For or While loop skipping an input

Question

Possible Duplicate:
Scanner issue when using nextLine after nextInt

I am creating a client program that needs to read both a String and an integer from my server. Depending on what integer it receives it adds some labels to the GUI. So far, my program reads the integer but skips the String. The following output is the output of my program when I try to write the integers to the program:

Server writes: 1
Server writes: 1
System prints: 1
System prints: j1
System prints: Name

The problem is that I am unable to write a String because it skips the String. How can I avoid this problem (note that I have also tried a for loop)

My code is as following:

int times = client.reciveCommando();
int o = 0;
System.out.println(times);

while (o != times) {
  int j = client.reciveCommando();
  System.out.println("j"+ j);
  String name = client.reciveString();
  System.out.println("Name " +name);
  createUser(j, name);
  o++;

}

The createUser method:

private void createUser(int j, String reciveChat) {
  if (j == 1) {
    chatPerson1.setVisible(true);
    lbl_Chatperson1_userName.setVisible(true);
    lbl_Chatperson1_userName.setText(reciveChat);
  } else if (j == 2) {
    lbl_chatPerson2.setVisible(true);
    lbl_userName2.setVisible(true);
    lbl_userName2.setText(reciveChat);
  } else {
    chatPerson3.setVisible(true);
    lbl_userName3.setVisible(true);
    lbl_userName3.setText(reciveChat);
  }
}

The client.reciveCommando method:

public int reciveCommando() throws IOException{
  Integer i = input.nextInt();
  return i;
}

The client.reciveString method:

public String reciveString(){
  String x = input.nextLine();
  return x;
}

Hope someone is able to help me with this :)

Thank you in advance.

Where do you increment o? (maybe I'm blind?) Your while loop will skip a string if times == 0. — belwood, Oct 09 '12 at 15:18
May be this is what you are looking for or it may help you. http://stackoverflow.com/questions/7056749/scanner-issue-when-using-nextline-after-nextint — Abubakkar, Oct 09 '12 at 15:48

score 1 · Answer 1 · answered Oct 09 '12 at 15:57

1

I don't see anywhere in the loop code where you are incrementing o or changing the value of times. So either the loop is being skipped altogether (ie: times = 0) or some other place in the code is modifying either the loop variable (o) or the loop condition (times) - very bad coding in either case.

Your loop variable/increment rules should be very clear in reading the loop and easily discernible what the start/stop conditions are without needing to read other methods/etc which may modify the values during loop iteration.

My immediate guess is that times = 0, or you would be in an endless loop.

answered Oct 09 '12 at 15:57

Eric B.

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Okay apprently my Copy paste failed at the end of the while loop it does o++; – Marc Rasmussen Oct 09 '12 at 16:26
Okay this is basicly what happens: First program starts, then it ask for times i set the times to 1(for example) then it ask me again for another number (this is the id of the client) i set that to 1. The system then prints 1, j1 , name. and if i try to set in a string i get an error because the thing that was posted wasnt an int (and it refers me to reciveCommando) which is an Integer. – Marc Rasmussen Oct 09 '12 at 16:30
I don't follow. "because the thing that was posted wasnt an int (and it refers me to reciveCommando)". What thing that was posted? – Eric B. Oct 09 '12 at 17:30
@ int j = client.reciveCommando(); the program ask for an integer then theprogram skips to "System.out.Println("name "+ name);" and repeats the loop(setting o to +1) instead of asking me for a String! so basicly it skips this line: String name = client.reciveString(); – Marc Rasmussen Oct 09 '12 at 17:46
What is `input`? Have you tried using a debugger? What happens when you step through each line of code? – Eric B. Oct 09 '12 at 17:59
Sadly i do not have a debugger. input is is the String that i should wait for from the server its the client.reciveString(); – Marc Rasmussen Oct 09 '12 at 21:17
In `receiveString()`, what is `input`? – Eric B. Oct 10 '12 at 01:11

score 0 · Accepted Answer · answered Oct 10 '12 at 11:14

i found the solution to my question it turned out to be quite simple!

first of all let me explain what i ment.

When i my program ran the while loop it basicly skipped the line where it should have recived an input from the server. i found that the reason it did that was that the input.nextLine(); was empty which makes sence when you read the api for input.nextLine();

Advances this scanner past the current line and returns the input that was skipped. This method returns the rest of the current line, excluding any line separator at the end. The position is set to the beginning of the next line. Since this method continues to search through the input looking for a line separator, it may buffer all of the input searching for the line to skip if no line separators are present. Returns: the line that was skipped

since the line that i tried to get was an empty line it would skip and set name to " ".

here is the full complete code for my program and it currently works:

The while loop:

            while (o != times) {
                int j = client.reciveCommando();
                System.out.println("j"+ j);
                    String name = client.reciveString();
                    System.out.println("Name " +name);
                    createUser(j, name);    
                o++;
            }

The client.reciveString();

    public String reciveString(){

    String x = input.next();
    return x;
}

The createUser();

    private void createUser(int j, String reciveChat) {
    if (j == 1) {
        chatPerson1.setVisible(true);
        lbl_Chatperson1_userName.setVisible(true);
        lbl_Chatperson1_userName.setText(reciveChat);

    }else if (j == 2) {
        lbl_chatPerson2.setVisible(true);
        lbl_userName2.setVisible(true);
        lbl_userName2.setText(reciveChat);
    }else if (j == 3){
        chatPerson3.setVisible(true);
        lbl_userName3.setVisible(true);
        lbl_userName3.setText(reciveChat);}

Thank you for all of your responses and i will be sure to vote you up :)

For or While loop skipping an input

2 Answers2