Random integer that is a multiple of 5?

Question

So far I got the program to tell me if the user's inputted letter is right or wrong, but how can I have the computer tell the user what slot the guessed letter (if correct) is in? I.e. the letter "e" is in the first and last slots.
This is what I have so far, but all it says is the previously used wrong letters and previously guessed right letters, but not in the order of the word.
          JOptionPane.showMessageDialog (null, "You have guessed all of the following letters incorrectly: \n" + guessedWrongLetters + "\n \nYou have guessed all of the following correct letters: \n" + guessedRightLetters);Also, How can I generate a random integer between one and 250 that is a multiple of five? I.e. 5, 10, 15, etc. I'm using the String.substring (+start+, end) method to choose a random word from a single variable "wordlist" that contains 50 five letter words.
This is what I have, but it just chooses at random letter for the first part of the substring, and adds five to that chosen letter for the second part. So it sometimes chooses something in the middle of a word.
compNumOne = 1 + randy.nextInt (245);
           compNumTwo = compNumOne + 5;
           itsComputersWord = itsComputersWordList.substring(compNumOne, compNumTwo);Here is the word list:
     private String itsComputersWordList = "firstbibleshoeshoneyearthgloveapplechildshipsboatsaortaacidsallowroadsburntbusesdocksbucksbuddydudesudderembedemberelveseggedentryebonydunesdoweldozeldoorsdiverdisksdirksdinerdualsfrothelbowfryerfrozefrownfretsfumedgullsneckshairyhabitlinenlilacmedal";}Edited by: Pluberus on Nov 20, 2008 4:17 PM - Fixed a mistake.

Answer 1

Pluberus wrote:
What about something like this:?
                indexHolder = itsComputersWord.indexOf(itsUsersLetter);
if (indexHolder == 0)
JOptionPane.showMessageDialog (null, "The letter you guessed is the first letter of the secret word.");
if (indexHolder == 1)
JOptionPane.showMessageDialog (null, "The letter you guessed is the second letter of the secret word.");
if (indexHolder == 2)
JOptionPane.showMessageDialog (null, "The letter you guessed is the third letter of the secret word.");
if (indexHolder == 3)
JOptionPane.showMessageDialog (null, "The letter you guessed is the fourth letter of the secret word.");               
if (indexHolder == 4)
JOptionPane.showMessageDialog (null, "The letter you guessed is the fifth letter of the secret word.");By the way, is there an easier way to do what I just did? instead of all the multiple ifs?
Also, since none of my words have a letter used 3 times. Could I use the lastIndexOf and the indexOf to tell the user what position their letter is in?Ouch. Are you considering doing that all the way to virtual infinity? Why not just show something like:
"The letter you guessed is in position " + (indexHolder + 1)
instead of converting numbers into english ordinal words. Unless you want to build that complexity into the system as well, for virtually no benefit.
Also, what if the letter isn't found, when indexHolder is -1? You should account for that as well.