random pass generator

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Kim
random pass generator
1 Answer(s)      8 years ago
Posted in : Java Beginners
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July 17, 2009 at 3:32 PM


Hi Friend,

Try the following code:

import java.io.*;
import java.util.*;
import java.net.*;

public class NamePass {
public static void main(String[] args)throws Exception {
FileReader inStream,inStream1;
BufferedReader in,in1;

String passwords="";
String pass="";
String name="";
Random r = new Random();
int[] integers = new int[20];
for (int i = 0; i < integers.length; i++) {
integers[i] = r.nextInt(10000000);
}
for (int i : integers) {
passwords+=Integer.toString(i)+"\n";
}

File myFile = new File("passwords.dat");
FileOutputStream outStream = new FileOutputStream(myFile);
PrintWriter out = new PrintWriter(outStream);
out.println(passwords);
out.close();
String p[]=new String[20];
String n[]=new String[20];

inStream1 = new FileReader(myFile);
in1 = new BufferedReader(inStream1);
String passwordData="";
for (int i=0;i<20;i++){
p[i]=in1.readLine();
pass+=p[i];
}

inStream= new FileReader("names.dat");
in = new BufferedReader(inStream);
for (int i=0;i<20;i++){
n[i]=in.readLine();
name+=n[i];
}


java.util.List<String> list = new ArrayList<String>();
for(int i = 0; i < n.length; i++) {
String line = n[i]+" "+p[i];
list.add(line);
}
System.out.println(list);
writeToFile(list, "NamePass.dat");
}
private static void writeToFile(java.util.List<String> list, String path) {
BufferedWriter out = null;
try {
File file = new File(path);
out = new BufferedWriter(new FileWriter(file));
for (String s : list) {
out.write(s);
out.newLine();
}
out.close();
} catch(IOException e) {}
}
}

Thanks

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