In Swift 3 this syntax has changed. 

What was 

`if let x = y, a = b where a == x {`

Is now

`if let x = y, let a = b, a == x {`

The justification is that each sub-clause of the `if ... {` is now an independent boolean test.

See the Xcode [Release notes][1] &amp; the [Swift Evolution proposal][2] for more info about this change.


  [1]: http://adcdownload.apple.com/Developer_Tools/Xcode_8_beta_4/Release_Notes_for_Xcode_8_beta_4.pdf
  [2]: https://github.com/apple/swift-evolution/blob/master/proposals/0099-conditionclauses.md

Example with two conditions 

    if let x = y, let a = b, a == x &amp;&amp; !x.isEmpty {

In xcode 9

    if let str = textField.text as String!, !str.isEmpty
    {
       params[key] = str
       TextFieldHelper.setup(textField: textField)
    }
    else
    { 
       TextFieldHelper.error(textField: textField)
    }

I have a MySQL query that includes &lt;&gt; in it. I don&#39;t know the exact usage of it.

    SELECT * FROM table_laef WHERE id = ? AND genre_type &lt;&gt; &#39;LIVE&#39;

P.S.: Im sorry for this basic syntax, since I have searched for this on Google. All they give is about &lt;=&gt;. Thanks anyway, guys!

What is the meaning of &lt;&gt; in mysql query?

I want to know what will happen to the users of my app that I used anonymous sign in method for them.


The Firebase documentation is really BAD and didn&#39;t explain everything and expect developer to find out himself. 
I found in its old version documentation that anonymous session will expires based on the expiration time has been set in Login &amp; Auth tab, but even there didn&#39;t mention this means just the session ends or it means that user id will remove also from my app users list or what EXACTLY happened?

I found [this answer][1] but it really is not acceptable. The number of anonymous users will grow very very fast if you do a web app and make every thing hard.
I even cannot see the number of my app users in my dashboard!!!!!
So, what should i do? should i develop a dashboard for my data myself or Firebase team should do it? At least for managing users i should have more power than just searching user with their email and when you use custom login you cannot do this also.


  [1]: https://stackoverflow.com/questions/37895385/delete-firebase-anonymous-users-after-a-while

What happens to Firebase anonymous users?

The Swift documentation at page 61 of the Swift manual hints to the possibility of using `where` to join an optional binding with a regular condition. Yet when I do it I have a warning suggesting me to substitute the `where` with a comma like in the following piece of code:

    if let geocodingError = error as? NSError where geocodingError.code == 2

Usage of where in if let assignment in Swift

The Swift documentation at page 61 of the Swift manual hints to the possibility of using <code>where</code> to join an optional binding with a regular condition. Yet when I do it I have a warning suggesting me to substitute the <code>where</code> with a comma like in the following piece of code:
<pre><code class="hljs language-swift">if let geocodingError = error as? NSError where geocodingError.code == 2
</code></pre>

    NSDictionary *dictionary = @{@&quot;A&quot; : @&quot;alfa&quot;,
                                 @&quot;B&quot; : @&quot;bravo&quot;,
                                 @&quot;C&quot; : @&quot;charlie&quot;,
                                 @&quot;D&quot; : @&quot;delta&quot;,
                                 @&quot;E&quot; : @&quot;echo&quot;,
                                 @&quot;F&quot; : @&quot;foxtrot&quot;};
    NSLog(@&quot;%@&quot;, dictionary.description);

prints out the following on the console:

    {
        A = alfa;
        B = bravo;
        C = charlie;
        D = delta;
        E = echo;
        F = foxtrot;
    }

---

    let dictionary: [String : String] = [&quot;A&quot; : &quot;alfa&quot;,
                                         &quot;B&quot; : &quot;bravo&quot;,
                                         &quot;C&quot; : &quot;charlie&quot;,
                                         &quot;D&quot; : &quot;delta&quot;,
                                         &quot;E&quot; : &quot;echo&quot;,
                                         &quot;F&quot; : &quot;foxtrot&quot;];
    print(dictionary)

prints out the following on the console:

    [&quot;B&quot;: &quot;bravo&quot;, &quot;A&quot;: &quot;alfa&quot;, &quot;F&quot;: &quot;foxtrot&quot;, &quot;C&quot;: &quot;charlie&quot;, &quot;D&quot;: &quot;delta&quot;, &quot;E&quot;: &quot;echo&quot;]

Is there a way in Swift to get it to pretty print dictionaries where each key-value pair occupies a new line?

