Ios 使用谓词-swift3的字典过滤器数组

我有一大堆字典

>[{"name": "John",
"address": 
{"home": "addr1", 
"work": "add2"}
},
{"name": "Anu",
"address": {"home": "addr1", 
"work": "add2"}
}]

我将它保存为用户默认值，如下所示-

让personsData1=[“name”：“John”，“address”：{“home”：“addr1”，“work”：“add2”}]作为[String:Any] 让personsData2=[“name”：“Anu”，“address”：{“home”：“addr1”，“work”：“add2”}]作为[String:Any]

var persons = [personsData, personsData1]
UserDefaults.standard.set(forKey: "persons") 

在另一个方法中检索并根据名称对其进行筛选。 让name=“约翰””

低于误差

无法使用类型为“（（任何？->Bool）”的参数列表调用“过滤器”

代码如下：-

func test () {

    let personData1 = ["name": "John", "addresses": ["home":"addr1", "work": "addr2"]] as [String : Any]
    let personData2 = ["name": "And", "addresses":  ["home":"addr1", "work": "addr2"]] as [String : Any]

    let persons = [personData1, personData2]
    (UserDefaults.standard.set(persons, forKey: "persons")

    print("Saved ----\(UserDefaults.standard.value(forKey: "persons"))")

    if let savedPersons = UserDefaults.standard.value(forKey: "persons") {
        let namePredicate = NSPredicate(format: "name like %@", name);

        var filteredArray: [[String:Any]] = savedPersons.filter { namePredicate.evaluate(with: $0) }

        print("names = \(filteredArray)")
    }
}

如果我尝试这样过滤-

let filteredArray = savedBrs.filter { $0["name"] == name }

获得不同的错误-

---

类型“Any”的值没有成员“filter”

和

NSPredicate

let arr = [["name":"Rego","address":["one":"peek","two":"geelo"]],["name":"pppp","address":["one":"peek","two":"geelo"]]]

let neededName = "Rego"

let pre = NSPredicate(format: "name == %@",neededName)

let result = arr.filter { pre.evaluate(with:$0) } 

print(result)

let result = arr.filter { $0["name"] as? String  == neededName }

没有预测的

NSPredicate

let arr = [["name":"Rego","address":["one":"peek","two":"geelo"]],["name":"pppp","address":["one":"peek","two":"geelo"]]]

let neededName = "Rego"

let pre = NSPredicate(format: "name == %@",neededName)

let result = arr.filter { pre.evaluate(with:$0) } 

print(result)

let result = arr.filter { $0["name"] as? String  == neededName }

---

使用

NSPredicate

let arr = [["name":"Rego","address":["one":"peek","two":"geelo"]],["name":"pppp","address":["one":"peek","two":"geelo"]]]

let neededName = "Rego"

let pre = NSPredicate(format: "name == %@",neededName)

let result = arr.filter { pre.evaluate(with:$0) } 

print(result)

let result = arr.filter { $0["name"] as? String  == neededName }

没有预测的

NSPredicate

let arr = [["name":"Rego","address":["one":"peek","two":"geelo"]],["name":"pppp","address":["one":"peek","two":"geelo"]]]

let neededName = "Rego"

let pre = NSPredicate(format: "name == %@",neededName)

let result = arr.filter { pre.evaluate(with:$0) } 

print(result)

let result = arr.filter { $0["name"] as? String  == neededName }

---

请在问题中包含过滤数组的代码。请在问题中包含过滤数组的代码。我这样声明字典-让arr1:[[String:Any]=[[“name”：“Rego”，“address”：[“one”：“peek”，“two”：“geelo”]]，[“name”：“pppp”，“address”：[“one”：“peek”，“two”：“geelo”]]我们不能使用Any吗？默认情况下，如果我保存并从用户默认值中检索此值，则此声明将给出[[String:Any]]，我将获得错误->读取use UserDefaults.standard.array（forKey:“persons”）时，类型为“Any”的值没有成员“filter”。谢谢我像正常值一样检索它我像这样声明字典-让arr1:[[String:Any]=[[“name”：“Rego”，“address”：[“one”：“peek”，“two”：“geelo”]，[“name”：“pppp”，“address”：[“one”：“peek”，“two”：“geelo”]]我们不能使用任何吗？此声明默认情况下会给出[[String:Any]]如果我保存并从用户默认值中检索此值，则在读取use UserDefaults.standard.array（forKey:“persons”）时，我会得到错误->类型为“Any？”的值没有成员“filter”。谢谢我像正常值一样检索它