Question

使用Swift Realm，我想从父母的角度来看一个孩子的主人。

简单来说，实体关系可以描述为：

Parent -< Child <- Player

使用给定的Parent，我想获得一个玩家（所有者）列表。

当我尝试查询时，我相信我的查询是错误的;

// Cut-down summary of Realm objects and relationships
public class Parent: Object {
  var children = List<Child>()
}
public class Child: Object {
  private let parents = LinkingObjects(fromType: Parent.self, property: "children")
  private let owners = LinkingObjects(fromType: Player.self, property: "children")
  var parent:Parent? {
      return self.parents.first
  }
  var owner: Player? {
      return self.owners.first
  }
}
public class Player : Object {
  public let children = List<Child>()
}



let realm = try! Realm()
self.createPlayers(amount: 5)
let playerList = realm.objects(EYPlayer.self)
XCTAssert(playerList.count == 5)


// In my tests:
let firstParent = realm.objects(Parent.self).first
let firstChild = firstParent?.children.first
let firstPlayer = realm.objects(Player.self).first

// Assign the first player with the first child
try! realm.write {
    firstPlayer?.children.append(firstChild!)
}

// give me a list of owners for a given parent
let owners = realm.objects(Parent.self).filter("ANY children.owners.@count > 0")
print (owners)

// Test fails; it says there are 0 results
XCTAssert(owners.count == 1, "Expected: 1, Found: \(owners.count)")

此测试失败：

XCTAssert(owners.count == 1, "Expected: 1, Found: \(owners.count)")

我希望有1位老板。

但相反，我得到0个拥有者。

我做错了什么？

如何使用NSPredicates和Realm获取给定孩子的所有者列表？

非常感谢

编辑：发现我没有添加任何玩家;添加玩家的摘要版本如上所示

Answer 1

我现在已经解决了这个问题。

没有玩家数据可以追加孩子;我已经在我的xctests

执行此操作后，对我的测试进行了一些调整后，查询似乎正在工作，并在我的记录器输出中进行了验证。

    let firstParent = realm.objects(Parent.self).first
    let firstPlayer = realm.objects(Player.self).first
    let firstChild = firstParent?.children.first

    XCTAssert(firstChild?.parent == firstParent)

    try! realm.write {
        firstPlayer!.children.append(firstChild!)
    }

    let listOfChildren = realm.objects(Parent.self).filter("ANY children.owners.@count > 0")
    XCTAssert(listOfChildren.count == 1)
    XCTAssert(firstPlayer?.children.count == 1)

    for child in listOfChildren {
        for (index, eng) in child.children.enumerated() {
            if (index == 0) {
                XCTAssert(eng.owner != nil)
            }
            XCTAssert(child.parent == loco)
        }
    }

测试通过，通过一些控制台日志，我能够验证我的期望。

问题已解决。

关闭。

领域 - 如何获得父子关系中实体所有者的列表？

1 个答案: