Thread: Determining the location of the file that initiates an application

Hi,

My question is, suppose I have an xml file say "myXml.xml" that can be stored in any directory on my PC and I set its default program (the one that will run when I double click myXml to "Myapplication.exe"
How do I determine the location of this file "myXml" when Myapplication starts, using java?
(Myapplication.exe is being programmed using java)

The reason I ask is that myXml would theoretically store application setting for Myapplication.exe and I would want to load them directly on start up if myApplication.exe is initiated by double click of an xml file

I understand how to use file objects and all that, I just dont know how to get either the location of this file "myXml" or its contents without hard coding its location within the application itself, cos as I said this file could be located virtually anywhere on the pc

Any help would be much appreciated!!!

Best regards
Michael

I'm a little confused by your statement that the .exe file is being programmed in java, since it's unusual for a Java program to be compiled into an .exe file, though not impossible. I'll let that go for now.

I don't know the answer, but I imagine it has something to do with the launch mechanism of the host operating system. Since double-clicking a file (or in some cases single-clicking) opens that file or launches it in its associated application, the filename and full path name of that file must be passed to the associated application. I assume that the filename is passed to the associated application by the OS using a mechanism similar to the command line arguments in a console-run Java program. You may need an OS-specific .bat file or script that accepts the selected .xml file name and passes it to the Java program as a parameter.

Keep searching and let us know what you find out.

Hi Greg, to clarify I am making a normal runnable jar file and then I am going to convert it into a .exe for my windows version just to make it neat on the platform.
Onto the problem at hand, my instinct is similar to what you have put forward here, but as of yet I have come up empty in my search for the answer

I would still appreciate anyone weighing in with some wise words, cos surely this is a problem that all applications that can load files needs to overcome at some point...

All help is greatly appreciated!!
Best regards
Michael

Ok, so far I have found a good half solution, java has System.getProperty("user.dir") which gives me the working directory of whatever initiated running my program. So the only thing I need is to get the actual file name that initiates my program, any ideas?

I also went and listed all the available system properties with:
Properties p = System.getProperties();
p.list(System.out);

But the actual filename that executes the application was not among them

Yes, that is what I am asking, if we are both understanding each other correctly.
So let me explain clearly, say my application "myApp.exe" is located at C:\Users\Michael\Desktop and I have set it to be the default program for xml files.
Now say that I have an xml file "myXml.xml" located at "C:\Users\Michael\Desktop\Development", therefore the full path of this file would be "C:\Users\Michael\Desktop\Development\myXml.xm l"

Now so far what I have found is that System.getProperty("user.dir") returns the current working directory and so if I double clicked myXml.xml this would return "C:\Users\Michael\Desktop\Development" as the current working directory.
The problem is I have no way so far of finding out the actual file name, in this case "myXml.xml"

Maybe I'm not understanding you, but the file-name doesn't get passed to the command line,
can you perhaps give me a small example of what you mean?

Thanks!

Hi Norm,
yes the string passed to main was empty, so im not sure how to get at the information, i.e. the name of the file?
Other than editing how the file association table, is there another way cos that seems like quite an impractical solution when distributing this application

Thanks for all your help so far, I'm quite out of my depth here so I appreciate it!

I'm not sure what you are asking, so I am just gonna Google a bit so I can give you an intelligent answer, ill reply shortly

Ok, so iv looked around a bit through my registry, do you mean this entry, Iv attached a png

1.jpg


I also looked through the FileExt list and there was no command line that I could see, I added a custom file association called .cmi and that is what is displayed here

2.jpg

Thanks!!

Let me know if you cant see the images cos they have come out really small on the site for some reason

Hi Norm, Ok I hope this is what you meant!
I took the file association that I have made and am using for my application and I also took the registry entry from the same location as you had yours

 
Windows Registry Editor Version 5.00
 
[HKEY_CLASSES_ROOT\.cmi]
@="cmi_auto_file"
 
[HKEY_CLASSES_ROOT\.cmi\PersistentHandler]
"OriginalPersistentHandler"="{00000000-0000-0000-0000-000000000000}"

Ok cool, this is whats currently in that file directory

Windows Registry Editor Version 5.00
 
[HKEY_CLASSES_ROOT\cmi_auto_file]
@=""
 
[HKEY_CLASSES_ROOT\cmi_auto_file\shell]
 
[HKEY_CLASSES_ROOT\cmi_auto_file\shell\open]
 
[HKEY_CLASSES_ROOT\cmi_auto_file\shell\open\command]
@="\"C:\\Users\\Michael\\Desktop\\CMIApplication.exe\" \"%1\""

Should I fix the quotes to be in the same format as yours are in the example you gave me?

I am working with a java program, but its a java program that I have bundled into a exe using Launch4j, I can change it to point to my jar file though, what do you suggest?

Thanks for your patience, I'm learning a lot!!

Hey Norm, thanks for your help so far,
my laptops screen died and is being repaired which is why I have been unable to reply till now

Can I ask what does \"%1\"" actually do in the registry command line? I know you mentioned it but I didnt really understand
Thanks so much!!

Ok cool, then as soon as I get my laptop back Ill do some experimenting, and ill report back with anything I discover!!
Thanks Norm