opening a Windows application

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4 total messages Started by JB Mon, 03 Jul 2006 16:59

#3360

Author: JB
Date: Mon, 03 Jul 2006 16:59

9 lines
1238 bytes

I have a file name stored in my application. I want to be able
to use this file name to open the file using the appropriate 
Windows application. (ie the application opened by double 
clicking on the file name in Windows Explorer for example.) I then want the opened application to run independently of my 
application and be closed as normal by the user.
Is this possible and if so how can I do this?
Your help will be much appreciated.
Thanks, JB

Re: opening a Windows application

#3361

Author: Ed Mulroy
Date: Mon, 03 Jul 2006 17:43

23 lines
1823 bytes

There are many functions with which you can do that.

Probably the simplest is WinExec.  An example is:
    WinExec("Notepad.exe", SW_SHOW);

Other functions that also can be used are CreateProcess, ShellExecute, 
system and the exec* set of functions.  WinExec, CreateProcess and 
ShellExecute are documented in the help for Windows and the others are in 
the help for C/C++.

.  Ed

> JB wrote in message
> news:44a94ce4$1@newsgroups.borland.com...
>
> I have a file name stored in my application. I want to be able
> to use this file name to open the file using the appropriate
> Windows application. (ie the application opened by double
> clicking on the file name in Windows Explorer for example.) I
> then want the opened application to run independently of my
> application and be closed as normal by the user.
> Is this possible and if so how can I do this?
> Your help will be much appreciated.

Re: opening a Windows application

#3362

Author: Hans Galema
Date: Mon, 03 Jul 2006 17:55

28 lines
1521 bytes

JB wrote:
> I have a file name stored in my application. I want to be able
> to use this file name to open the file using the appropriate 
> Windows application. 


   AnsiString FileName = ..........;

   if ( ! FileExists ( FileName ) )
      {
      ShowMessage (  FileName + "\n\nnot found.");

      return;
      }

   HINSTANCE hinstance = ShellExecute (
      NULL
      , "open"
      , FileName.c_str()
      , NULL
      , NULL
      , SW_SHOWNORMAL
      );

    if ( (int)hinstance < 32 )
      ShowMessage ( FileName + "\n\ncould not be shellexecuted");

Hans.

Re: opening a Windows application

#3367

Author: JB
Date: Tue, 04 Jul 2006 10:04

34 lines
1726 bytes

Thanks, ShellExecute solves my problem. 
Do I assume that with the first parameter set to NULL the 
executed application is independent of my application?
Thanks again.  
JB

Hans Galema <notused@notused.nl> wrote:
>
>
>   AnsiString FileName = ..........;
>
>   if ( ! FileExists ( FileName ) )
>      {
>      ShowMessage (  FileName + "\n\nnot found.");
>
>      return;
>      }
>
>   HINSTANCE hinstance = ShellExecute (
>      NULL
>      , "open"
>      , FileName.c_str()
>      , NULL
>      , NULL
>      , SW_SHOWNORMAL
>      );
>
>    if ( (int)hinstance < 32 )
>      ShowMessage ( FileName + "\n\ncould not be shellexecuted");

>JB wrote:
>> I have a file name stored in my application. I want to be >> able to use this file name to open the file using the 
>> appropriate Windows application.

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