RealmSwift + Multiple Predicate

时间:2015-12-07 11:37:23

标签: swift filter realm predicate

我不知道如何根据用户选择组合谓词。以下是我的数据概述。

Person.swift

class Person: Object {

   let id = RealmOptional<Int>()
   dynamic var name = ""
   dynamic var console = ""

   override static func primaryKey() -> String {
      return "id"
}

示例json数据

{
 "id": 822,
 "name": "Ron",
 "console": "XBox"
},
{
 "id": 823,
 "name": "Jenny"
 "console": "Playstation 4"
}
例如,我有很多数据和控制台包括“XBox”,“Playstation 4”,“Wii”,“Sega”或“PS Vita”。我创建了一个复选框过滤选项,允许用户选择他们想要过滤的选项,并显示拥有哪个控制台的人的姓名。

例如,如果他们同时选择“XBox”和“Playstation 4”,我如何预测它们并显示人名的结果?目前我只能展示一个这样的选项。

var player: Results<Person>!

func filter(sender: AnyObject) {
    let realm = try! Realm()
    self.player = realm.objects(Person).filter("%K = %@", "console", "XBox")
}

2 个答案:

答案 0 :(得分:15)

您不应该动态制作NSPredicate查询字符串。以编程方式创建NSCompoundPredicate要容易得多(也更安全)。您可以将NSPredicate传递给Realm的RealmCollectionType.filter(...)方法。以下是获取拥有Xbox或Playstation 4的所有Person的示例:

// could use .AndPredicateType here too, depending on what you want
let query = NSCompoundPredicate(type: .OrPredicateType, subpredicates: [NSPredicate(format: "console = 'Xbox'"), NSPredicate(format: "console = 'Playstation 4')])
let player = realm.objects(Person).filter(query)

虽然在您的情况下,如果您想要IN语义,最好使用OR查询:

let selectedConsoles = ["Xbox", "Playstation 4"]
let player = realm.objects(Person).filter("console IN %@", selectedConsoles)

答案 1 :(得分:0)

快捷键5

多个谓词

组合谓词与NSCompoundPredicate:

let predict = NSPredicate.init(format: "console == %@", "Xbox")
let predict2 = NSPredicate.init(format: "console == %@", "Playstation 4")
let query = NSCompoundPredicate(type: NSCompoundPredicate.LogicalType.or, subpredicates: [predict,predict2])
let player = realm.objects(Person).filter(query)