How to generate a random numeric string + random char string and combine?

Question

What I'm trying to do is to generate a random string of numbers E.G 2645237 and one char in a string in the range of A-Z E.G. W and combine the two strings to make 2645237W. I can generate a random number no problem. What I'm stuck on is: 1. Generating a random Char as a string. 2. Combining the two strings to create one string. To be clear what it's for is a school assignment to achieve some extra credit in my marking. Like always I'm not looking for the full answer. Some pseudo-code or a working example would be fine but I'd like the final "A-HA!" moment to be my own doing. A final parameter. This end result (the one string) would need to be a generated 50 times differently (I can do this) and then used as a sort of password. (Meant to replicate a PPS number, the added char is the bit that has my whole class stumped).

I'm not looking to cheat my way to a coded answer, just stuck on this problem (We've all been there)

• Software Engineer: Clue: Look at ascii characters 65 onwards...
• takendarkk: The char data type is actually represented by an integer. Find the integer range of the characters you want, generate a random int in that range, convert it to a char. From there turning an int and a char into strings and combining them should be easy.
• user2965503: Hey Engineer, Could you point me in the right direction to where I can learn ascii chars, I'v only been learning for about 3 months in a slow paced class with a below average tutor (Shows his work and expects us to learn just from the raw code).
• Tim B: +1 for asking for pointers but not final answer. :)
• takendarkk: http://www.asciitable.com/. Just to start you out - 65 = A, 66 = B, etc...

Answer 1

You can generate a random character simply by doing 'a' (or 'A' for upper case) and then generating a random number from 0 to 25 and adding that to it. i.e. 'a'+3 is 'd'. Note the use of a single quote character to say this is a char literal as opposed to the double quote for a String literal.

That random character can then be appended to the string. StringBuilder would do it for you easily, I'm not sure off hand what the String + operator will do with it.

• user2965503: So say for example I put in this pseudo-code (turned into proper code in a IDE naturally).. String letter = 'A'+(RNG); Example: String letter = 'A'+(14); Would the output become another letter? The output on this example would be O, right?
• Tim B: Yes. Char addition changes the value of the char, exactly the same as int 65+14 gives 79, 'A'+14 gives letter 79. This is different from + in String which concatenates strings.
• user2965503: Okay I used your method for the char ('A'+3). Then I made a 7 digit number (Hardcoded for now to save time) And combined them using Rakesh KR's method (myChar+""+Num) as this was a serious problem for me previously. Might I also ask one more thing if its not too much trouble ? I'll put it into a seperate comment.
• user2965503: I make the String 4534343W and put the code in a loop (Creating 50 strings and printing them to the bottom. I need the 50 numbers to be in an array. Do I create the array in the same method as the random number generator or do I declare it in the Main? Keeping in mind I will need to call the array into another method allowing a user to check if there code matches one of the 50.
• user2965503: I'd upvote you for the help (Greatly appreciated) but I need another 7 rep to do so sadly

Answer 2

Try,

Random rn = new Random();
int range = 9999999 - 1000000 + 1; 
int randomNum = rn.nextInt(range) + 1000000; // For 7 digit number
System.out.println(randomNum);
Random rc = new Random();
char c = (char)(rc.nextInt(26) + 'A');
System.out.println(c);
String str = randomNum+""+c; 
System.out.println(str);

str prints like 1757217Y

• user2965503: Okay that actually makes me slightly mad, not that you gave me the code (A working example is great!) But I had something very similar to that code yet I could not combine the two variables into the one String. Error said char cannot be dereferenced
• user2965503: Hey Rakesh could you explain how this line in your code works? (char)(rc.nextInt(26) + 'A'); I understand that rc.nextInt(26) adds a value 1-25 to 'A' to give a new letter but how does rc.nextInt() work? I'm gonna guess that it creates a random int for the next int in the range of 26?
• Rakesh KR: @user2965503 Read docs.oracle.com/javase/7/docs/api/java/util/…
• user2965503: @RakeshKR Thanks, I was having trouble finding a source for the info :)

Answer 3

To generate the letter and append on your number sequence:

String msg1 = "random number sequence";
Random gen = new Random();
char c = (char) (65 + gen.nextInt(26));
StringBuilder sb = new StringBuilder();
sb.append(msg1);
sb.append(c);
String result = sb.toString();
System.out.println(result);

By the way, 65 is the ascii code of the letter 'A' and gen.nextInt(26) generates a number between 0 and 25, ie, we have a range between 65 and 90 which are the letters' A'-'Z' in ascii table