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Return multiple arrays to the main method

My program is suppose to take a text file, read the first four names, create a random number between 1-4, and then assign the names to 4 different teams based on what the random number was. For instance, if the number was 3, then the first name would go to team 3, second name to team 4, etc. etc.(repeat process until there are no more names) I believe I have all of the code for that correct, the problem is I can't figure out how to return all the names I have put into the arrays that were brought into the method. Here is my code:
public static void main(String args[]) throws IOException
{
BufferedReader girlFile = new BufferedReader(new FileReader("girls40.txt"));
PrintWriter teamFile = new PrintWriter(new FileWriter("xxxxxxx-teamlist.txt"));
String team1[] = new String[20];
String team2[] = new String[20];
String team3[] = new String[20];
String team4[] = new String[20];
int n;
n = loadTeams(team1,team2,team3,team4,girlFile);
girlFile.close();
teamFile.close();
}
public static String[] loadTeams(String team1[],String team2[],String team3[],String team[],BufferedReader girlFile)
{
int n;
int random;
String name1;
String name2;
String name3;
String name4;
while((name1=girlFile.readLine())!=null)
{
name2=girlFile.readLine();
name3=girlFile.readLine();
name4=girlFile.readLine();
random = 1 + (int)(Math.random() * 4);
if(random==1)
{
team1[n]=name1;
team2[n]=name2;
team3[n]=name3;
team4[n]=name4;
}
if(random==2)
{
team1[n]=name4;
team2[n]=name1;
team3[n]=name2;
team4[n]=name3;
}
if(random==3)
{
team1[n]=name3;
team2[n]=name4;
team3[n]=name1;
team4[n]=name2;
}
if(random==4)
{
team1[n]=name2;
team2[n]=name3;
team3[n]=name4;
team4[n]=name1;
}
n++;
}
return team1[],team2[],team3[],team4[];
}`
The main method was given to me, so it cannot be changed.

If there is more code in main method than you've posted here. You'll have to mention what is the variable n and how is it being used else follow the answer.
main Method can't be changed
In your main method,
int n;
n = loadTeams(team1,team2,team3,team4,girlFile);
girlFile.close();
teamFile.close();
} // End of Main Method
You have not used returned value n for nothing. So it really doesn't matter what you return from method loadTeams() as long as it is an int.
Also, here loadTeams() returns an String[] which can't be assigned be int n, you'll have to change return type of loadTeams() to int as
public static int loadTeams(String team1[],String team2[],String team3[],String team[],BufferedReader girlFile) {
/*
...
*/
return 0; // whatever, it isn't being used
}
This the solution if you can't change the main method.

The call to loadTeams() expects a return value of type int. Not an array or multiple arrays. If you can't change the main method then loadTeams should return an integer.
// ...
int n;
n = loadTeams(team1,team2,team3,team4,girlFile);
// ...

you don't have to return anything, arrays created in main() will be passed to your method by reference, you can fill them there, and after execution of your method, values will be kept in these arrays

The loadTeams should return an int and not String[].
There is no need to return arrays. Changes made in the arrays in the loadTeams methods will be reflected back to the array in main method.

how to make more than condition in toString method

I want to list all names that end with "Reda" and ignore case sensitivity, I have tried the condition in the toString method at the bottom, but it would not print any thing.
public class Customer {
public static void main(String[] args) throws IOException {
File a = new File("customer.txt");
FileWriter v = new FileWriter(a);
BufferedWriter b = new BufferedWriter(v);
PrintWriter p = new PrintWriter(b);
human Iman = new human("Iman", 5000);
human Nour = new human("Nour", 3500);
human Redah = new human("Redah", 0);
human iman = new human("iman", 200);
human MohamedREDA = new human("MohamedREDA", 3000);
human Mohamed_Redah = new human("Mohamed Redah", 2000);
human[] h = new human[6];
h[0] = Iman;
h[1] = Nour;
h[2] = Redah;
h[3] = iman;
h[4] = MohamedREDA;
h[5] = Mohamed_Redah;
p.println(Iman);
p.println(Nour);
p.println(Redah);
p.println(iman);
p.println(MohamedREDA);
p.println(Mohamed_Redah);
p.flush();
}
}
class human {
public String name;
public double balance;
public human(String n, double b) {
this.balance = b;
this.name = n;
}
#Override
public String toString() {
if (name.equalsIgnoreCase("Reda") && (name.equalsIgnoreCase("Reda"))) {
return name + " " + balance;
} else
return " ";
}
}

Please avoid putting condition in toString method. Remove the condition there
public String toString() {
return name + " " + balance;
}
and change your logic in Customer class
human[] h = new human[6];
h[0] = Iman;
h[1] = Nour;
h[2] = Redah;
h[3] = iman;
h[4] = MohamedREDA;
h[5] = Mohamed_Redah;
for (int i = 0; i < h.length; i++) {
if (h[i].name.toLowerCase().endsWith("reda")) { // condition here
p.println(h[i]);
}
}
And make use of loops do not duplicate the lines of code.Every where you are manually writing the lines.
Check Java String class and use required methods to add condition.