Is there a way to pretty print Swift dictionaries to the console?

I need all Apple Emojis.&lt;br&gt; I can get all the emojis and put them into a *String* by copying them from the site [getemoji][1] but in my app i need the emojis in the right order **as images**.&lt;br&gt;&lt;br&gt;
Is there a nice way to convert the emojis I copy into a String to a *UIImage*?&lt;br&gt;
Or a better solution to get all the Apple emojis in the right order?


  [1]: http://getemoji.com

Convert Apple Emoji (String) to UIImage

I am trying to add a `UITapGesture` to a UIButton so it will trigger a function when tapped. I am using `Swift 3` and is getting some error:

&gt;Terminating app due to uncaught exception &#39;NSInvalidArgumentException&#39;, reason: &#39;-[SwiftRunner.ViewController tapBlurButton]: unrecognized selector sent to instance 0x149e07610&#39;

This is roughly what I have:

    // Swift 3
    import UIKit
    class ViewController {

       @IBOutlet weak var qsBlurButton: UIButton!       

       override func viewDidLoad() {
          super.viewDidLoad()
          
          let tapGesture = UITapGestureRecognizer(target: self, action: Selector((&quot;tapBlurButton&quot;)))
          qsBlurButton.addGestureRecognizer(tapGesture)
       }
    
       func tapBlurButton(sender: UITapGestureRecognizer) {
          print(&quot;Please Help!&quot;)
       }
    }

Swift3 iOS - How to make UITapGestureRecognizer trigger function

I have been only recently working with Xcode 8 beta 4 but it has been working fine up until now. The list of simulators is not showing for any of my projects. I already tried Creating a new Scheme and it did not work. How do I fix this?

xcode8 list of simulators not showing

I just started learning iOS development, cannot find how to make simple rounded button. I find resources for old versions. Do I need to set a custom background for a button? In Android, I would just use a 9 patch, but I know iOS does not have this capability.

[![enter image description here][1]][1]


  [1]: http://i.stack.imgur.com/kpJ2X.png

How to make a simple rounded button in Storyboard?

In `Objective-C` it was sometimes useful to use static string constants to define alternate API keys (for example to differentiate between RELEASE and DEBUG keys for analytics packages, like MixPanel, Flurry or Crashlytics):

    #if DEBUG
    static NSString *const API_KEY = @&quot;KEY_A&quot;;
    #else
    static NSString *const API_KEY = @&quot;KEY_B&quot;;
    #endif

and then...

    [Analytics startSession:API_KEY];


How does this translate to Swift, since the Swift compiler no longer uses a preprocessor?

Swift: how to use PREPROCESSOR Flags (like `#if DEBUG`) to implement API keys?

When I&#39;m declaring variables as `weak` in Swift, I sometimes get the error message from Xcode: 

&gt; &#39;weak&#39; may only be applied to class and class-bound protocol types

or

&gt; &#39;weak&#39; must not be applied to non-class-bound &#39;SomeProtocol&#39;; consider adding a protocol conformance that has a class bound

I&#39;m wondering why the keyword `weak` can only applied to class and class-bound protocol types? What is the reason behind this requirement?

Why can the keyword &quot;weak&quot; only be applied to class and class-bound protocol types

Given:

    typealias Action = () -&gt; ()
    
    var action: Action = { }
    
    func doStuff(stuff: String, completion: @escaping Action) {
        print(stuff)
        action = completion
        completion()
    }
    
    func doStuffAgain() {
        print(&quot;again&quot;)
        action()
    }
    
    doStuff(stuff: &quot;do stuff&quot;) { 
        print(&quot;swift 3!&quot;)
    }
    
    doStuffAgain()

Is there any way to make the `completion` parameter (and `action`) of type `Action?` and also keep `@escaping` ?

