I have developed a character sorter and I want to prompt the user to enter a new string every time after the string is sorted. The issue I am having is that the scanner keeps scanning the first input of the user. If I use scanner.next(), it does not counts the white space at the end of the input, which is not the solution.
This is the start of the while loop. Once the code finishes, it starts again from "inputText".
inputText:
while(keepSorting){
System.out.print("Please input the string to be sorted: " );
c.settingInput(s.nextLine());
System.out.println(" ");
int option;
more code
.... }
Scanner sc= new Scanner(System.in);
while(true){
String a = sc.nextLine();
System.out.println(a);
}
If you want to know more about Scanner then read this documentation.
I need to take input in a single line of String which is a combination of String and integers.
I have tried to take the input like this
Scanner in = new Scanner(System.in);
String stringWithoutSpaces=in.nextLine();
But, scanner reads only the first character.
Input String:A 10,B 10,C 10,D 10
Required String:A10,B10,C10,D10
I need this input in a one single line using scanner class.
From what I understand, you are trying to receive the string from the user without storing all the white spaces. If your goal is to remove all the white spaces from the user input string, you can use replaceAll.
Code:
Scanner in = new Scanner(System.in);
System.out.print("Enter input: ");
String input = in.nextLine();
String noWhiteSpaces = input.replaceAll(" ", "");
System.out.println(noWhiteSpaces);
Console:
Enter input: A 10,B 10,C 10,D 10
A10,B10,C10,D10
import java.util.*;
public class Solution {
public static void main(String args[]) {
System.out.println("Hello world");
Scanner scanner=new Scanner(System.in);
String aItems = scanner.nextLine();
System.out.println(aItems);
}
}
This is working fine. I have Provided same input as mentioned by you and it is taking it as one string.Share your complete code or match from this one.
Why nextLine() method doesn't work? I mean, I can not enter any sentence after the second scan call because the program runs to the end and exits.
Input: era era food food correct correct sss sss exit
Should I use another Scanner object?
import java.util.*;
public class Today{
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
String str="";
String exit="exit";
System.out.println("Please enter some words : ");
while(true){
str=scan.next();
if(str.equalsIgnoreCase(exit)) break;
System.out.println(str);
}
System.out.println("Please enter a sentnce : ");
String sentence1 = scan.nextLine();
System.out.println("the word you entered is : " + sentence1);
}
}
What Scanner#nextLine does is to
Advance 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.
Since your input is era era food food correct correct sss sss exit you read inside the while every word with Scanner#next, so when Scanner#nextLine is called it returns "" (empty string) because there is nothing left of that line. That's why you see the word you entered is : (at the begging of the text is the empty string).
If you would have used this input: era era food food correct correct sss sss exit lastWord you would have seen the word you entered is : lastWord
The only thing you need to do in order to fix this is to call scan.nextLine(); first to move to the next line for the new input the user is going to provide and then get the new word with Scanner#nextLine() like this:
Scanner scan = new Scanner(System.in);
String str="";
String exit="exit";
System.out.println("Please enter some words : ");
while(true){
str=scan.next();
if(str.equalsIgnoreCase(exit)) break;
System.out.println(str);
}
scan.nextLine(); // consume rest of the string after exit word
System.out.println("Please enter a sentnce : ");
String sentence1 = scan.nextLine(); // get sentence
System.out.println("the word you entered is : " + sentence1);
Demo: https://ideone.com/GbwBds
What is the main difference between next() and nextLine()?
My main goal is to read the all text using a Scanner which may be "connected" to any source (file for example).
Which one should I choose and why?
I always prefer to read input using nextLine() and then parse the string.
Using next() will only return what comes before the delimiter (defaults to whitespace). nextLine() automatically moves the scanner down after returning the current line.
A useful tool for parsing data from nextLine() would be str.split("\\s+").
String data = scanner.nextLine();
String[] pieces = data.split("\\s+");
// Parse the pieces
For more information regarding the Scanner class or String class refer to the following links.
Scanner: http://docs.oracle.com/javase/7/docs/api/java/util/Scanner.html
String: http://docs.oracle.com/javase/7/docs/api/java/lang/String.html
next() can read the input only till the space. It can't read two words separated by a space. Also, next() places the cursor in the same line after reading the input.
nextLine() reads input including space between the words (that is, it reads till the end of line \n). Once the input is read, nextLine() positions the cursor in the next line.
For reading the entire line you can use nextLine().
From JavaDoc:
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
next(): Finds and returns the next complete token from this scanner.
nextLine(): Advances this scanner past the current line and returns the input that was skipped.
So in case of "small example<eol>text" next() should return "small" and nextLine() should return "small example"
The key point is to find where the method will stop and where the cursor will be after calling the methods.
All methods will read information which does not include whitespace between the cursor position and the next default delimiters(whitespace, tab, \n--created by pressing Enter). The cursor stops before the delimiters except for nextLine(), which reads information (including whitespace created by delimiters) between the cursor position and \n, and the cursor stops behind \n.
