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Author Topic: How to create an object by its "string" class name ?  (Read 11496 times)

izd

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How to create an object by its "string" class name ?
« on: March 07, 2017, 06:40:01 pm »
Hi everybody !  :D

I would like to found out an easy way to create an object by its class name from a string.

It should look like :

Code: Pascal  [Select][+][-]
  1. var
  2.   obj: TObject;
  3.  
  4.   obj := BuildObject( 'TButton', Self );

Of course, the NOT easy way is to cast the class name...

nb: For now, my interest is about controls (cf. Self)

Thank you in advance  :)

Thaddy

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Re: How to create an object by its "string" class name ?
« Reply #1 on: March 07, 2017, 06:57:07 pm »
This is not a scripting language but probably the RTTI lovers  :-* will give you an elaborate  8-) answer on how exactly you can do that. >:(  O:-) (And it's a bad idea)
Due to censorship, I changed this to "Nelly the Elephant". Keeps the message clear.

marcov

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Re: How to create an object by its "string" class name ?
« Reply #2 on: March 07, 2017, 07:08:56 pm »

Afaik VCL classes are registered, so something like

getclass('TButton').create(selfl);

izd

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Re: How to create an object by its "string" class name ?
« Reply #3 on: March 07, 2017, 07:11:51 pm »
Thank you mates  :)

@marcov: Unfortunately, its not works...

Thaddy

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Re: How to create an object by its "string" class name ?
« Reply #4 on: March 07, 2017, 07:29:29 pm »
They are only registered when used somewhere. Otherwise you have to register them by hand. (code reduction)
Due to censorship, I changed this to "Nelly the Elephant". Keeps the message clear.

izd

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Re: How to create an object by its "string" class name ?
« Reply #5 on: March 07, 2017, 07:33:12 pm »
This is what I thought after searching the web...  :'(

sky_khan

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Re: How to create an object by its "string" class name ?
« Reply #6 on: March 07, 2017, 08:29:36 pm »
Notice that, this is not thread-safe but it provides you a way to use classes you want, without interfering streaming of components

Code: Pascal  [Select][+][-]
  1. unit ControlReg;
  2.  
  3. {$mode objfpc}{$H+}
  4.  
  5. interface
  6.  
  7. uses
  8.   Classes, SysUtils, Controls, fgl;
  9.  
  10. type
  11.   TClassMap = specialize TFPGMap<string,TControlClass>;
  12.  
  13.   { TControlRegistry }
  14.  
  15.   TControlRegistry = class(TClassMap)
  16.   public
  17.     procedure RegisterControl(aClass:TControlClass);
  18.     procedure RegisterControls(aClassArr : array of TControlClass);
  19.     function CreateControl(const aClassName:string;aOwner:TComponent):TControl;
  20.   end;
  21.  
  22.  
  23.   function ControlRegistry : TControlRegistry;
  24.  
  25. implementation
  26. uses
  27.   rtlconsts;
  28.  
  29. var
  30.   FReg : TControlRegistry;
  31.  
  32. function ControlRegistry : TControlRegistry;
  33. begin
  34.   if FReg=nil then
  35.     FReg :=TControlRegistry.Create;
  36.   Result:=FReg;
  37. end;
  38.  
  39. { TControlRegistry }
  40.  
  41. procedure TControlRegistry.RegisterControl(aClass: TControlClass);
  42. begin
  43.   Add(aClass.ClassName,aClass);
  44. end;
  45.  
  46. procedure TControlRegistry.RegisterControls(aClassArr: array of TControlClass);
  47. var
  48.   aClass : TControlClass;
  49. begin
  50.   for aClass in aClassArr do
  51.     RegisterControl(aClass);
  52. end;
  53.  
  54. function TControlRegistry.CreateControl(const aClassName: string;
  55.   aOwner: TComponent): TControl;
  56. var
  57.   Ndx : Integer;
  58. begin
  59.   Ndx:=IndexOf(aClassName);
  60.   if Ndx>=0 then
  61.     Result:=Data[Ndx].Create(aOwner)
  62.   else
  63.     raise EClassNotFound.CreateFmt(SClassNotFound, [AClassName]);
  64. end;
  65.  
  66. initialization
  67. finalization
  68.   FReg.Free;
  69. end.
  70.  


