swift中类的属性列表

Question

注意：在here上针对目标c发布了类似的问题，但我想在swift中实现它。

我有一个像swift这样声明的类：

import UIKit

class EachDayCell : UITableViewCell
{

    @IBOutlet var dateDisplayLabel : UITextField
    @IBOutlet var nameDisplayLabel : UITextField

    @IBAction func goToPendingItems(sender : AnyObject) {
    }
    @IBAction func showDateSelectionPicker(sender : AnyObject) {
    }

    init(style: UITableViewCellStyle, reuseIdentifier: String!)
    {
        super.init(style: style, reuseIdentifier: reuseIdentifier)
    }
}

现在我想在swift中获取一个数组：dateDisplayLabel，nameDisplayLabel。

我怎样才能做到这一点？

Answer 1

使用Mirror

这是一个纯粹的Swift解决方案，但有一些限制：

protocol PropertyNames {
    func propertyNames() -> [String]
}

extension PropertyNames
{
    func propertyNames() -> [String] {
        return Mirror(reflecting: self).children.flatMap { $0.label }
    }
}

class Person : PropertyNames {
    var name = "Sansa Stark"
    var awesome = true
}

Person().propertyNames() // ["name", "awesome"]

限制：

• 返回Objective-C对象的空数组

• 不会返回计算属性，即：

  var favoriteFood: String { return "Lemon Cake" }
  

• 如果self是一个类的实例（例如，一个结构），则不会报告其超类的属性，即：

  class Person : PropertyNames {
      var name = "Bruce Wayne"
  }
  
  class Superhero : Person {
      var hasSuperpowers = true
  }
  
  Superhero().propertyNames() // ["hasSuperpowers"] — no "name"
  

  您可以使用superclassMirror()解决此问题，具体取决于您的行为。

使用class_copyPropertyList

如果你正在使用Objective-C对象，你可以使用这种方法：

var count = UInt32()
let classToInspect = NSURL.self
let properties : UnsafeMutablePointer <objc_property_t> = class_copyPropertyList(classToInspect, &count)
var propertyNames = [String]()
let intCount = Int(count)
for var i = 0; i < intCount; i++ {
    let property : objc_property_t = properties[i]
    guard let propertyName = NSString(UTF8String: property_getName(property)) as? String else {
        debugPrint("Couldn't unwrap property name for \(property)")
        break
    }

    propertyNames.append(propertyName)
}

free(properties)
print(propertyNames)

如果classToInspect为NSURL，则输出到控制台：

  

["pathComponents", "lastPathComponent", "pathExtension", "URLByDeletingLastPathComponent", "URLByDeletingPathExtension", "URLByStandardizingPath", "URLByResolvingSymlinksInPath", "dataRepresentation", "absoluteString", "relativeString", "baseURL", "absoluteURL", "scheme", "resourceSpecifier", "host", "port", "user", "password", "path", "fragment", "parameterString", "query", "relativePath", "hasDirectoryPath", "fileSystemRepresentation", "fileURL", "standardizedURL", "filePathURL"]

这不适用于游乐场。只需将NSURL替换为EachDayCell（或重复使用与扩展名相同的逻辑），它就可以正常工作。

Answer 2

这是另一个版本。我认为这很简单和纯粹。

Swift 2.0

protocol Reflectable {
  func properties()->[String]
}

extension Reflectable
{
    func properties()->[String]{
        var s = [String]()
        for c in Mirror(reflecting: self).children
        {
            if let name = c.label{
                s.append(name)
            }
        }
        return s
    }
}


class Test:Reflectable
{
    var name99:String = ""
    var name3:String = ""
    var name2:String = ""
}

Test().properties()

Swift 1.2

class Reflect:NSObject {
    func properties()->[String]
    {
        let m = reflect(self)
        var s = [String]()
        for i in 0..<m.count
        {
            let (name,_)  = m[i]
            if name == "super"{continue}
            s.append(name)
        }
        return s
    }
}

class Test:Reflect
{
    var name99:String = ""
    var name3:String = ""
    var name2:String = ""
}

Test().properties()

Answer 3

我将bolivia's代码转换为Swift 4.此函数接受NSObject并返回对象的键的字典以及该键的类型。

请注意，类型有点难看。对于原始属性，引擎返回一个字母标识符（如bool为B，int为i等）但对于Obj-C类型，它返回@"NSString"之类的内容。看来这对我来说真的只是一个调试功能我不介意。如果您不想弄乱字典，可以取消注释print行并将其转储到控制台。 String(cString:cAttr)还包含许多有用的信息，包括属性是否可变，它是引用样式等等。 For more info on this here's Apple's documentation.

func getKeysAndTypes(forObject:Any?) -> Dictionary<String,String> {
        var answer:Dictionary<String,String> = [:]
        var counts = UInt32();
        let properties = class_copyPropertyList(object_getClass(forObject), &counts);
        for i in 0..<counts {
            let property = properties?.advanced(by: Int(i)).pointee;

            let cName = property_getName(property!);
            let name = String(cString: cName)

            let cAttr = property_getAttributes(property!)!
            let attr = String(cString:cAttr).components(separatedBy: ",")[0].replacingOccurrences(of: "T", with: "");
            answer[name] = attr;
            //print("ID: \(property.unsafelyUnwrapped.debugDescription): Name \(name), Attr: \(attr)")
        }
        return answer;
    }

Answer 4

Swift 3.1

let contorller = UIActivityViewController(activityItems: [url], applicationActivities: nil)
var count: UInt32 = 0
guard let properties = class_copyPropertyList(object_getClass(contorller), &count) else {
    return
}
for index in 0...count {
    let property1 = property_getName(properties[Int(index)])
    let result1 = String(cString: property1!)
    print(result1)
}

Answer 5

您也可以像这样在 NSObject 上进行扩展：

extension NSObject {
    var properties: [Mirror.Child] {
        Mirror(reflecting: self).children.compactMap { $0 }
    }

    var propertyNames: [String] {
        properties.compactMap { $0.label }
    }

    var propertyValues: [Any] {
        properties.map { $0.value }
    }
}