Opening a URL from an iOS FireMonkey application

Q: How do you open a URL in Safari on the iPhone using FireMonkey?
A: You use UIApplication.sharedApplication.openUrl

Below is a very simple unit that will do the code. Just create an iOS app with one button and add the following code to the unit.

unit Unit1;

{$IFDEF FPC}
{$mode objfpc}{$H+}
{$modeswitch objectivec1}
{$ENDIF}

interface

uses
SysUtils, Types, UITypes, Classes, Variants, FMX_Types, FMX_Controls, FMX_Forms,
FMX_Dialogs
{$IFDEF FPC}
, iPhoneAll
{$ENDIF}
;

type
TForm1 = class(TForm)
Button1: TButton;
procedure Button1Click(Sender: TObject);
private
{ Private declarations }
public
{ Public declarations }
end;

var
Form1: TForm1;

implementation

{$R *.lfm}

procedure TForm1.Button1Click(Sender: TObject);
begin
// Open a URL
{$IFDEF FPC}
UIApplication.sharedApplication.openUrl(NSUrl.URLWithString(NSSTR(PChar('http://blogs.embarcadero.com/ao'))));
{$ENDIF}
end;

end.