gpt4 book ai didi

swift - 如何获取与 coreData 中的记录相关的关系

转载 作者:行者123 更新时间:2023-11-28 14:06:21 25 4
gpt4 key购买 nike

我是 coredata 的新手,我想获取与记录相关的属性。我使用以下代码插入一条记录。

let entityDescription = NSEntityDescription.entity(forEntityName: "Person", in: mainContext!)
let newPerson = NSManagedObject(entity: entityDescription!,insertInto: mainContext)
let entityAddress = NSEntityDescription.entity(forEntityName: "Address", in: mainContext!)
let newAddress = NSManagedObject(entity: entityAddress!,insertInto: mainContext)
let otherAddress = NSManagedObject(entity: entityAddress!, insertInto: mainContext)

newPerson.setValue("Bart", forKey: "first")
newPerson.setValue("Jacobs", forKey: "last")
newPerson.setValue(44, forKey: "age")
newAddress.setValue("Main Street", forKey: "street")
newAddress.setValue("Boston", forKey: "city")
newPerson.setValue(NSSet(object: newAddress), forKey: "addresses")

otherAddress.setValue("5th Avenue", forKey:"street")
otherAddress.setValue("New York", forKey:"city")
// Add Address to Person
let addresses = newPerson.mutableSetValue(forKey: "addresses")
addresses.add(otherAddress)



// Set First and Last Name


do {
try newPerson.managedObjectContext?.save()
}
catch
{
print("Couldnot do this op \(error)")
}

我的CoreData关系如下enter image description here

然后我使用以下代码获取用户

let fetchRequest = NSFetchRequest<NSFetchRequestResult>()

let entityDescription = NSEntityDescription.entity(forEntityName: "Person", in: mainContext!)
fetchRequest.entity = entityDescription
fetchRequest.predicate = NSPredicate(value: true)
do{
let result = try self.mainContext?.fetch(fetchRequest)
for personTest in result!
{
print(personTest)
}
if ((result?.count)! > 0)
{
let person = result![0] as! NSManagedObject
print(person.value(forKey: "first"))
}

}
catch
{
let fetchError = error as NSError
print(fetchError)
}

我的问题是如何获取与当前人相关的地址

最佳答案

您不获取它们,而是通过addresses 属性访问它们

let adresses = personTest.adresses

关于swift - 如何获取与 coreData 中的记录相关的关系,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52956620/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com