Code: Pascal  [Select][+][-]
  1. uses ControlReg;
  2.  
  3. procedure TForm1.Button1Click(Sender: TObject);
  4. begin
  5.   with ControlRegistry.CreateControl('TEdit',Self) do
  6.     Parent:=Panel1;
  7. end;
  8.  
  9. constructor TForm1.Create(aOwner: TComponent);
  10. begin
  11.   inherited Create(aOwner);
  12.   ControlRegistry.RegisterControls([TMemo,TButton,TEdit,TLabel]);
  13. end;                                  
  14.  

Remy Lebeau

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Re: How to create an object by its "string" class name ?
« Reply #7 on: March 07, 2017, 08:34:10 pm »

Afaik VCL classes are registered, so something like

getclass('TButton').create(selfl);

GetClass() returns a general purpose TPersistentClass (aka 'class of TPersistent').  You have to type-cast it in order to call the virtual TComponent.Create() constructor:

Code: [Select]
var
  Btn: TButton;
begin
  Btn := TButton(TComponentClass(GetClass('TButton')).Create(Self));
end;

Alternatively:

Code: [Select]
type
  TButtonClass = class of TButton;

var
  Btn: TButton;
begin
  Btn := TButtonClass(GetClass('TButton')).Create(Self);
end;

Just make sure the desired classes are registered at runtime, either by the RTL itself when initializing packages, or explicitly in code:

Code: [Select]
RegisterClass(TDesiredControlHere);
« Last Edit: March 09, 2017, 08:29:09 pm by Remy Lebeau »
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bylaardt

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Re: How to create an object by its "string" class name ?
« Reply #8 on: March 07, 2017, 11:16:20 pm »
No magic:
Code: Pascal  [Select][+][-]
  1. function BuildControl(Const ObjectClass:String;ObjectOwner:TComponent):TwinControl
  2. begin
  3.   case LowerCase(ObjectClass) of
  4.     'tbutton' : result:=TButton.create(ObjectOwner);
  5.     'tlabel'  : result:=TLabel.create(ObjectOwner);
  6.     'tedit'   : result:=TEdit.create(ObjectOwner);
  7.     'tmemo'   : result:=TMemo.create(ObjectOwner);
  8.     // and so on
  9.   else
  10.      raise Exception.create(ObjectClass+' not allowed in BuilControl');
  11.   end;
  12. end;
  13.  

izd

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Re: How to create an object by its "string" class name ?
« Reply #9 on: March 08, 2017, 07:18:48 pm »
So.....no magic....(sic)

Code: Pascal  [Select][+][-]
  1. function BuildControl( const pClassName: String ): TObject;
  2. begin
  3.   case pClassName of
  4.     'TButton'  : Result := TButton.Create( nil );
  5.     .../... and so on .../...
  6.   end;
  7. end;

Thank you all friends  :)

ps: if someone found out the way to do so, I'll take it  ;)

Ñuño_Martínez

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Re: How to create an object by its "string" class name ?
« Reply #10 on: March 09, 2017, 11:33:28 am »
May be you can use an array of a record to store the classes.  Something like:

Code: Pascal  [Select][+][-]
  1.   TYPE
  2.   { Any component. }
  3.     ClassType = CLASS OF TComponent;
  4.  
  5.   { Stores class information. }
  6.     TClassListItem = RECORD
  7.       Name: STRING;
  8.       TheClass: ClassType;
  9.     END;
  10.  
  11.   VAR
  12.     ClassList: ARRAY [...] OF TClassListItem;
  13.  
  14.   BEGIN
  15.   { Add class information. }
  16.     ClassList[0].Name := 'TButton';
  17.     ClassList[0].TheClass := TButton;
  18. ...
  19.   { Searchs "ClassName" and creates "TheObject" }
  20.     FOR Ndx := LOW (ClassList) TO HIGH (ClassList) DO
  21.       IF ClassName = ClassList[Ndx].ClassName THEN
  22.         TheObject := ClassList[Ndx].TheClass.Create (SELF)
  23.   END;
  24.  
I didn't test actually so may be it doesn't work at all.
« Last Edit: March 09, 2017, 11:37:35 am by Ñuño_Martínez »
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marcov

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Re: How to create an object by its "string" class name ?
« Reply #11 on: March 09, 2017, 12:26:08 pm »
If you use LCL you might as well use the fact that such array/tlist already exist, see Remy's answer.

Ñuño_Martínez

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Re: How to create an object by its "string" class name ?
« Reply #12 on: March 11, 2017, 06:42:55 pm »
If you use LCL you might as well use the fact that such array/tlist already exist, see Remy's answer.
I see.  Didn't knew that.
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