如何在Swift中获取选择器名称作为对象的属性？

Question

在使用rest JSON-API时，在整个应用程序中保持属性使用的一致性很方便，并且在此示例中调用服务器时也是如此：

class Picture: NSManagedObject {

    @NSManaged var copyright: String
    @NSManaged var likes: NSNumber
}


func postToServer() {
   let newPicture = Picture(copyright: "haaakon", likes:433)

   let arguments = @{  
       newPicture.copyright.selectorName : newPicture.copyright,
       newPicture.likes.selectorName : newPicture.likes 
   }

   println(arguments)
}

println应给出以下输出：

{ "copyright": "haaakon", "likes": "433" }

在给定属性上创建扩展(或其他解决方案)时，返回的基本相同于：

NSStringFromSelector(Selector(newPicture.copyright)) // returns "copyright"
newPicture.copyright.selectorName                    // returns "copyright"

Answer 1

试试这个：

extension Picture{
    func dictionary() -> Dictionary<String, AnyObject> {
        var outCount: CUnsignedInt = 0
        let properties: UnsafeMutablePointer<objc_property_t> = class_copyPropertyList(self.dynamicType, &outCount)

        var result: Dictionary<String, AnyObject> = Dictionary()

        for var i = 0; i < Int(outCount); i++ {
            var property : NSString! = NSString(CString: property_getName(properties[i]), encoding: NSUTF8StringEncoding)
            result[property] = self.valueForKey(property)
        }

        return result
    }
}