Java. How to convert random line of characters to a string?

Question

I'm new to java. I have started about 3 days ago. I want to make a line of random characters and to put them into a string. Thank you.

    import java.util.Random;
      public class test{
      public static void main (String[]args){

final String alphabet = "abcdefghigklmnopqrstuvwxyz";
final int N = alphabet.length();
Random r = new Random();

for (int i = 0; i < 50; i++) {
    String s = alphabet.charAt(r.nextInt(N));
    // System.out.println(alphabet.charAt(r.nextInt(N)));
}}}

Compilation completed. The following files were not compiled: 1 error found: File: C:\Users\Arnas\Desktop\Java\hack.java [line: 10] Error: incompatible types required: java.lang.String found: char — Aras, Aug 22 '13 at 11:51
First rule of programming: read the error message. Second rule of [Java] programming: read the [Java]doc. What does the documentation say about String.charAt()? What does it return? — JB Nizet, Aug 22 '13 at 11:51
A question that was asked multiple times in stackoverflow http://stackoverflow.com/questions/41107/how-to-generate-a-random-alpha-numeric-string — AurA, Aug 22 '13 at 11:51
I guess you want to use something like a StringBuilder which you append in your for loop. After the loop you can call toString on that builder and you have your String built. What you are doing is to create a new String with exactly one character in every run through your loop. — Matthias, Aug 22 '13 at 11:52
Another similar question http://stackoverflow.com/questions/2863852/how-to-generate-a-random-string-in-java — AurA, Aug 22 '13 at 11:52

score 7 · Accepted Answer · answered Aug 22 '13 at 11:52

7

Easiest way would be to use a StringBuilder or StringBuffer (syntax is the same).

StringBuilder sb = new StringBuilder();
for (int i = 0; i < 50; i++) {
    sb.append(alphabet.charAt(r.nextInt(N)));
}
String s = sb.toString();

answered Aug 22 '13 at 11:52

hermansen

158
3

+1 for using StringBuilder – DarthVader Aug 22 '13 at 11:53
upvote for StringBuilder – Prasad Kharkar Aug 22 '13 at 11:55
Thank you. It worked but I had to add if statement that if sb length is over 30 sb would be reseted and then I used Mikkel Løkke method. – Aras Aug 22 '13 at 12:09
I'm not familliar with StringBuilder yet. Thank you anyway. – Aras Aug 22 '13 at 12:09

score 1 · Answer 2 · answered Aug 22 '13 at 11:52

1

String s = "";
for (int i = 0; i < 50; i++) {
    s += alphabet.charAt(r.nextInt(N));
}

answered Aug 22 '13 at 11:52

drewdles

244
1
10

score 1 · Answer 3 · answered Aug 22 '13 at 11:52

1

Append your random characters to StringBuilder that acts as a text buffer.

StringBuilder sb = new StringBuilder();

for (int i = 0; i < 50; i++) {
    sb.append(alphabet.charAt(r.nextInt(N)));
}

System.out.println(sb);

answered Aug 22 '13 at 11:52

Ravi K Thapliyal

51,095
9
76
89

score 1 · Answer 4 · answered Aug 22 '13 at 11:53

1

Use StringBuilder

StringBuilder sb = new StringBuilder()
for (int i = 0; i < 50; i++) {
    sb.append(alphabet.charAt(r.nextInt(N)))
}
String s = sb.toString()

answered Aug 22 '13 at 11:53

Mikkel Løkke

3,710
23
37

score 1 · Answer 5 · answered Aug 22 '13 at 11:53

1

The simplest way would be to just add the char with the addition-operator ("+") to the string. Its function is overloaded, so that should work.

As you get more comfortable with Java, you should start using StringBuilder or StringBuffer.

answered Aug 22 '13 at 11:53

The13thToast

65
8

score 1 · Answer 6 · answered Aug 22 '13 at 11:54

Use a StringBuilder for adding characters to build a string -

import java.util.Random;

public class test{

    public static void main (String[]args){
        final String alphabet = "abcdefghigklmnopqrstuvwxyz";
        final int N = alphabet.length();
        Random r = new Random();
        StringBuilder builder = new StringBuilder();

        // make strings with 50 characters
        for (int i = 0; i < 50; i++) {
            builder.append(alphabet.charAt(r.nextInt(N)));
        }

        System.out.println(builder.toString();
    }
}

Java. How to convert random line of characters to a string?

6 Answers6