Lakshan Dissanayake

Tim Holloway wrote:Actually, I think there's supposed to be a "working directory" directive for that desktop link file, so that would eliminate the need for the "cd" command in the Exec directive.

Note that "/opt" is supposed to be a shared system directory, so it's not er, optimal, for use as a single-user app.

In fact, if you want multiple users to have the ability to use the same app, you can put the executable in /opt - preferably a subdirectory of /opt, but you should use the Java Preferences system if you need per-user files.

Probably even better would be to put the Java app under /usr/local (or /usr/local/bin*), use Java Preferences for per-user data, and put any common data files under /var/lib. And while you're at it, add a shell script to run the java -jar command so that the Exec is more Linux-friendly.


*Note that there are attempts being made to reduce the number of "bin" directories in Linux, so check current recommended practices in the LSB.

Thanks Tom!

Campbell Ritchie wrote:. . . and what is wrong with keeping your .jar somewhere in /home? Your desktop link shou‍ld work nicely however.

found a working solution with the help of a friend (Rob Brew)

[Desktop Entry]
Version=1.0
Name=My Software
Exec=bash -c 'cd "/opt/myJavaProgram" && java -jar myJar.jar'
Terminal=false
Type=Application


Thanks Rob & Campbell Ritchie!

Campbell Ritchie wrote:Why are you running your work from /opt rather from a home directory?

I need to create a desktop shortcut entry and here is how its looks like

[Desktop Entry]
Version=1.0
Name=My Software
Exec=java -jar /opt/myJavaProgram/myJar.jar
Terminal=false
Type=Application

anyway Thanks Campbell Ritchie!

Rob Spoor wrote:That's because . is the directory where the program was started from - /opt/myJavaProgram in your first example, and your home directory (~) in the second.

Thanks Rob!  

But i was looking for something like to force java or program to consider relative path from it's jar location.

Anyway thanks for your time Rob

-supun

Please note

if i run the jar file using java -jar myJar.jar it works fine
and it will NOT work if i use java -jar /opt/myJavaProgram/myJar.jar

supun@supunz /opt/myJavaProgram $ java -jar myJar.jar // this works fine

supun@supunz ~ $ java -jar /opt/myJavaProgram/myJar.jar //this is not

Hi folks, i made a java programs which uses some files for its execution

... File report = new File("./reports/myReport.jrxml"); ...

I created a jar file and i put the jar file and above mention report file in /opt/myJavaProgram/

my directory tree is like follows
/opt/myJavaProgram/myJar.jar
/opt/myJavaProgram/reports/myReport.jrxml

now is the question
if i run the jar file using java -jar myJar.jar it works fine
and it will NOT work if i use java -jar /opt/myJavaProgram/myJar.jar

i guess that is because working directory related problem. but i cannot figure out how to get rid of it

any idea to solve that?
Thanks in advance!

-supun

Rob Spoor wrote:There is no such thing as a Class object for a generic type. There is only Response.class for the raw Response type. However, you can still do what you want with a little bit of cheating:
   @SuppressWarnings("unchecked")    public SpiceRequetImpl(){        super((Class<Response<RESULT>>) (Class<?>) Response.class);    }
You can't cast Response.class directly to a Class<Response<RESULT>>, but the intermediate cast solves that.[/quote

Thank you very much Rob!
You're life saver

Hi folks,

I'm doing some robospice stuff in android and got strucked in class type retrival

public class SpiceRequest<E>{    public SpiceRequest(Class<E> clazz){        //...    } } public class Response<T>{    private Response(){        //...    }    //... } public class SpiceRequetImpl<RESULT> extends SpiceRequest<Response<RESULT>>{        public SpiceRequetImpl(){        super(Response<RESULT>.class); //error here    } }

i also tried following to use instead for Response and there was no luck because Response doesn't have a public constructor

public class ResponseImpl extends Response<T>{    public ResponseImpl(){        super(); // error due to no access to parent constructor        //...    } }
Any help would be greatly appriciated!

Thanks!
-supun

Campbell Ritchie wrote:That is no different from <T> String or <T> int before the method name. It is a generic method and you can read about it in the Java® Tutorials.

Thanks Campbell Ritchie!

Hi folks,

I dig some tutorials and found a strange method
public <T> void doWork(List<T> param){ //do some stuff }

what is special about <T> void?
I can understand meaning of following.
public T doWork(List<T> param){ //do some stuff }

Thanks in advance!

It seems like getSupportFragmentManager() is null.
It's bit hard to make a guess.
If you can send source code, I hope I'll be able to help you to solve this problem.
please check your private messages for my email address.

-Supun

why did you use something =="1"

If you want to check the equality of two strings you should use equals method

Pay attention to following example

public class Example{ public static void main(String[] args) { String str1 = "123"; String str2 = "123"; String str3 = new String("123"); String str4 = new String("123"); System.out.println(str1 == str2); //prints true <------------------- position 1 System.out.println(str3 == str4); //prints false System.out.println(str1 == str4); //prints false System.out.println(str1.equals(str2)); //prints true System.out.println(str3.equals(str4)); //prints true System.out.println(str1.equals(str4)); //prints true } }

find out WHY position 1 gives true
Google it
You will find the answer, trust me ;)

Hi folks!,

I am new to reporting in android.
I need to print a report(like sales bill).
I did some research and found something relative with JasperMobile.
Note: I can get a print using byte array with bluetooth printer.

But No idea how to use it.

Please suggest me tips to achieve my goal.

Thanks in advance!

-Supun

Campbell Ritchie wrote: . . .
So, how are you going to use ZERO in class Bar. You either have to use the old‑fashioned technique of writing Foo.ZERO, or write plain simple ZERO, which requires a static import.

At least I think it does; I have never tried. Try it and tell us what happens.

Yeah that works. but in my problem public class AbstractControllerImpl extends AbstractController so no need of static import.

anyway thanks for your help.

Campbell Ritchie wrote:You would still need a static import even if the classes are in the same package.

Why? there is no need of that.