Thread: Cant type at login box!?!?

The method that deals with keystrokes is processLoginScreenInput()

This is the important part of the method, past all the clicking shit.

Code:

			int l1 = readChar(-796); //read keystrokes
			if (l1 == -1) { //no new keystrokes
				break;
			}
			boolean validChar = false; //is the character valid?
			for (int i2 = 0; i2 < validUserPassChars.length(); i2++) { //check and see
				if (l1 != validUserPassChars.charAt(i2)) {
					continue; //it's not valid, break the loop
				}
				validChar = true; //it's valid, proceed
				break;
			}

and

Code:

				if (loginScreenCursorPos == 0) {
					if ((l1 == 8) && (myUsername.length() > 0)) { //backspace
						myUsername = myUsername.substring(0, myUsername.length() - 1);
					}
					if ((l1 == 9) || (l1 == 10) || (l1 == 13)) { //enter or return
						loginScreenCursorPos = 1;
					}
					if (validChar) {
						myUsername += (char) l1; //add the keystroke
					}
					if (myUsername.length() > 12) {
						myUsername = capitalize(myUsername.substring(0, 11)); //make sure it's not too long
					}
				}

Some of that is unique to my client, but the concept is the same.

The drawing part is handled in drawLoginScreen()

Code:

			chatTextDrawingArea.method389(true, c / 2 - 90, 0xffffff, "Username: " + myUsername + ((loginScreenCursorPos == 0) & (loopCycle % 40 < 20) ? "@[email protected]|" : ""), j);
			j += 15;
			chatTextDrawingArea.method389(true, c / 2 - 88, 0xffffff, "Password: " + TextClass.passwordAsterisks(myPassword) + ((loginScreenCursorPos == 1) & (loopCycle % 40 < 20) ? "@[email protected]|" : ""), j);
			j += 15;

Or something similar to that.

You're gonna wanna watch the variable loginScreenState, but basically make sure you're processing the login stage that you're drawing.

No that wasnt my problem, but i got it fixed. Thanks for givin me some info though