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this is your issue.
You've got your loop in the wrong place. As it s you will only ever cause the option window to choose the planet to appear once

:d

try to get it outputting:
aa
ab
ac
...
az
ba
...
bz

You define 'myuser' and 'mypass', but in the connection string you don't pass them in.


"jdbc:sqlserver://empiremain.empirevoip.com:3306/SQLEXPRESS/DB_Accounts;user=myuser;password=mypass;"
this...

1. Use code tags - nobody is going to look at the code if it's not in a readable format.
2. What exactly is the problem you are having?
3. If an error is being thrown, provide the stack trace as...

Ah understood.
Put that bit of code inside the try {} statement that way any exception will be handled by the catch statements.

What the parseLong(String str) method does is convert a number in string format to a Long number - i.e. Long.parseLong("123") equals 123 as a long number.

Passing in non-numeric input to the...

Well a long is a number - passing "sdfg" to it isn't going to work is it?

2nd page and still no error message.
You've been asked many times - either post the error message or nobody will be able to help you.

In the error trace it tells you the problem is the input string "sdfg" and that the line thats throwing the error is line 115 of CountryUserInput.

Which line is this?
I think it is because you...

You'll still need to include the check that the argument is not empty. I left it out for clarity on the important points

calculatorEngine calcEngine = new calculatorEngine(this);
button0.addActionListener(calcEngine);
button1.addActionListener(calcEngine);
button2.addActionListener(calcEngine);...

You are along the right lines.
You do need to use a for loop. What i would do is use the for loop at the start, then add the first character to the end...




public String rotateLeft(String s)...

exception handling is done via try/catch statements.
for example:



try {
..do the thing that could cause the error to be thrown
}
catch (NumberFormatException e) {
....handle the error...

Take it one line at a time - after each line, write down what the variable is.
If you are still struggling, you could always type it all into the computer and run the program yourself.

use Eclipse as your IDE - it gives handy hints as to how you can resolve such issues.
it's very handly when you are just starting out

If you get a not found error like that it is usually a missing import (as is the case here).
I'd recommend using an IDE like Eclipse as it highlights possible solutions to simple problems like these.

//Base Case 1: End of Maze
if(maze[r][c] == 'E')
{
total++;
maze[r][c] = PATH;
System.out.println(maze);
return;
}

...

Dude give us the stack trace of the error.
Most people aren't going to run your code to replicate the error, so if you post the stack trace you are more likely to get a response.

In java a single = sign assigns a value to a variable - i.e. i=1 sets the variable i to the value 1.
i == 1 is an equality checker, which checks that the variable has the same value as the object on...

You can tell from the stack trace that the GZIP class is attempting to read the header, so it doesn't look like there is anything wrong with your code.
Are you certain the file is a valid file? How...

I'm not sure if what i am asking about is possible, but here goes....

If i have a simple GUI with a textfield 'address line1' and i have an address object with a string variable for 'line1', is it...

If you single left click on the wee yellow icon you will get a suggested solution to the problem in Eclipse.

you are telling the calculator to push/pop an item in the calculate method, but the calculator is instantiated to null in the constructor and it isn't set to be anything else elsewhere.

you need...

Your method can either return null, or it can return a primitive type.
Your method returns an int(shown by public int eval....), therefore you must use the keywork "return" followed by a variable...