Changing the type gives the following error:

 &gt; @escaping attribute only applies to function types

Removing the `@escaping` attribute, the code compiles and runs, but doesn&#39;t seem to be correct since the `completion` closure is escaping the scope of the function.


Swift optional escaping closure parameter

I have some error like this &quot;Expression implicitly coerced from `String?` to `Any`&quot; this is my code :

    func application(_ application: UIApplication, didFinishLaunchingWithOptions launchOptions: [UIApplicationLaunchOptionsKey: Any]?) -&gt; Bool {
        // Override point for customization after application launch.
        
        FIRApp.configure()
        FIRAuth.auth()?.signIn(withEmail: &quot;myemail@gmail.com&quot;, password: &quot;mypassword&quot;, completion: { (user, error) in
            if (error != nil) {
                print(user?.email)
            }else {
                print(error?.localizedDescription)
            }
        })
        
        return true
    }

Error in this line 

    print(user?.email)

And

    print(error?.localizedDescription)

Please help me

Expression implicitly coerced from &#39;String?&#39; to Any

I&#39;m trying to make something like this:

     private String getStringIfObjectIsPresent(Optional&lt;Object&gt; object){
            object.ifPresent(() -&gt;{
                String result = &quot;result&quot;;
                //some logic with result and return it
                return result;
            }).orElseThrow(MyCustomException::new);
        }

This won&#39;t work, because ifPresent takes Consumer functional interface as parameter, which has void accept(T t). It cannot return any value. Is there any other way to do it ?


Java 8 optional: ifPresent return object orElseThrow exception

Since beta 8.3, zillions warnings &quot;String interpolation produces a debug description for an optional value; did you mean to make this explicit?&quot; appeared in my code.

For example, the warning popped in the following situation up, where options could lead to nil:

    let msg = &quot;*** Error \(options[&quot;taskDescription&quot;]): cannot load \(sUrl) \(error)&quot;

As previously designed, it was ok for me (and the compiler) the optionals to be interpolated as &#39;nil&#39;. But compiler changed its mind.

What the compiler suggests is to add a String constructor with description as follows:

    let msg = &quot;*** Error \(String(describing: options[&quot;taskDescription&quot;])): cannot load \(sUrl) \(error)&quot;

Obviously, the results is explicit but also very very cumbersome in my opinion. Is there a better option? Do I have to fix all those warning or better wait for the next beta?

[![Screenshot for description][1]][1]


  [1]: https://i.stack.imgur.com/llT41.png

How to solve &quot;String interpolation produces a debug description for an optional value; did you mean to make this explicit?&quot; in Xcode 8.3 beta?

I do user and invitation validation using the Optional facility

    @DeleteMapping(&quot;/friends/{username}&quot;)
    public
    HttpEntity&lt;Boolean&gt; removeFriend(
            @ApiParam(value = &quot;The user&#39;s name&quot;, required = true) @PathVariable String username
    ) {
        Long fromId = authorizationService.getUserId();

        return userService.findByUsername(username)
                .map(user -&gt; {
                     return friendshipService.findFriendship(fromId, user.getId())
                            .map(friendship -&gt; {
                                friendshipService.removeFriendship(friendship);

                                friendship.setToId(friendship.getFromId());
                                friendship.setFromId(friendship.getToId());

                                friendshipService.removeFriendship(friendship);

                                return ResponseEntity.ok(true);
                            }).orElseGet(() -&gt; ResponseEntity.notFound().build());
                }).orElseThrow(() -&gt; new ResourceNotFoundException(&quot;User not found&quot;));
However, `IntelliJ` is colouring my grey `return`, But when I remove the `return`, it highlights to me that there is no `return`.

Could someone explain how it works and what is it all about?

