Get class name of UIViewController in swift - ios

How do you get the class name of a UIViewController class in Swift?
In Objective-C, we can do something like this:
self.appDelegate = (shAppDelegate *)[[UIApplication sharedApplication] delegate];
UIViewController *last_screen = self.appDelegate.popScreens.lastObject ;
if(last_screen.class != self.navigationController.visibleViewController.class){
//.......
}
but in Swift I tried:
let appDelegate = UIApplication.sharedApplication().delegate as AppDelegate
let last_screen = appDelegate.popScreens?.lastObject as UIViewController
Can't do this.
if last_screen.class != self.navigationController.visibleViewController.class {
//....
}
no class method of UIViewController i.e last screen

To know your class name you can call something like this:
var className = NSStringFromClass(yourClass.classForCoder)

The cleanest way without needing to know the name of your class is like this.
let name = String(describing: type(of: self))

A simple way in swift 3 is to write the below code:
for instances:
let className = String(describing: self)
for classes:
let className = String(describing: YourViewController.self)

Expanding on juangdelvalle's answer.
I added this as an extension so that it's reusable and easier to call from any view controller. Also in some cases NSStringFromClass in Swift returns a string in the format like this:
< project name >.viewControllerClassName.
This extension property is modified to get rid of the project name prefix and return only the class name.
extension UIViewController {
var className: String {
NSStringFromClass(self.classForCoder).components(separatedBy: ".").last!
}
}

Swift 4
Suppose we have class with name HomeViewController. Then you can get name of class with the following code:
let class_name = "\(HomeViewController.classForCoder())"
The classForCoder() method returns AnyClass object (name of your class) which we convert to string for user.

Here is a swift3 version of isuru's answer.
extension UIViewController {
var className: String {
return NSStringFromClass(self.classForCoder).components(separatedBy: ".").last!;
}
}

Swift 5 solution:
extension NSObject {
var className: String {
return String(describing: type(of: self))
}
class var className: String {
return String(describing: self)
}
}
USAGE:
class TextFieldCell: UITableVIewCell {
}
class LoginViewController: UIViewController {
let cellClassName = TextFieldCell.className
}

The property is called dynamicType in Swift.

Use String.init(describing: self.classForCoder)
example:
let viewControllerName = String.init(describing: self.classForCoder)
print("ViewController Name: \(viewControllerName)")

How about:
extension NSObject {
static var stringFromType: String? {
return NSStringFromClass(self).components(separatedBy: ".").last
}
var stringFromInstance: String? {
return NSStringFromClass(type(of: self)).components(separatedBy: ".").last
}
}

We can also do: String(describing: Self.self) in Swift 5.1.

Related

Swift - Passing custom objects as a parameter

I started learning Swift today and in my first test app I am getting this error:
TestClass is not convertible to AnotherClass
The following is the TestClass:
class TestClass : NSObject {
var parameter1 : String = ""
var parameter2 : String = ""
override init() {
super.init()
}
func createJob(parameter1: String, parameter2: String) -> TestClass {
self.parameter1 = parameter1
self.parameter2 = parameter2
return self;
}
}
And this is the AnotherClass:
class AnotherClass: NSObject {
private struct internalConstants {
static let test1 = "testData"
static let test2 = "testData2"
}
var current : String
override init() {
self.current = internalConstants.test1
super.init()
}
func executeTask(testClass : TestClass) {
if testClass.parameter1 == "abc" {
return;
}
}
}
And this is the ViewController where I am getting the compiler error:
class ViewController: UIViewController {
override func viewDidLoad() {
super.viewDidLoad()
let obj = TestClass()
AnotherClass.executeTask(obj)
}
}
AnotherClass.executeTask line is giving the compiler error.
The obj variable sent as a parameter on this line is highlighted by Xcode with the error
"TestClass is not convertible to AnotherClass".
In C# or Objective C it is allowed to pass custom objects as a parameter to another methods. How can I do it in Swift?
Let's correct first the TestClass. This is how you should init a class like that:
class TestClass : NSObject {
....
init(parameter1: String, parameter2: String) {
....
}
}
Much simpler. Now, going back to your problem,
"TestClass is not convertible to AnotherClass".
Take a look at it again. The line you've mentioned in your question. You are trying to do this:
let obj = TestClass()
AnotherClass.executeTask(obj)
This line, AnotherClass.executeTask(obj), is giving you an error because indeed executeTask() is an instance method. You could do three ways for that.
add static keyword to the func executeTask... So it becomes like this: static func executeTask(testClass : TestClass) {
Instead of static keyword, you could add class. It becomes like so: class func executeTask(....
OR, better if you just instantiate the AnotherClass. Make a new object of AnotherClass. How to instantiate? You tell me. But here:
let anotherClass = AnotherClass()
Either implement executeTask as a class function
class func executeTask(testClass : TestClass) {
if testClass.parameter1 == "abc" {
return;
}
}
or instantiate AnotherClass in vieweDidLoad
let obj = TestClass()
let another = AnotherClass()
another.executeTask(testClass: obj)
Note the slightly different call to executeTask with the argument name.
And there is really no reason for you to subclass NSObject as I see it.
I think it's best to keep is simple. Create an instance of AnotherClass inside of ViewController.
class ViewController: UIViewController {
// Create an instance of AnotherClass which lives with ViewController.
var anotherClass = AnotherClass()
override func viewDidLoad() {
super.viewDidLoad()
let obj = TestClass()
// Use the instance of AnotherClass to call the method.
anotherClass.executeTask(testClass: obj)
}
}

Removing WKWebView Accesory bar in Swift

I am trying for a few days now to get this converted into Swift without really having much background with it.
This is what I've got so far ... and I have been looking on google not really knowing what to search for in order to be more specific. Can you please shed some light on what I'm doing wrong ? Thanks
Update:
I have aded the objective-c tag just so more people that are related to this thread may be able to see it and hopefully get an answer.
For those who are still looking, the WebKit team updated WKWebView (iOS 13+) so that you can subclass it to remove/update the input accessory view:
https://trac.webkit.org/changeset/246229/webkit#file1
In Swift, I subclassed it, and returned nil. Worked as expected. I hope it helps.
FYI: I checked the docs, and it doesn't mention not to subclass WKWebView, so subclassing is allowed.
import WebKit
class RichEditorWebView: WKWebView {
var accessoryView: UIView?
override var inputAccessoryView: UIView? {
// remove/replace the default accessory view
return accessoryView
}
}
You can find a working version of it here: https://github.com/cbess/RichEditorView/commits/master
Michael Dautermann answer has got everything right, but in order to hide the accessory bar you need to swizzle the method inputAccessoryView() of UIView Class with the inputAccessoryView() of the _NoInputAccessoryView class. I have just added the couple of extra lines to the code which does this job of method swizzling.
First you'll need a fake class to swap with
final class FauxBarHelper: NSObject {
var inputAccessoryView: AnyObject? { return nil }
}
Then create this method in your controller class
/// Removes the keyboard accessory view from the web view
/// Source: http://stackoverflow.com/a/32620344/308315 / http://stackoverflow.com/a/33939584/308315
func _removeInputAccessoryView(webView: UIWebView) {
var targetView: UIView? = nil
for view in webView.scrollView.subviews {
if String(describing: type(of: view)).hasPrefix("WKContent") {
targetView = view
}
}
guard let target = targetView else { return }
let noInputAccessoryViewClassName = "\(target.superclass!)_NoInputAccessoryView"
var newClass: AnyClass? = NSClassFromString(noInputAccessoryViewClassName)
if newClass == nil {
let targetClass: AnyClass = object_getClass(target)
newClass = objc_allocateClassPair(targetClass, noInputAccessoryViewClassName.cString(using: String.Encoding.ascii)!, 0)
}
let originalMethod = class_getInstanceMethod(FauxBarHelper.self, #selector(getter: FauxBarHelper.inputAccessoryView))
class_addMethod(newClass!.self, #selector(getter: FauxBarHelper.inputAccessoryView), method_getImplementation(originalMethod), method_getTypeEncoding(originalMethod))
object_setClass(target, newClass)
}
HTH ;)
Here's a slightly safer (no unsafe unwraps) version that works with Swift 4 and (at least) iOS 9 trough 12.
fileprivate final class InputAccessoryHackHelper: NSObject {
#objc var inputAccessoryView: AnyObject? { return nil }
}
extension WKWebView {
func hack_removeInputAccessory() {
guard let target = scrollView.subviews.first(where: {
String(describing: type(of: $0)).hasPrefix("WKContent")
}), let superclass = target.superclass else {
return
}
let noInputAccessoryViewClassName = "\(superclass)_NoInputAccessoryView"
var newClass: AnyClass? = NSClassFromString(noInputAccessoryViewClassName)
if newClass == nil, let targetClass = object_getClass(target), let classNameCString = noInputAccessoryViewClassName.cString(using: .ascii) {
newClass = objc_allocateClassPair(targetClass, classNameCString, 0)
if let newClass = newClass {
objc_registerClassPair(newClass)
}
}
guard let noInputAccessoryClass = newClass, let originalMethod = class_getInstanceMethod(InputAccessoryHackHelper.self, #selector(getter: InputAccessoryHackHelper.inputAccessoryView)) else {
return
}
class_addMethod(noInputAccessoryClass.self, #selector(getter: InputAccessoryHackHelper.inputAccessoryView), method_getImplementation(originalMethod), method_getTypeEncoding(originalMethod))
object_setClass(target, noInputAccessoryClass)
}
}
This code snippet should get you over your issue:
class _NoInputAccessoryView: NSObject {
func removeInputAccessoryViewFromWKWebView(webView: WKWebView) {
// make sure to make UIView an optional here...
var targetView: UIView? = nil
for view in webView.scrollView.subviews {
if String(view.dynamicType).hasPrefix("WKContent") {
targetView = view
}
}
// only optionals can be nil
if targetView == nil {
return
}
let noInputAccessoryViewClassName = "\(targetView!.superclass)_NoInputAccessoryView"
var newClass : AnyObject? = NSClassFromString(noInputAccessoryViewClassName)
if newClass == nil {
let uiViewClass : AnyClass = object_getClass(targetView!)
newClass = objc_allocateClassPair(uiViewClass, noInputAccessoryViewClassName.cStringUsingEncoding(NSASCIIStringEncoding)!, 0)
}
}
You can also use "String(view.dynamicType)" to get the class name of the object you're looking at, as I noticed via this answer as I was researching the way to solve your problem.
Using hasPrefix like that in both Objective-C and Swift is really hacky and perhaps a better way of hiding the keyboard could be found for production code?

Get Swift class name in "class func" method

I have a static method in Swift
class BaseAsyncTask: WebServiceClient {
class func execute(content : [String:AnyObject], cancelled:CustomBool)
{
// Print class name (BaseAsyncTask) here
}
}
And I want to know how to get the class name inside that method. I tried
self.dynamicType
but that gives error (I suppose because of the self inside a class function)
There are different methods to do that, if your method inherits from NSObject you can expose it to objective-c and do something like that.
#objc(BaseAsyncTask)
class BaseAsyncTask: WebServiceClient {
class func execute(content : [String:AnyObject], cancelled:CustomBool)
{
println("Class \(NSStringFromClass(self))")
}
}
For pure SWIFT introspection check here about MirrorType
I've found also this extension credits to ImpactZero
public extension NSObject{
public class var nameOfClass: String{
return NSStringFromClass(self).components(separatedBy: ".").last!
}
public var nameOfClass: String{
return NSStringFromClass(type(of: self)).components(separatedBy: ".").last!
}
}
[Xcode 8]
Alex suggested me that in the Xcode 8 version this code shows a warning. To avoid that we should prefix the method like that:
#nonobjc class var className: String{
return NSStringFromClass(self).components(separatedBy: ".").last!
}
You can use string interpolation to print self:
let className = "\(self)"
Sample code:
class BaseAsyncTask: WebServiceClient {
class func execute(content : [String:AnyObject], cancelled: CustomBool)
{
let className = "\(self)"
print(className)
}
}
class AnotherAsyncTask : BaseAsyncTask {
}
BaseAsyncTask.execute([:], cancelled: true) // prints "BaseAsyncTask"
AnotherAsyncTask.execute([:], cancelled: true) // prints "AnotherAsyncTask"
Another way to do this, when you don't have an instance of the class is this.
Swift 4
print(String(describing:BaseAsyncTask.self))
Swift 2
print(String(BaseAsyncTask))
Inspired here.
Get class name of object as string in Swift

