XCODE 9.2 insert and hyperlink - xcode9.2

I am new to Xcode and downloaded the last 9.2 version, I am trying to insert a link to the web in a button and is not functioning. Here is the code.
link not working
override func viewDidLoad() {
super.viewDidLoad()
// Do any additional setup after loading the view, typically from a nib.
}
override func didReceiveMemoryWarning() {
super.didReceiveMemoryWarning()
// Dispose of any resources that can be recreated.
}
#IBAction func button(_ sender: Any) {
if let url = URL(string: "https://www.youtube.com/watch?v=UPfA3mUfvGA&t=90s") {
UIApplication.shared.canOpenURL(url)
UIApplication.shared.open(url, options: [:], completionHandler:nil)
}
}
}
When you press Guorlin Premium button it takes you to a website
Now, when I press the Guorlin Premium button is not working, it does not take me to the website.

Just press option+shift+command+k and press Ok to clean the folder and try to run it on a real device instead of the simulator.
It definitely works, I had a similar problem with a similar code and it worked for me. No change in code is needed.

Related

How do I add the SendQuote Action to the Mobile App

I am trying to add the Send Quote action from the actions menu in the web app to the mobile app. This is on the sales quote screen (CR304500). I followed the documentation from the Mobile Framework guide, and I was able to add other buttons. The code I'm using for the mobile screen is
update screen CR304500 {
update container "QuoteSummary" {
add recordAction "SendQuote" {
behavior = Void
}
}
}
Thank you in advance.
Turns out because it was in the Actions Folder, I had to enter it as follows.
update screen CR304500 {
update container "QuoteSummary" {
add recordAction "SendQuoteActionsFolder" {
behavior = Void
}
}
}

How to hide keyboard when touch on another control (such as button) Swift 2.0

I've read this page: Close iOS Keyboard by touching anywhere using Swift
to find out how to hide keyboard, it's good when i touch on view, but when i touch on button it not work, does anyone know?
Paste this in your controller:
let tap: UITapGestureRecognizer = UITapGestureRecognizer(target: self, action: #selector(YourViewController.dissmissKeyboard))
view.addGestureRecognizer(tap)
#objc func dissmissKeyboard() {
view.endEditing(true)
}
This is rather an hack, but it works pretty well and the keyboard even hides when tapped on background. I'm not sure if it works out for your problem with the button, but have a try!
EDIT: Try to add the UITapGestureRecognizer to your button?
in viewDidLoad add code:
self.hideKeyboardWhenTappedAround()
Then add an extension to your ViewController:
extension UIViewController {
func hideKeyboardWhenTappedAround() {
let tap: UITapGestureRecognizer = UITapGestureRecognizer(target: self, action: "dismissKeyboard")
view.addGestureRecognizer(tap)
}
func dismissKeyboard() {
view.endEditing(true)
}
}
Works really well for me.
SWIFT 4
Add this to your viewDidLoad.
let endEditingTapGesture = UITapGestureRecognizer(target: view, action: #selector(UIView.endEditing(_:)))
endEditingTapGesture.cancelsTouchesInView = false
view.addGestureRecognizer(endEditingTapGesture)

