Start program in user folder without knowing user name

Question

I have no idea if I can just run this code and it will work but here:

public void actionPerformed(ActionEvent e) {
        try
        {
        Runtime rt = Runtime.getRuntime();
        Process p = rt.exec("c:\\Users\\Toby\\AppData\\Roaming\\.minecraft\\minecraft.exe");
        }

        catch(Exception a)
        {

        }
    }

This is linked to a button, and it launches the minecraft launcher. If I want to give it to my friends, what do I have to do so that it doesn't look for the user 'Toby', but instead looks for their home folder? Sorry if its confusing!

Thanks for the fast answers! Its worked, thank you so much.

user1650305
– user1650305

2012-09-08 19:21:50 +00:00
Commented Sep 8, 2012 at 19:21 — user1650305
– user1650305, Commented Sep 8, 2012 at 19:21

jefflunt · Accepted Answer · 2012-09-08 20:20:18Z

5

I think you're looking for the user.home property. There's a list of properties available here: http://www.mindspring.com/~mgrand/java-system-properties.htm

So, your code would be changed to:

Process p = rt.exec(System.getProperty("user.home") + \\AppData\\Roaming\\.minecraft\\minecraft.exe");

edited Sep 8, 2012 at 20:20

answered Sep 8, 2012 at 18:48

jefflunt

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1 Comment

jefflunt Over a year ago

Well, it's not an endorsement of the website. I found another resource and updated my answer.

Sujay · Accepted Answer · 2012-09-08 19:25:25Z

4

If you want to keep consistency with other version of Windows, I would suggest using the System.getenv("APPDATA") instead of adding to System.getProperty("user.home") because Roaming folder is not present in older version of Windows

So your code would change to something like this:

Process p = rt.exec(System.getenv("APPDATA") + ".minecraft\\minecraft.exe")

edited Sep 8, 2012 at 19:25

answered Sep 8, 2012 at 19:06

Sujay

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Comments

Kumar Vivek Mitra · Accepted Answer · 2012-09-08 18:49:12Z

0

Try System.getProperty("user.home")

String mhome = System.getProperty("user.home"))+"";

    public void actionPerformed(ActionEvent e) {
            try
            {
            Runtime rt = Runtime.getRuntime();
            Process p = rt.exec(mhome+"\\AppData\\Roaming\\.minecraft\\minecraft.exe");
            }

            catch(Exception a)
            {

            }
        }

answered Sep 8, 2012 at 18:49

Kumar Vivek Mitra

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