Question

无法识别的选择器到实例名称。

我想从section创建partion数组。我试图在swift 2中这样做，但我无法让它工作。

     var currentCollation : UILocalizedIndexedCollation!
        var sections: [Section] {
            let selector: Selector = "name"

            let users: [User] = array.map { name in
                let a = name["fullName"] as? String
                let b = name["email"] as! String
                let c = name["mobile"] as! String
                let d = name["img"] as! String
                let user = User(name: a! )
                user.email = b
                user.mobile = c
                user.img = d


                user.section = UILocalizedIndexedCollation.current().section(for: user, collationStringSelector:"name")
                return user
            }

            var sections = [Section]()
            for _ in 0..<currentCollation.sectionIndexTitles.count {
                sections.append(Section())
            }
            for user in users {
                sections[user.section!].addUser(user: user)

            }


            print(sections)
            for section in sections {

               print(section.users)
                var user = section.users as? User
                print(user?.name)
            section.users = self.currentCollation.sortedArray(from: section.users, collationStringSelector: "name") as! [User]
            }
            return sections

        }

@objc class User: NSObject {
    let name: String
    var section: Int?
    var img: String?
    var email: String?
    var mobile : String?

    init(name: String) {
        self.name = name
    }

}
 class Section {
    var users: [User] = []
    func addUser(user: User) {
        self.users.append(user)
    }
}

Answer 1

不要在Swift中使用字符串作为选择器。使用#selector构造：

let selector: Selector = #selector(name)

但是，只有当前的类是NSObject且具有Objective-C函数＆＃34; name＆＃34;没有参数：

class Foo: NSObject {
    @objc func name() {
        print("In \(#function)")
    }
}

（实际上，你可以为不能从NSObject继承的对象创建一个选择器，但你不能在这样的选择器上使用像perform()这样的函数，所以它不是很有用）

你似乎没有一个名为＆＃34; name＆＃34;的功能。在您的班级中，类型为@objc。

未记录的选择器名称

1 个答案: