NSPredicate(format: "contactName == %@", name))
for contact in contacts
{
    print ("Contact name = \(contact.contactName), no = \(contact.contactNo)")
}

deleteRecords(Contact.self, search: NSPredicate(format: "contactName == %@", name))

recs = recordsInTable(Contact.self)
print ("Contacts table has \(recs) records")

saveDatabase()

Answer 5

这是迁移到Swift 3.0的最简单方法,只需添加即可 <Country>

(测试和工作)

let request = NSFetchRequest<Country>(entityName: "Country")