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swift中具有重复实体值的核心数据

core-data swift ios

naga · 技术社区 · 7 年前

我正在解析Json中的一些数据并将其保存到coredata中,在将核心数据提取到tableview中后显示工作正常,tableview正在重复显示所有值中如何避免重复值我尝试了很多方法,但没有找到方法

Json格式

{
    "contacts": [
        {
                "id": "c200",
                "name": "Ravi Tamada",
                "email": "ravi@gmail.com",
                "address": "xx-xx-xxxx,x - street, x - country",
                "gender" : "male",
                "phone": {
                    "mobile": "+91 0000000000",
                    "home": "00 000000",
                    "office": "00 000000"
                }
        },
        {
                "id": "c201",
                "name": "Johnny Depp",
                "email": "johnny_depp@gmail.com",
                "address": "xx-xx-xxxx,x - street, x - country",
                "gender" : "male",
                "phone": {
                    "mobile": "+91 0000000000",
                    "home": "00 000000",
                    "office": "00 000000"
                }
        },
        {
                "id": "c202",
                "name": "Leonardo Dicaprio",
                "email": "leonardo_dicaprio@gmail.com",
                "address": "xx-xx-xxxx,x - street, x - country",
                "gender" : "male",
                "phone": {
                    "mobile": "+91 0000000000",
                    "home": "00 000000",
                    "office": "00 000000"
                }
        }
  ]
}

当我获取显示重复值的“name”时

这些是保存和获取代码

func getfetchdata()
{

    let fetchRequest = NSFetchRequest<NSFetchRequestResult>(entityName: "Contacts")


    do{
        let fetchResults = try self.context.fetch(fetchRequest) as? [Contacts]
    if fetchResults!.count > 0 {

        for bEntity in fetchResults! {

            let employeeEntity =  bEntity

            print(employeeEntity.name as Any)

            TableviewData.append(ContactsDataVal.init(name: employeeEntity.name!,
                                                      id: employeeEntity.id!, email: employeeEntity.email!, gender: employeeEntity.gender!, address: employeeEntity.address!))

        }

      print("data values already")

        }}
    catch let error as NSError
    {
        print(error)
    }
}
func getdata()
{

    let url = URL(string: "https://api.androidhive.info/contacts/")

    URLSession.shared.dataTask(with: url!) { (Data, response, error) in
       do
        {
            let data = try JSONSerialization.jsonObject(with: Data!, options: JSONSerialization.ReadingOptions.allowFragments)as! [String:AnyObject]

            let arraydata = data["contacts"]as! [[String:Any]]


            for arravalues in arraydata
            {

                let entityDescription = NSEntityDescription.entity(forEntityName: "Contacts", in:self.context)

                let favouriteObj = Contacts(entity: entityDescription!, insertInto: self.context)

                favouriteObj.name = arravalues["name"] as? String

                favouriteObj.id = arravalues["id"] as? String

                favouriteObj.email = arravalues["email"] as? String

                favouriteObj.gender = arravalues["gender"] as? String

                favouriteObj.address = arravalues["address"] as? String

                do {
                    try self.context.save()
                }

            }

        }catch let error as NSError{

            print("error",error)
        }
    }
        .resume()
  }

如何避免核心数据中的重复值,并在tableview中显示正确的数据

2 回复 | 直到 7 年前

vadian 7 年前

首先在 getdata 不要在循环的每次迭代中保存上下文,请在循环后保存一次。

为了避免重复,请从核心数据中提取所有联系人,将其映射到名称,并检查数组是否包含收到的名称

func getdata()
{

    let url = URL(string: "https://api.androidhive.info/contacts/")
    let names : [String]
    do {
        let fetchRequest = NSFetchRequest<NSFetchRequestResult>(entityName: "Contacts")
        let fetchResults = try self.context.fetch(fetchRequest)
        names = fetchResults.map{ $0.name }
    } catch {
        names = []
    }

    URLSession.shared.dataTask(with: url!) { (data, response, error) in
        if error != nil { print(error!); return }
        do {
            let data = try JSONSerialization.jsonObject(with: data!) as! [String:Any]
            let arraydata = data["contacts"] as! [[String:Any]]
            for arravalues in arraydata
            {
                guard let name = arravalues["name"] as? String, !names.contains(name) else { continue }
                let entityDescription = NSEntityDescription.entity(forEntityName: "Contacts", in:self.context)
                let favouriteObj = Contacts(entity: entityDescription!, insertInto: self.context)
                favouriteObj.name = name
                favouriteObj.id = arravalues["id"] as? String
                favouriteObj.email = arravalues["email"] as? String
                favouriteObj.gender = arravalues["gender"] as? String
                favouriteObj.address = arravalues["address"] as? String
            }
            try self.context.save()
        } catch {
            print("error",error)
        }
    }
    .resume()
}

备注:

使用通用fetch请求返回的核心数据获取总是一 非可选 的数组 NSManagedObject 成功时指定为泛型类型的子类。
从不使用检查空数组 foo.count > 0 使用 !foo.isEmpty
Swift 3+中的JSON字典总是 [String:Any] 而不是 [String:AnyObject]
处理中的潜在错误 dataTask 完成区块。
将完成块中的第一个参数命名为小写( data )避免与类型混淆 Data 。
省略 options 中的参数 jsonObject(with 因此,显然是一种集合类型。

Joakim Danielson 7 年前

可能可以对fetch请求执行类似的操作 How to get distinct results from a single field in Core Data (Swift 4) 但另一种选择是,只需从获取结果中创建一个集合,即可删除重复项:

let fetchSet = Set(fetchResults)

并在集合上迭代