For example, consider the following illustration:
|23_24_25_26_27\n
| -> the current cursor position
_ -> whitespace
stream -> Bold (the information got by the calling method)
See what happens when you call these methods:
nextInt()
read 23|_24_25_26_27\n
nextDouble()
read 23_24|_25_26_27\n
next()
read 23_24_25|_26_27\n
nextLine()
read 23_24_25_26_27\n|
After this, the method should be called depending on your requirement.
What I have noticed apart from next() scans only upto space where as nextLine() scans the entire line is that next waits till it gets a complete token where as nextLine() does not wait for complete token, when ever '\n' is obtained(i.e when you press enter key) the scanner cursor moves to the next line and returns the previous line skipped. It does not check for the whether you have given complete input or not, even it will take an empty string where as next() does not take empty string
public class ScannerTest {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int cases = sc.nextInt();
String []str = new String[cases];
for(int i=0;i<cases;i++){
str[i]=sc.next();
}
}
}
Try this program by changing the next() and nextLine() in for loop, go on pressing '\n' that is enter key without any input, you can find that using nextLine() method it terminates after pressing given number of cases where as next() doesnot terminate untill you provide and input to it for the given number of cases.
next() and nextLine() methods are associated with Scanner and is used for getting String inputs. Their differences are...
next() can read the input only till the space. It can't read two words separated by space. Also, next() places the cursor in the same line after reading the input.
nextLine() reads input including space between the words (that is, it reads till the end of line \n). Once the input is read, nextLine() positions the cursor in the next line.
import java.util.Scanner;
public class temp
{
public static void main(String arg[])
{
Scanner sc=new Scanner(System.in);
System.out.println("enter string for c");
String c=sc.next();
System.out.println("c is "+c);
System.out.println("enter string for d");
String d=sc.next();
System.out.println("d is "+d);
}
}
Output:
enter string for c
abc def
c is abc
enter string for d
d is def
If you use nextLine() instead of next() then
Output:
enter string for c
ABC DEF
c is ABC DEF
enter string for d
GHI
d is GHI
In short: if you are inputting a string array of length t, then Scanner#nextLine() expects t lines, each entry in the string array is differentiated from the other by enter key.And Scanner#next() will keep taking inputs till you press enter but stores string(word) inside the array, which is separated by whitespace.
Lets have a look at following snippet of code
Scanner in = new Scanner(System.in);
int t = in.nextInt();
String[] s = new String[t];
for (int i = 0; i < t; i++) {
s[i] = in.next();
}
when I run above snippet of code in my IDE (lets say for string length 2),it does not matter whether I enter my string as
Input as :- abcd abcd or
Input as :-
abcd
abcd
Output will be like
abcd
abcd
But if in same code we replace next() method by nextLine()
Scanner in = new Scanner(System.in);
int t = in.nextInt();
String[] s = new String[t];
for (int i = 0; i < t; i++) {
s[i] = in.nextLine();
}
Then if you enter input on prompt as -
abcd abcd
Output is :-
abcd abcd
and if you enter the input on prompt as
abcd (and if you press enter to enter next abcd in another line, the input prompt will just exit and you will get the output)
Output is:-
abcd
From javadocs
next() Returns the next token if it matches the pattern constructed from the specified string.
nextLine() Advances this scanner past the current line and returns the input that was skipped.
Which one you choose depends which suits your needs best. If it were me reading a whole file I would go for nextLine until I had all the file.
From the documentation for Scanner:
A Scanner breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
From the documentation for next():
A complete token is preceded and followed by input that matches the delimiter pattern.
Just for another example of Scanner.next() and nextLine() is that like below :
nextLine() does not let user type while next() makes Scanner wait and read the input.
Scanner sc = new Scanner(System.in);
do {
System.out.println("The values on dice are :");
for(int i = 0; i < n; i++) {
System.out.println(ran.nextInt(6) + 1);
}
System.out.println("Continue : yes or no");
} while(sc.next().equals("yes"));
// while(sc.nextLine().equals("yes"));
Both functions are used to move to the next Scanner token.
The difference lies in how the scanner token is generated
next() generates scanner tokens using delimiter as White Space
nextLine() generates scanner tokens using delimiter as '\n' (i.e Enter
key presses)
A scanner breaks its input into tokens using a delimiter pattern, which is by default known the Whitespaces.
Next() uses to read a single word and when it gets a white space,it stops reading and the cursor back to its original position.
NextLine() while this one reads a whole word even when it meets a whitespace.the cursor stops when it finished reading and cursor backs to the end of the line.
so u don't need to use a delimeter when you want to read a full word as a sentence.you just need to use NextLine().
public static void main(String[] args) {
// TODO code application logic here
String str;
Scanner input = new Scanner( System.in );
str=input.nextLine();
System.out.println(str);
}
I also got a problem concerning a delimiter.
the question was all about inputs of
enter your name.
enter your age.
enter your email.
enter your address.
The problem
I finished successfully with name, age, and email.
When I came up with the address of two words having a whitespace (Harnet street) I just got the first one "harnet".
The solution
I used the delimiter for my scanner and went out successful.
Example
public static void main (String args[]){
//Initialize the Scanner this way so that it delimits input using a new line character.