Is there a way to dynamically instantiate a class type

I would like to do something on the lines of (pseudo code):
let class_name = "MyClass"
let RealClass = class_name()
I am familiar with http://ijoshsmith.com/2014/06/05/instantiating-classes-by-name-in-swift/
You can Use NSClassFromString to get class then cast to the type you want.
For example:
var dynamicClass = NSClassFromString("Test") as? Test.Type
if dynamicClass != nil{
var instance = dynamicClass!()
instance.test()
}
The class Test
#objc(Test)
class Test{
func test() {
println("dynamic")
}
required init() {
}
}

Swift language NSClassFromString

How to achieve reflection in Swift Language?
How can I instantiate a class
[[NSClassFromString(#"Foo") alloc] init];
You must put #objc(SwiftClassName) above your swift class.
Like:
#objc(SubClass)
class SubClass: SuperClass {...}
This is the way I init derived UIViewController by class name
var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass()
More information is here
In iOS 9
var className = "YourAppName.TestViewController"
let aClass = NSClassFromString(className) as! UIViewController.Type
let viewController = aClass.init()
Less hacky solution here: https://stackoverflow.com/a/32265287/308315
Note that Swift classes are namespaced now so instead of "MyViewController" it'd be "AppName.MyViewController"
Deprecated since XCode6-beta 6/7
Solution developed using XCode6-beta 3
Thanks to the answer of Edwin Vermeer I was able to build something to instantiate Swift classes into an Obj-C class by doing this:
// swift file
// extend the NSObject class
extension NSObject {
// create a static method to get a swift class for a string name
class func swiftClassFromString(className: String) -> AnyClass! {
// get the project name
if var appName: String? = NSBundle.mainBundle().objectForInfoDictionaryKey("CFBundleName") as String? {
// generate the full name of your class (take a look into your "YourProject-swift.h" file)
let classStringName = "_TtC\(appName!.utf16count)\(appName)\(countElements(className))\(className)"
// return the class!
return NSClassFromString(classStringName)
}
return nil;
}
}
// obj-c file
#import "YourProject-Swift.h"
- (void)aMethod {
Class class = NSClassFromString(key);
if (!class)
class = [NSObject swiftClassFromString:(key)];
// do something with the class
}
EDIT
You can also do it in pure obj-c:
- (Class)swiftClassFromString:(NSString *)className {
NSString *appName = [[NSBundle mainBundle] objectForInfoDictionaryKey:#"CFBundleName"];
NSString *classStringName = [NSString stringWithFormat:#"_TtC%d%#%d%#", appName.length, appName, className.length, className];
return NSClassFromString(classStringName);
}
I hope this will help somebody !
UPDATE: Starting with beta 6 NSStringFromClass will return your bundle name plus class name separated by a dot. So it will be something like MyApp.MyClass
Swift classes will have a constructed internal name that is build up of the following parts:
It will start with _TtC,
followed by a number that is the length of your application name,
followed by your application name,
folowed by a number that is the length of your class name,
followed by your class name.
So your class name will be something like _TtC5MyApp7MyClass
You can get this name as a string by executing:
var classString = NSStringFromClass(self.dynamicType)
Update In Swift 3 this has changed to:
var classString = NSStringFromClass(type(of: self))
Using that string, you can create an instance of your Swift class by executing:
var anyobjectype : AnyObject.Type = NSClassFromString(classString)
var nsobjectype : NSObject.Type = anyobjectype as NSObject.Type
var rec: AnyObject = nsobjectype()
It's almost the same
func NSClassFromString(_ aClassName: String!) -> AnyClass!
Check this doc:
https://developer.apple.com/library/prerelease/ios/documentation/Cocoa/Reference/Foundation/Miscellaneous/Foundation_Functions/#//apple_ref/c/func/NSClassFromString
I was able to instantiate an object dynamically
var clazz: NSObject.Type = TestObject.self
var instance : NSObject = clazz()
if let testObject = instance as? TestObject {
println("yes!")
}
I haven't found a way to create AnyClass from a String (without using Obj-C). I think they don't want you to do that because it basically breaks the type system.
