从 NSDictionary/字典中删除值 == -1.0 的所有项目

Question

我正在将数据从一种格式迁移到另一种格式，以便对我的一个iOS应用程序进行大型更新。这意味着当用户更新其应用程序时，它会将其数据迁移到新位置。这个新位置有更严格的规则并使用适当的做法(基本上当我第一次创建应用程序时，我是一个菜鸟，目前数据库中有不良做法，这个新位置仅用于良好的数据(。

我让一切都很好地迁移，除了我需要做最后一件事之外，一切都可以正常工作。我保存到数据库的大型字典(它使用 firebase fyi(在应该有一个 nil 值的地方存储了 -1.0(我知道这很糟糕(。我需要做的是遍历整个字典并删除该键的值为 -1.0 的任何键。

字典的类型是[AnyHashable：Any]，这是Firebase使用的。

到目前为止，我已经尝试过了。

if let data = dataDictionary as? [AnyHashable : Any] {
let foundItems = data.filter { $0.value as? Double == -1.0 }
print(foundItems)
}

计划是遍历找到的项目数组，并从包含它的数据字典中删除任何键。

数据字典如下所示：

"-KpIdh_TQMG4fyfFgkdt" =                 {
assignments =                     {
"-KpIgH6uN19OpcuedYe1" =                         {
assignmentGoal = "-1";
assignmentName = "Information System Proposal";
assignmentResult = 100;
assignmentWeight = 5;
};
"-KpIgJnFlC6fhgS0NWxF" =                         {
assignmentGoal = "-1";
assignmentName = "Information System";
assignmentResult = "-1";
assignmentWeight = 35;
};
"-KpIgOGSAwg_VSpDWhWR" =                         {
assignmentGoal = "-1";
assignmentName = "Process Analysis";
assignmentResult = "-1";
assignmentWeight = 30;
};
"-KpIgPhu_3Zxw36xt3O4" =                         {
assignmentGoal = "-1";
assignmentName = Labs;
assignmentResult = "-1";
assignmentWeight = 10;
};
"-KpIgQoFEdRnLlMAq2VN" =                         {
assignmentGoal = "-1";
assignmentName = Exam;
assignmentResult = "-1";
assignmentWeight = 20;
};
};
paperColor = 22;
paperGoal = 95;
paperName = "Systems Analysis";
};
};
semesterCode = 17S2;
semesterGoal = 90;
semesterName = "2017 Semester Two";
};
};

这就是它应该的样子

"-KpIdh_TQMG4fyfFgkdt" =                 {
assignments =                     {
"-KpIgH6uN19OpcuedYe1" =                         {
assignmentName = "Information System Proposal";
assignmentResult = 100;
assignmentWeight = 5;
};
"-KpIgJnFlC6fhgS0NWxF" =                         {
assignmentName = "Information System";
assignmentWeight = 35;
};
"-KpIgOGSAwg_VSpDWhWR" =                         {
assignmentName = "Process Analysis";
assignmentWeight = 30;
};
"-KpIgPhu_3Zxw36xt3O4" =                         {
assignmentName = Labs;
assignmentWeight = 10;
};
"-KpIgQoFEdRnLlMAq2VN" =                         {
assignmentName = Exam;
assignmentWeight = 20;
};
};
paperColor = 22;
paperGoal = 95;
paperName = "Systems Analysis";
};
};
semesterCode = 17S2;
semesterGoal = 90;
semesterName = "2017 Semester Two";
};
};

Answer 1

我的解决方案是映射(迭代(每个父键，并将其子节点分配给可变子节点。然后测试每个子属性(assignmentGoal，assignmentName等(的字符串为"-1"和整数的-1。如果是匹配项，请从子项中删除该键：值对。

然后将子节点分配回父节点。

let result: [Any] = dict.map { dictionary in
var d = dictionary
var child = d.value as Dictionary
if let goal = child["assignmentGoal"] as? String, goal == "-1" {
child.removeValue(forKey: "assignmentGoal")
}
if let name = child["assignmentName"] as? String, name == "-1" {
child.removeValue(forKey: "assignmentName")
}
if let result = child["assignmentResult"] as? Int, result == -1 {
child.removeValue(forKey: "assignmentResult")
}
if let weight = child["assignmentWeight"] as? Int, weight == -1 {
child.removeValue(forKey: "assignmentWeight")
}
d.value = child
return d
}
//show the output
for item in result {
print(item)
}

和结果输出

(key: "-KpIgOGSAwg_VSpDWhWR", 
value: ["assignmentName": "Process Analysis System", "assignmentWeight": 30])
(key: "-KpIgQoFEdRnLlMAq2VN",
value: ["assignmentName": "Exam", "assignmentWeight": 20])
(key: "-KpIgJnFlC6fhgS0NWxF",
value: ["assignmentName": "Information System", "assignmentGoal": "3", "assignmentWeight": 35])
(key: "-KpIgPhu_3Zxw36xt3O4",
value: ["assignmentName": "Labs", "assignmentWeight": 10])
(key: "-KpIgH6uN19OpcuedYe1",
value: ["assignmentName": "Information System Proposal", "assignmentResult": 100, "assignmentWeight": 5])