Just some simple Java question

[Niner49]

I want my program to ask the user if they would like to play another game of rock, paper, scissors. However, I want it to only accept (y for yes or n for no). At this point all I have been able to do is create a infinite loop, or just make the program not compile altogether. I have converted the char to a string, and attempted to use the .equals method but no such look.

I know it's a really simple question, but any suggestion would appreciated

[BigDaddyLH]

Post your best attempt (just the user prompting, not the rock-paper-scissors.

[Niner49]

Post your best attempt (just the user prompting, not the rock-paper-scissors.

[CODE=java]//Ask the user if he/she would like to play another game
System.out.prin t("\n\nWould you like to play another game ('y' for yes, 'n' for no): ");
cntrl = kb.next().charA t(0);
con = new Character(cntrl ).toString();

//A if statement that determines the whether the user wants to play another game
while(con.equal s('y') || con.equals('Y') || con.equals('n') || con.equals('N') != true)
{
System.out.prin t("Sorry, you must enter y for yes, or n for no: ");
cntrl = kb.next().charA t(0);
}

if(con.equals(' y') || con.equals('Y') )
{
ansr = true;
comTalley = 0;
useTalley = 0;
rslt = 0;
roundNum = 1;
gameNum++;
System.out.prin tln("\n\n\n\n\n ");
}

else if(con.equals(' n') || con.equals('N') )
{
ansr = false;
}[/CODE]


I had it working when the program didn't care what you entered to end, but I just want it to accept 'n' to end the loop that i have it in.

[BigDaddyLH]

[CODE=Java] cntrl = kb.next().charA t(0);[/CODE]

1. what if the user just hits return? I suspect kb.next() would return "" and so charAt(0) would be out-of-bounds!

2. You are confusing String and char. Stick with chars, which are simpler and because they are a primitive type, should be compared with == and !=

[CODE=Java]if (cntrl == 'y') {
...
if (cntrl != 'n') {
[/CODE]

[Niner49]

[CODE=Java] cntrl = kb.next().charA t(0);[/CODE]

1. what if the user just hits return? I suspect kb.next() would return "" and so charAt(0) would be out-of-bounds!

2. You are confusing String and char. Stick with chars, which are simpler and because they are a primitive type, should be compared with == and !=

[CODE=Java]if (cntrl == 'y') {
...
if (cntrl != 'n') {
[/CODE]

I'll give it a try, thanks for the advice. Sometimes I have problems with over complicating the simplest of things

[r035198x]

I'll give it a try, thanks for the advice. Sometimes I have problems with over complicating the simplest of things

There is a golden rule called KISS.