UIView element not updating

I need to get data from a API I’ve used debug prinln’s and these print almost instantly including the information gotten from the API, but when I try to display this info to the user in the ViewController this takes around 30 seconds to display.
I will get the information from the API here
func httpGet(request: NSURLRequest!, callback: (NSData?, String?) -> Void) {
var session = NSURLSession.sharedSession()
var task = session.dataTaskWithRequest(request){
(data, response, error) -> Void in
if error != nil {
callback(nil, error.localizedDescription)
} else {
callback(data, nil)
}
}
task.resume()
}
I invoke the request and return a string for debugging
func loadSchedule() {
httpGet(request) {
(data, error) -> Void in
onComplete(“Milo Cesar”)
}
And here I will display the data to the user
#IBAction func settingsButtonPress(sender: AnyObject) {
println("Settings Button Pressed: Invoking Schedule")
Schedule().loadSchedule(){
(scheduleData) in
println("Found Schedule: Setting Title to \(scheduleData)")
self.settingsButton.setTitle(scheduleData, forState: UIControlState.Normal)
}
}
The whole process from the “Settings Button Pressed: Invoking Schedule” statement in my console till I get “Found Schedule: Settings Title to Milo Cesar” takes less than a second.
Since the name already gets displayed in that console log I think that the information has been loaded. Though it takes up to 35 seconds for the Button to change it’s title.
Why does it take so long for my UIButton to update it’s title after the data has been retrieved?
Are you sure you are updating its title in main thread? try to wrap it up in GCD
dispatch_async(dispatch_get_main_queue()){
self.settingsButton.setTitle(scheduleData, forState: UIControlState.Normal)
}
EDIT
it fixes it because everything UI related must be performed by main thread, its a restriction that comes directly from Apple. You can use background threads for long taking tasks(download data, perform long taking calculations or sort large arrays) but everything that needs to be displayed on your screen must be performed by main thread(tableview reload, animations, inserting text into label).
Also here is a great tutorial on GCD and its functionality: http://www.raywenderlich.com/60749/grand-central-dispatch-in-depth-part-1

VS 2012 Start UI Windows from VSPackage

i've started to work with the VS2012 extensibility possibilities. I did the first few Walkthroughs and now I'm trying get further on. What I'm trying is pretty easy I guess... I'm trying to build a simply vspackage which starts an UI window. Actually i do not find any howto or sample code.
Do you have some links with further information about doing something like that ?
Thanks for you help..
Iki
You can find initial information here.
Here is my code for menu item:
/// <summary>
/// Initialization of the package; this method is called right after the package is sited, so this is the place
/// where you can put all the initialization code that rely on services provided by VisualStudio.
/// </summary>
protected override void Initialize()
{
Debug.WriteLine ("Entering Initialize() of: {0}", this);
base.Initialize();
// Add our command handlers for menu (commands must exist in the .vsct file)
OleMenuCommandService mcs = GetService(typeof(IMenuCommandService)) as OleMenuCommandService;
if ( null != mcs )
{
// Create the command for the menu item.
CommandID menuCommandID = new CommandID(GuidList.guidPackageProject, (int)PkgCmdIDList.Impl);
OleMenuCommand menuItem = new OleMenuCommand(MenuItemCallback, menuCommandID);
mcs.AddCommand( menuItem );
}
}
/// <summary>
/// This function is the callback used to execute a command when the a menu item is clicked.
/// See the Initialize method to see how the menu item is associated to this function using
/// the OleMenuCommandService service and the MenuCommand class.
/// </summary>
private void MenuItemCallback(object sender, EventArgs e)
{
MyForm form = new MyForm();
form.ShowDialog(); // Here your form is opening
}
I have been searching for a solution to this recently as I also needed to start a WPF form from a VSPackage. I have got things working after a couple of hours searching various topics on this and some good ol' trial and error.
I had an existing WPF-Project in a separate solution, which had to be merged into a VSPackage. Here's the steps to get this working:
Create a new Solution of Project type 'Visual Studio Package'
Make sure you select the 'Tool Window' option in the VS Package
Wizard (see the image below)
Now that the Solution has been created, add the already existing
WPF-Project to it (Right-Click 'Solution', Add->Existing Project) NOTE: It might be wise to copy the WPF-project to the Solution folder prior to adding it to the Solution.
Make sure you create a reference to the WPF-Project from your
VSPackage-Project and (if necessary) edit the namespaces of the WPF-Project to meet those of the VSPackage-Project, or the other way around.
Your Solution will now look something like this:
Now, you need to edit MyToolWindow.cs:
// Original:
base.Content = new MyControl();
// Change to:
base.Content = new MainWindow();
Make the following changes to VSPackage1Package.cs (or whatever your *Package.cs file is called)
// Original
private void ShowToolWindow(object sender, EventArgs e)
{
// Get the instance number 0 of this tool window. This window is single instance so this instance
// is actually the only one.
// The last flag is set to true so that if the tool window does not exists it will be created.
ToolWindowPane window = this.FindToolWindow(typeof(MyToolWindow), 0, true);
if ((null == window) || (null == window.Frame))
{
throw new NotSupportedException(Resources.CanNotCreateWindow);
}
IVsWindowFrame windowFrame = (IVsWindowFrame)window.Frame;
Microsoft.VisualStudio.ErrorHandler.ThrowOnFailure(windowFrame.Show());
}
// Change to:
private void ShowToolWindow(object sender, EventArgs e)
{
// Get the instance number 0 of this tool window. This window is single instance so this instance
// is actually the only one.
// The last flag is set to true so that if the tool window does not exists it will be created.
//ToolWindowPane window = this.FindToolWindow(typeof(MyToolWindow), 0, true);
//if ((null == window) || (null == window.Frame))
//{
// throw new NotSupportedException(Resources.CanNotCreateWindow);
//}
//IVsWindowFrame windowFrame = (IVsWindowFrame)window.Frame;
//Microsoft.VisualStudio.ErrorHandler.ThrowOnFailure(windowFrame.Show());
MainWindow mainwin = new MainWindow();
mainwin.Show();
}
If you get no build errors, you should be fine.
To test if your WPF-form opens, Press 'Start' to run the VSPackage in a new 'Experimental' Visual Studio instance. If everything went OK, you will find and should be able to run your WPF-from from the View->Other Windows menu.
If you don't see your VSPackage listed in the menu, close your 'Experimental' Visual Studio instance. Then Clean en Build your Solution and press 'Start' again. It should show up now.

