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How can I get the name of the application from inside a package.

Posted on 2006-07-22
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Last Modified: 2010-03-31
All,

I have a Log object inside a utilities package that is shared between several applications. How can it find the name of the application it is in so it can name its log file correctly?

Paul
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Question by:PaulCaswell
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12 Comments
 
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Expert Comment

by:CEHJ
ID: 17160778
You probably would have to pass it the name of your app
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LVL 16

Author Comment

by:PaulCaswell
ID: 17160787
I looked in System.getProperty and there doesnt seem to be one there. I used to use       

String myName = this.getClass().getName();
fName = myName.split("\\.")[0]+".log";

but that doesnt work any more as its now in its own package.

Paul
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LVL 86

Expert Comment

by:CEHJ
ID: 17160794
That would in any case only produce 'Log.log' wouldn't it?
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LVL 2

Expert Comment

by:RoyalNepal
ID: 17160807
Yeah..you need to write the name of the application to get the expected name when you do..

>>String myName = this.getClass().getName();
>>>fName = myName.split("\\.")[0]+".log";

otherwise it will give only gives you Log.log


0
 
LVL 16

Author Comment

by:PaulCaswell
ID: 17160888
Could I collect the name of the class at the bottom of the stack? I believe there is a way of walking the stack isnt there?

Paul
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LVL 86

Expert Comment

by:CEHJ
ID: 17160899
You could, but only if the main class is at the bottom will you get what you want
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LVL 16

Author Comment

by:PaulCaswell
ID: 17160908
>>only if the main class is at the bottom
I can assume that as this is a utility program. How do I get an instance of my own stack to walk?

Paul
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Accepted Solution

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CEHJ earned 125 total points
ID: 17160917
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LVL 16

Author Comment

by:PaulCaswell
ID: 17160982
Thanks guys! This seems to work! Or at least, it gets the name of the file containing the main method, which should be enough for me.

        // Who is at the bottom of the stack?
        Throwable stack = new Exception();
        StackTraceElement [] calls = stack.getStackTrace();
      fName = calls[calls.length-1].getFileName().split("\\.")[0];

Thanks!

Paul
0
 
LVL 86

Expert Comment

by:CEHJ
ID: 17160989
:-)
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LVL 86

Expert Comment

by:CEHJ
ID: 17161009
You can actually do

calls[calls.length-1].getClassName()
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LVL 16

Author Comment

by:PaulCaswell
ID: 17161043
Yeah but that will give me 'package.class' so I will get the package name when I split at the '.' which will usually be the same (except for the case). This way, if user changes the name of the file, the log file name will change too so I prefer this way. Thanks for the heads-up though.

Paul
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