Scanner s = new Scanner(System.in).useDelimiter("\n");
System.out.println("Enter Your Name: ");
String name = s.next();
System.out.println("Enter Your Age: ");
int age = s.nextInt();
System.out.println("Enter Your E-mail: ");
String email = s.next();
System.out.println("Enter Your Address: ");
String address = s.next();
System.out.println("Name: "+name);
System.out.println("Age: "+age);
System.out.println("E-mail: "+email);
System.out.println("Address: "+address);
}
The basic difference is next() is used for gettting the input till the delimiter is encountered(By default it is whitespace,but you can also change it) and return the token which you have entered.The cursor then remains on the Same line.Whereas in nextLine() it scans the input till we hit enter button and return the whole thing and places the cursor in the next line.
**
Scanner sc=new Scanner(System.in);
String s[]=new String[2];
for(int i=0;i<2;i++){
s[i]=sc.next();
}
for(int j=0;j<2;j++)
{
System.out.println("The string at position "+j+ " is "+s[j]);
}
**
Try running this code by giving Input as "Hello World".The scanner reads the input till 'o' and then a delimiter occurs.so s[0] will be "Hello" and cursor will be pointing to the next position after delimiter(that is 'W' in our case),and when s[1] is read it scans the "World" and return it to s[1] as the next complete token(by definition of Scanner).If we use nextLine() instead,it will read the "Hello World" fully and also more till we hit the enter button and store it in s[0].
We may give another string also by using nextLine(). I recommend you to try using this example and more and ask for any clarification.
The difference can be very clear with the code below and its output.
public static void main(String[] args) {
List<String> arrayList = new ArrayList<>();
List<String> arrayList2 = new ArrayList<>();
Scanner input = new Scanner(System.in);
String product = input.next();
while(!product.equalsIgnoreCase("q")) {
arrayList.add(product);
product = input.next();
}
System.out.println("You wrote the following products \n");
for (String naam : arrayList) {
System.out.println(naam);
}
product = input.nextLine();
System.out.println("Enter a product");
while (!product.equalsIgnoreCase("q")) {
arrayList2.add(product);
System.out.println("Enter a product");
product = input.nextLine();
}
System.out.println();
System.out.println();
System.out.println();
System.out.println();
System.out.println("You wrote the following products \n");
for (String naam : arrayList2) {
System.out.println(naam);
}
}
Output:
Enter a product
aaa aaa
Enter a product
Enter a product
bb
Enter a product
ccc cccc ccccc
Enter a product
Enter a product
Enter a product
q
You wrote the following products
aaa
aaa
bb
ccc
cccc
ccccc
Enter a product
Enter a product
aaaa aaaa aaaa
Enter a product
bb
Enter a product
q
You wrote the following products
aaaa aaaa aaaa
bb
Quite clear that the default delimiter space is adding the products separated by space to the list when next is used, so each time space separated strings are entered on a line, they are different strings.
With nextLine, space has no significance and the whole line is one string.
I have the following Java code:
Scanner input = new Scanner(System.in);
try {
System.out.println("Enter your name>>");
String name = input.next();
System.out.println("Enter your year of birth>>");
int age = input.nextInt();
System.out.println("Name: " + name);
System.out.println("Year of Birth: " + age);
System.out.println("Age: " + (2012 - age));
} catch (InputMismatchException err) {
System.out.println("Not a number");
}
When I enter a name with a space (e.g. "James Peterson"), I get the next line output correctly (Enter your year of birth) and then an InputMismatchException immediately. The code works with a single name without spaces. Is there something I'm missing?
System.out.println("Enter your name>>");
String name = input.next();
System.out.println("Enter your year of birth>>");
int age = input.nextInt();
Scanner, breaks its input into tokens using a delimiter pattern, which by default matches whitespace.
When you enter "James Peterson", Scanner#next() takes "James" as a token and assigns it to String name and then you do a Scanner#nextInt() which takes "Peterson" as the next token, but it is not something which can be cast to int and hence the InputMismatchException asnextInt() will throw an InputMismatchException if the next token does not match the Integer regular expression, or is out of range.
Actually the problem in your code has been clearly pointed out. In your case, there are two typical ways to achieve that:
One
Using useDelimiter method to delimite the input and after each input, you need to hit the Enter.
Sets this scanner's delimiting pattern to a pattern constructed from the specified String.
In your case, you need to
Scanner sc = new Scanner(System.in);
sc.setDelimiter("\\n");
// hit the "Enter" after each input for the field;
Two
Also you can achieve the same result using readLine
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.
In this way, you can do
Scanner sc = new Scanner(System.in);
System.out.println("Enter your name>>");
String name = sc.nextLine();
...
System.out.println("Enter your year of birth>>");
int age = sc.nextInt();
P.S.
Actually you can achieve it in a more restrictive way, controlling the pattern using next(Pattern pattern), if you need it.
int age = sc.next(Pattern.compile("\\d+{1,3}"));
In this case, if the input DOES NOT match the pattern, it will throw InputMismatchException as you were trying to nextInt while the input is a string.
When I enter a name with a space (e.g. "James Peterson"), I get the next line output correctly (Enter your year of birth>>) and then an InputMismatchException immediately.