String redahname = ("Redah").toLowerCase(); //put your h[0] instead of ("Redah")
if(name.endsWith("redah")){ //IMPORTANT TO BE IN LOWER CASE, (it is case insenitive this way)
//your code here if it ends with redag
System.out.println(redahname);
} //if it does not end with "redah" it wont out print it!
You can use this, but can you please explain your question more? What exactly do you need?

try this
#Override
public String toString() {
if (name.toLowerCase().endsWith("reda"))) {
return name + " " + balance;
} else
return " ";
}

String.equals() is not what you want as you're looking for strings which ends with "Reda" instead of those equal to "Reda". Using String.match or String.endsWith together with String.toLowerCase will do this for you. The following is the example of String.match:
public class Reda {
public static void main(String[] args) {
String[] names = {"Iman", "MohamedREDA", "Mohamed Redah", "reda"};
for (String name : names) {
// the input to matches is a regular expression.
// . stands for any character, * stands for may repeating any times
// [Rr] stands for either R or r.
if (name.matches(".*[Rr][Ee][Dd][Aa]")) {
System.out.println(name);
}
}
}
}
and its output:
MohamedREDA
reda
and here is the solution using endsWith and toLowerCase:
public class Reda {
public static void main(String[] args) {
String[] names = {"Iman", "MohamedREDA", "Mohamed Redah", "reda"};
for (String name : names) {
if (name.toLowerCase().endsWith("reda")) {
System.out.println(name);
}
}
}
}
and its output:
MohamedREDA
reda

You shouldn't put such condition in toString() method cause, it's not properly put business application logic in this method.
toString() is the string representation of an object.
What you can do, is putting the condition before calling the toString() , or making a helper method for this.
private boolean endsWithIgnoringCase(String other){
return this.name.toLowerCase().endsWith(other.toLowerCase());
}

None of your humans are called, ignoring case, Reda, so your observation of no names printed is the manifestation of properly working logic.
Your condition is redundant: you perform the same test twice:
name.equalsIgnoreCase("Reda") && (name.equalsIgnoreCase("Reda"))
If you need to match only the string ending, you should employ a regular expression:
name.matches("(?i).*reda")
toString is a general-purpose method defined for all objects. Using it the way you do, baking in the business logic for just one special use case, cannot be correct. You must rewrite the code so that toString uniformly returns a string representation of the object.

java,returning null value from method

I need to return finalString value for input operator name.
where,internalPrestring is fixed for specific operator,internalDigit would be retrieved from getting operator name.then all of'em would be added to finalString.
but it is giving null, i can't understand the problem
import java.io.*;
import java.lang.*;
class CallManager
{
public static final String postString = "#";
StringBuilder stringBuilder;
String internalPreString;
String preString;
String middleString;
String finalString;
String operatorName;
int internalDigit;
//needs to set oprator name
public void setOperatorName( String getMeFromPreferences)
{
operatorName = getMeFromPreferences;
System.out.println("I got it " + operatorName);
}
//afeter having operator name need to set inrernal digit for each operator
public void setOperatorBasedInternalDigit(int getIntegerForOperator)
{
internalDigit = getIntegerForOperator;
System.out.println("I got it too " + internalDigit);
}
//it needs to get string from ocr
public void setString( String getMeFromOCR )
{
middleString = getMeFromOCR;
}
//preString creator for differnet operator
public String getOperatorBasedPreString(String operatorName)
{
if(operatorName.equals("Airtel"))
internalPreString = "787";
else if(operatorName.equals("Banglalink"))
internalPreString = "123";
else if(operatorName.equals("Grameen"))
internalPreString = "555";
else if(operatorName.equals("Robi"))
internalPreString = "111";
else if(operatorName.equals("TeleTalk"))
internalPreString = "151";
stringBuilder.append("*").append(internalPreString).append("*");
preString = stringBuilder.toString();
return preString;
}
//get operator name and retrive midlle string's digit size from it
public int getOperatorBasedInternalDigit( String operatorName)
{
if(operatorName.matches("^Airtel | Grameen | Robi$"))
internalDigit = 16;
else if(operatorName.matches("^Banglalink$"))
internalDigit = 14;
else if(operatorName.matches("^TeleTalk$"))
internalDigit = 13;
return internalDigit;
}
//check operator-based digit number with input middle string as a number then retrive final string
public String getString( String toBeInserted, int inetrnalDigit)
{
if(toBeInserted.length() == internalDigit)
{
int counter = 0;
char [] insertHere = new char[internalDigit];
for(int verifier = 0; verifier < internalDigit; verifier ++)
{
insertHere[verifier] = toBeInserted.charAt(verifier);
if(!Character.isDigit(insertHere[verifier]))
break;
counter ++;
}
if(counter == internalDigit)
{
stringBuilder.append(preString).append(toBeInserted).append(postString);
finalString = stringBuilder.toString();
//to see what i've got finally as input for using this call manager method.it would be removed too
System.out.println(finalString);
return finalString;
}
else
{
//this printing could be used in main program
System.out.println("number is less or more than desired ..... INVALID SCAN");
System.out.println(middleString);
//here i will call the method for scan the card again
//
//
return middleString;
}
}
else
{
//this printing could be used in main program
System.out.println("number is less or more than desired ..... INVALID SCAN");
System.out.println(middleString);
//here i will call the method for scan the card again
//
//
return middleString;
}
}
}
//tester class that CallManager works rightly or not
class CallManagerDemo
{
public static void main(String args[]) throws IOException
{
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter name of Operator");
CallManager clm = new CallManager();
clm.setOperatorName("Banglalink");
System.out.println(clm.internalPreString);
System.out.println(clm.preString);
}
}