Problems with LWUIT in J2ME on Nokia E72

Well, I'm developing a app in my cellphone that is going to connect to my PC, the problem is that everytime that I return a URLRequest to the cellphone, it shows the previous Form on the screen and not de actual one, for example this is what goes in my actionListener:
public void actionPerformed(ActionEvent ae) {
if (ae.getCommand() == guiaUtil.cSelecionar()) {
LoginRemote loginRemote = new LoginRemote();
try {
//This is the request, returns true or false, does not affect the form
loginRemote.login(tLogin.getText(), tPassword.getText());
} catch (Exception e) {
GuiaUtil.error(e);
return;
}
guiaUtil.mainApp().startMenu();
}
}
Then in the "guiaUtil.mainApp().startMenu()" I have this
public void startMenu() {
if (itemsMenu == null) {
itemsMenu = new List();
itemsMenu.setWidth(320);
itemsMenu.addItem("Sincronize Spots");
itemsMenu.addItem("Find Spots");
itemsMenu.addItem("Work");
itemsMenu.setFocus(true);
this.addComponent(itemsMenu);
this.addCommandListener(this);
this.addCommand(guiaUtil.cSelect());
Form form = new Form();
form.addComponent(itemsMenu);
}
form.show();
}
Anyway, after the request returns, it shows my Login form again, instead of showing the Menu List
Maybe what is going is that you are getting an exception, treating it with GuiaUtil.error and returning from actionPerformed without calling startMenu.
I would move guiaUtil.mainApp().startMenu() inside the try/catch block.
Not sure what happens in loginRemote.login(tLogin.getText(), tPassword.getText());
If you access the network, I would put that part in a different thread.
Inform the main thread by some kind of callback when the "remote login" is done,
you can show the menuForm from the edt then.
You have to put the following code outside the if condition.
Form form = new Form();
form.addComponent(itemsMenu);
You are having two form object. one inside if and another one outside of if. Object created inside the loop will loses the scope inside if. You are showing form object outside if. That's why, menu list screen was not displayed.

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