For swift2, I created a very simple extension to do this more quickly
https://github.com/damienromito/NSObject-FromClassName
extension NSObject {
class func fromClassName(className : String) -> NSObject {
let className = NSBundle.mainBundle().infoDictionary!["CFBundleName"] as! String + "." + className
let aClass = NSClassFromString(className) as! UIViewController.Type
return aClass.init()
}
}
In my case, i do this to load the ViewController I want:
override func viewDidLoad() {
super.viewDidLoad()
let controllers = ["SettingsViewController", "ProfileViewController", "PlayerViewController"]
self.presentController(controllers.firstObject as! String)
}
func presentController(controllerName : String){
let nav = UINavigationController(rootViewController: NSObject.fromClassName(controllerName) as! UIViewController )
nav.navigationBar.translucent = false
self.navigationController?.presentViewController(nav, animated: true, completion: nil)
}
This will get you the name of the class that you want to instantiate. Then you can use Edwins answer to instantiate a new object of your class.
As of beta 6 _stdlib_getTypeName gets the mangled type name of a variable. Paste this into an empty playground:
import Foundation
class PureSwiftClass {
}
var myvar0 = NSString() // Objective-C class
var myvar1 = PureSwiftClass()
var myvar2 = 42
var myvar3 = "Hans"
println( "TypeName0 = \(_stdlib_getTypeName(myvar0))")
println( "TypeName1 = \(_stdlib_getTypeName(myvar1))")
println( "TypeName2 = \(_stdlib_getTypeName(myvar2))")
println( "TypeName3 = \(_stdlib_getTypeName(myvar3))")
The output is:
TypeName0 = NSString
TypeName1 = _TtC13__lldb_expr_014PureSwiftClass
TypeName2 = _TtSi
TypeName3 = _TtSS
Ewan Swick's blog entry helps to decipher these strings: http://www.eswick.com/2014/06/inside-swift/
e.g. _TtSi stands for Swift's internal Int type.
In Swift 2.0 (tested in the Xcode 7.01) _20150930
let vcName = "HomeTableViewController"
let ns = NSBundle.mainBundle().infoDictionary!["CFBundleExecutable"] as! String
// Convert string to class
let anyobjecType: AnyObject.Type = NSClassFromString(ns + "." + vcName)!
if anyobjecType is UIViewController.Type {
// vc is instance
let vc = (anyobjecType as! UIViewController.Type).init()
print(vc)
}
xcode 7 beta 5:
class MyClass {
required init() { print("Hi!") }
}
if let classObject = NSClassFromString("YOURAPPNAME.MyClass") as? MyClass.Type {
let object = classObject.init()
}
string from class
let classString = NSStringFromClass(TestViewController.self)
or
let classString = NSStringFromClass(TestViewController.classForCoder())
init a UIViewController class from string:
let vcClass = NSClassFromString(classString) as! UIViewController.Type
let viewController = vcClass.init()
I am using this category for Swift 3:
//
// String+AnyClass.swift
// Adminer
//
// Created by Ondrej Rafaj on 14/07/2017.
// Copyright © 2017 manGoweb UK Ltd. All rights reserved.
//
import Foundation
extension String {
func convertToClass<T>() -> T.Type? {
return StringClassConverter<T>.convert(string: self)
}
}
class StringClassConverter<T> {
static func convert(string className: String) -> T.Type? {
guard let nameSpace = Bundle.main.infoDictionary?["CFBundleExecutable"] as? String else {
return nil
}
guard let aClass: T.Type = NSClassFromString("\(nameSpace).\(className)") as? T.Type else {
return nil
}
return aClass
}
}
The use would be:
func getViewController(fromString: String) -> UIViewController? {
guard let viewController: UIViewController.Type = "MyViewController".converToClass() else {
return nil
}
return viewController.init()
}
I think I'm right in saying that you can't, at least not with the current beta (2). Hopefully this is something that will change in future versions.
You can use NSClassFromString to get a variable of type AnyClass but there appears to be no way in Swift to instantiate it. You can use a bridge to Objective C and do it there or -- if it works in your case -- fall back to using a switch statement.
Apparently, it is not possible (anymore) to instantiate an object in Swift when the name of the class is only known at runtime. An Objective-C wrapper is possible for subclasses of NSObject.
At least you can instantiate an object of the same class as another object given at runtime without an Objective-C wrapper (using xCode Version 6.2 - 6C107a):