Statement lambda can be replaced with expression lambda

The [`ObjectiveC.swift`](https://github.com/apple/swift/blob/master/stdlib/public/SDK/ObjectiveC/ObjectiveC.swift#L228) file from the standard library contains the following few lines of code around line 228:

    extension NSObject : Equatable, Hashable {
      /// ...
      open var hashValue: Int {
        return hash
      }
    }

What does `open var` mean in this context, or what is the `open` keyword in general?

What is the &#39;open&#39; keyword in Swift?

is there any way to know the state of my application if it is in background mode or in foreground . Thanks 

Detect if the application in background or foreground in swift

I have a protocol:

    enum DataFetchResult {
        case success(data: Data)
        case failure
    }
    
    protocol DataServiceType {
        func fetchData(location: String, completion: (DataFetchResult) -&gt; (Void))
        func cachedData(location: String) -&gt; Data?
    }

With an example implementation:

        /// An implementation of DataServiceType protocol returning predefined results using arbitrary queue for asynchronyous mechanisms.
        /// Dedicated to be used in various tests (Unit Tests).
        class DataMockService: DataServiceType {
            
            var result      : DataFetchResult
            var async       : Bool = true
            var queue       : DispatchQueue = DispatchQueue.global(qos: .background)
            var cachedData  : Data? = nil
            
            init(result : DataFetchResult) {
                self.result = result
            }
            
            func cachedData(location: String) -&gt; Data? {
                switch self.result {
                case .success(let data):
                    return data
                default:
                    return nil
                }
            }
            
            func fetchData(location: String, completion: (DataFetchResult) -&gt; (Void)) {
                
                // Returning result on arbitrary queue should be tested,
                // so we can check if client can work with any (even worse) implementation:
                
                if async == true {
                    queue.async { [weak self ] in
                        guard let weakSelf = self else { return }
                        
                        // This line produces compiler error: 
                        // &quot;Closure use of non-escaping parameter &#39;completion&#39; may allow it to escape&quot;
                        completion(weakSelf.result)
                    }
                } else {
                   completion(self.result)
                }
            }
        }

The code above compiled and worked in Swift3 (Xcode8-beta5) but does not work with beta 6 anymore. Can you point me to the underlying cause?

Closure use of non-escaping parameter may allow it to escape

I&#39;d like to return a `UIViewController` conforming to `MyProtocol` from a method, so I&#39;m using the method signature:

    func myMethod&lt;T where T : UIViewController, T : MyProtocol&gt;() -&gt; T {

First thing I don&#39;t understand: if `myMethod` returns e.g. a `MyViewController` which has to following signature, I have to force cast it:

    class MyViewController: UIViewController, MyProtocol

I cannot simply `return MyViewController()` but I need to cast it like this: `return MyViewController() as! T` - why is this necessary?

And the second thing: how can I use this method somewhere? I cannot simply say

    let x = myMethod() as? UIViewController

as I get the error

    Generic parameter &#39;T&#39; could not be inferred

How can I achieve something like this? If I cast it to `MyViewController` it works, but I would like to avoid that of course.

**EDIT: Example**

    class MyViewController : UIViewController, MyProtocol {
    }
    
    protocol MyProtocol {
    }
    
    func myMethod&lt;T&gt;() -&gt; T where T : UIViewController, T : MyProtocol {
        return MyViewController() as! T // why is the cast necessary?
    }

ok, I do get one part, but why is the cast to `T` necessary? `MyViewController` is a subclass of `UIViewController` and conforms to the protocol, so no cast should be necessary, right?

Content Type	Original Author	Original Content on Stackoverflow
Question	Fabrizio Bartolomucci	View Question on Stackoverflow
Solution 1 - Ios	Grimxn	View Answer on Stackoverflow
Solution 2 - Ios	Alex	View Answer on Stackoverflow
Solution 3 - Ios	luhuiya	View Answer on Stackoverflow

Usage of where in if let assignment in Swift

Ios Problem Overview

Ios Solutions

Solution 1 - Ios

Solution 2 - Ios

Solution 3 - Ios

What happens to Firebase anonymous users?

What is the meaning of <> in mysql query?

Attributions