You are having only four lines that deals with your CallManager class:
CallManager clm = new CallManager();
clm.setOperatorName("Banglalink");
System.out.println(clm.internalPreString);
System.out.println(clm.preString);
The reason why you are getting null :
You are using a default constructor right and there processing is
done in it. So this is not a problem
Now on next line you call setOperator method which has this code:
public void setOperatorName( String getMeFromPreferences)
{
operatorName = getMeFromPreferences;
System.out.println("I got it " + operatorName);
}
Now here you are only setting thw variable operatorName and nothing else. So all other variables are null as you not doing any processing or something that will initialize them to something.
So when you print clm.internalPreString and clm.preString you get null as they are not initialized. But try printing clm.operatorName and it will print the operator name that you passed and was initialzed inside your method setOperatorName.
So as you have defined so many method inside your class, use them so that all the variables are set as per your logic
UPDATE
public void setOperatorName( String getMeFromPreferences)
{
operatorName = getMeFromPreferences;
//call any methods for example and use the values returned from the method by storing it inside a variable
String mystring = getOperatorBasedPreString(String operatorName)
}

You have never set the values for the variables for those you are getting the NULL value.The terms get/set must be used where an attribute is accessed directly.Read Java Programming Style GuideLines For more clarity.Use appropriate getters and setter for getting and setting the value like you have done for operatorName.

Don't you think that you should call any of the function instead of the string variables using object.
You are just calling one function that is
public void setOperatorName(String getMeFromPreferences) {
operatorName = getMeFromPreferences;
System.out.println("I got it " + operatorName);
}
You are calling default constructor with out any variable setting there,
You had not initialized any String you are calling form object.
I think you should call any of the function e-g
public int getOperatorBasedInternalDigit(String operatorName)
OR
public String getString(String toBeInserted, int inetrnalDigit)
Then you will get some string as you are expecting ...
Hope this will help you.

Behavior of return statement in catch and finally

public class J {
public Integer method(Integer x)
{
Integer val = x;
try
{
return val;
}
finally
{
val = x + x;
}
}
public static void main(String[] args)
{
J littleFuzzy = new J();
System.out.println(littleFuzzy.method(new Integer(10)));
}
}
It will return "10".
Now I just replace Return type Integer to StringBuilder and Output was changed.
public class I {
public StringBuilder method(StringBuilder x)
{
StringBuilder val = x;
try
{
return val;
}
finally
{
val = x.append("aaa");
}
}
public static void main(String[] args)
{
I littleFuzzy = new I();
System.out.println(littleFuzzy.method(new StringBuilder("abc")));
}
}
OutPut is "abcaaa"
So, Anybody can explain me in detail.?
what are the differences.?