class Test : NSObject {}
var test1 = Test()
var test2 = test1.dynamicType.alloc()
In Swift 2.0 (tested in the beta2 of Xcode 7) it works like this:
protocol Init {
init()
}
var type = NSClassFromString(className) as? Init.Type
let obj = type!.init()
For sure the type coming from NSClassFromString have to implement this init protocol.
I expect it is clear, className is a String containing the Obj-C runtime name of the class which is by default NOT just "Foo", but this discussion is IMHO not the major topic of your question.
You need this protocol because be default all Swift classes don't implement an init method.
Looks like the correct incantation would be...
func newForName<T:NSObject>(p:String) -> T? {
var result:T? = nil
if let k:AnyClass = NSClassFromString(p) {
result = (k as! T).dynamicType.init()
}
return result
}
...where "p" stands for "packaged" – a distinct issue.
But the critical cast from AnyClass to T currently causes a compiler crash, so in the meantime one must bust initialization of k into a separate closure, which compiles fine.
I use different targets, and in this case the swift class is not found. You should replace CFBundleName with CFBundleExecutable. I also fixed the warnings:
- (Class)swiftClassFromString:(NSString *)className {
NSString *appName = [[NSBundle mainBundle] objectForInfoDictionaryKey:#"CFBundleExecutable"];
NSString *classStringName = [NSString stringWithFormat:#"_TtC%lu%#%lu%#", (unsigned long)appName.length, appName, (unsigned long)className.length, className];
return NSClassFromString(classStringName);
}
Isn't the solution as simple as this?
// Given the app/framework/module named 'MyApp'
let className = String(reflecting: MyClass.self)
// className = "MyApp.MyClass"
Also in Swift 2.0 (possibly before?) You can access the type directly with the dynamicType property
i.e.
class User {
required init() { // class must have an explicit required init()
}
var name: String = ""
}
let aUser = User()
aUser.name = "Tom"
print(aUser)
let bUser = aUser.dynamicType.init()
print(bUser)
Output
aUser: User = {
name = "Tom"
}
bUser: User = {
name = ""
}
Works for my use case
Try this.
let className: String = String(ControllerName.classForCoder())
print(className)
I have implemented like this,
if let ImplementationClass: NSObject.Type = NSClassFromString(className) as? NSObject.Type{
ImplementationClass.init()
}
Swift 5, easy to use, thanks to #Ondrej Rafaj's
Source code:
extension String {
fileprivate
func convertToClass<T>() -> T.Type? {
return StringClassConverter<T>.convert(string: self)
}
var controller: UIViewController?{
guard let viewController: UIViewController.Type = convertToClass() else {
return nil
}
return viewController.init()
}
}
class StringClassConverter<T> {
fileprivate
static func convert(string className: String) -> T.Type? {
guard let nameSpace = Bundle.main.infoDictionary?["CFBundleExecutable"] as? String, let aClass = NSClassFromString("\(nameSpace).\(className)") as? T.Type else {
return nil
}
return aClass
}
}
Call like this:
guard let ctrl = "ViewCtrl".controller else {
return
}
// ctrl do sth
A page jump example shown here, the hope can help you!
let vc:UIViewController = (NSClassFromString("SwiftAutoCellHeight."+type) as! UIViewController.Type).init()
self.navigationController?.pushViewController(vc, animated: true)
// Click the Table response
tableView.deselectRow(at: indexPath, animated: true)
let sectionModel = models[(indexPath as NSIndexPath).section]
var className = sectionModel.rowsTargetControlerNames[(indexPath as NSIndexPath).row]
className = "GTMRefreshDemo.\(className)"
if let cls = NSClassFromString(className) as? UIViewController.Type {
let dvc = cls.init()
self.navigationController?.pushViewController(dvc, animated: true)
}
Swift3+
extension String {
var `class`: AnyClass? {
guard
let dict = Bundle.main.infoDictionary,
var appName = dict["CFBundleName"] as? String
else { return nil }
appName.replacingOccurrences(of: " ", with: "_")
let className = appName + "." + self
return NSClassFromString(className)
}
}
Here is a good example:
class EPRocks {
#require init() { }
}
class EPAwesome : EPRocks {
func awesome() -> String { return "Yes"; }
}
var epawesome = EPAwesome.self();
print(epawesome.awesome);

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