Just because integer in immutable so after method returns even if value is changed in method it does not reflect, and does reflect in StringBuilder Object
EDIT:
public class J {
public String method(String x) {
String val = x;
try {
return val;
} finally {
val = x + x;
}
}
public static void main(String[] args) {
J littleFuzzy = new J();
System.out.println(littleFuzzy.method("abc"));
}
}

The principal operations on a StringBuilder are the append and insert methods, which are overloaded so as to accept data of any type. Each effectively converts a given datum to a string and then appends or inserts the characters of that string to the string builder. The append method always adds these characters at the end of the builder; the insert method adds the characters at a specified point.
For example, if z refers to a string builder object whose current contents are "start", then the method call z.append("le") would cause the string builder to contain "startle", whereas z.insert(4, "le") would alter the string builder to contain "starlet".
In general, if sb refers to an instance of a StringBuilder, then sb.append(x) has the same effect as sb.insert(sb.length(), x). Every string builder has a capacity. As long as the length of the character sequence contained in the string builder does not exceed the capacity, it is not necessary to allocate a new internal buffer. If the internal buffer overflows, it is automatically made larger.
Instances of StringBuilder are not safe for use by multiple threads. If such synchronization is required then it is recommended that StringBuffer be used.
In above method, finally block is calling everytime.

When an object is passed, the copy of its reference gets passed and you can change the contents if it is mutable.

Java class Anagrams

Im new to the java programming language and need help writing a class Anagrams that prints the permutations of words in a sentence. Example: red car -> red car, car red. This is what i have written so far and i think im on the right track and even though my code is not finished, i would at least like to get it to run.
import javax.swing.JOptionPane;
public class Anagrams
{
private String x;
private char[] xarray;
private String[] words;
public void Anagrams(String phrase1)
{
x = phrase1;
}
public void printPerms()
{
int perms = 0;
xarray = x.toCharArray();
for (int i = 0; i < x.length(); i++)
{
if(xarray[i] == ' ') perms = perms + 1;
}
words = x.split(" ");
for (int i = 0; i < perms; i++)
{
System.out.println(words[i]);
}
}
public void main(String args[])
{
String phrase1 = JOptionPane.showInputDialog("Enter phrase 1.");
Anagrams(phrase1);
printPerms();
}
}
This is the error i get when i try to run.
Exception in thread "main" java.lang.NoSuchMethodError: main
Right now im just trying to get my program to run not print out the permutations. I think i can figure that out once it at least print something out. Can someone tell me why it doesnt run and how do you get input from the user like c++ cin>>, if there is another way other than JOptionPane.
Thanks

A main method needs to be static.
How about this:
public static void main(String args[])
{
String phrase1 = JOptionPane.showInputDialog("Enter phrase 1.");
new Anagrams(phrase1).printPerms();
}

Even After Declaring your main method as static you may or may not be required to make all other methods as static(If calling methods dirctly without use of objects make methods as static).Because a static method can call or use only static data memebers or methods.
And in your code because you have defined all the methods in the same class which contains main method you need to make other methods also as static.

The method should return true if the two arguments are anagrams of each other, false if they are not.
For example, anagram(“glob”, “blog”) would return true;and anagram(“glob”, “blag”) false.
Assumes that the input strings will contain only letters and spaces. Treat upper- and lower-case letters as identical, and ignore spaces.
<br/> Uses the following algorithm:
<ul> <li> clean input strings from spaces and convert to lower case
</li> <li>convert to char array and sort them
</li> <li>if sorted arrays are identical, words are anagrams
</li></ul>
*/
public static boolean anagram(String str1, String str2)
{
//handle nulls
if(str1==null && str2==null)
return true;
else if( (str1==null && str2!=null) || (str2==null && str1!=null) )
return false;
//clean input strings from spaces and convert to lower case
String s1 = str1.replace(" ", "").toLowerCase();
String s2 = str2.replace(" ", "").toLowerCase();
//convert to char array and sort them
char[] cArr1 = s1.toCharArray();
char[] cArr2 = s2.toCharArray();
java.util.Arrays.sort(cArr1);
java.util.Arrays.sort(cArr2);
//if sorted arrays are identical, words are anagrams
s1 = new String(cArr1);
s2 = new String(cArr2);
return s1.equals(s2);
}
public static void main(String[] args)
{
//test: anagram(“glob”, “blog”) would return true; anagram(“glob”, “blag”) false.
System.out.println("anagram(“glob”, “blog”):"+(anagram("glob", "blog")));
System.out.println("anagram(“glob”, “blag”):"+(anagram("glob", "blag")));
}

You are missing static in:
public void main(String args[])
The main method needs to be static.
Also you are calling printPerms from main directly (without an object) so it must be made static as well or call them on a Anagram class object.
You are missing the new keyword while creating the object:
Anagrams(phrase1);
printPerms();
try
new Anagrams(phrase1).printPerms();
Also there is no Anagram class constructor that takes a String. What you have is a method named Anagram as you've specified the return type.
public void Anagrams(String phrase1) {
drop the void.