How to skip whitespaces and take input in Single line using Scanner - java

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.

Related

Removing Spaces from Scanner (System.in) Input

Right now I'm working on taking an equation, in "infix" notation and remove any spaces within that string prior to performing the rest of my program. Right now, without any spaces in the string, I receive the correct "Postfix" equation in return. But for some reason I can't seem to remove the spaces of a string that was entered using new Scanner(System.in); prior to performing my "Postfix" method(s). Here is the main method of my file:
public static void main(String[] args){
Scanner keyboard = new Scanner(System.in);
System.out.print("Enter the equation you'd like to evaluate: ");
InfixToPostFix postFixString = new InfixToPostFix();
String infix = keyboard.next();
String newInfix = infix;
//String newInfix = "9 * 5.3";
//String deleteSpaceInInfix = newInfix.replace(" ", "");
//System.out.println("deleteSpaceInInfix: " + deleteSpaceInInfix);
System.out.println("Postfix representation: " + postFixString.InfixToPostfix(infix));
}
Now I've noted out three lines that I tried to test while noting out the lines using the scanner information. In doing so, the result of the lines commented out is: 9*5.3 as expected. So I believe that it is something with the Scanner String object.
The way that you see this method now, when 9 * 5.3 is entered produces only 9. Everything after the first space is dropped.
I've tried to look up possible causes for this problem I'm not understanding and looked it up in the API documentation but haven't seen anything.
I'd appreciate any information in helping me better understand why my Scanner object (String infix = keyboard.next(); in this instance) is being treated differently than a normal String newInfix = "9 * 5.3; object is?
The default delimiter of Scanner is whitespace. Use nextLine() if you want to read an entire line:
Scanner keyboard = new Scanner(System.in);
String infix = keyboard.nextLine();
infix = infix.replace(" ", "");
System.out.println(infix);
Note: There's also hasNextLine() to check if there is another line in the input.

Java String - How to get chars until and after space

I'm load to variable string using:
Scanner scanner = new Scanner(System.in);
x = scanner.nextLine();
String always looks like: "Random Example". I want to grab first word (before space) for one variable and second word (after space) to next one variable. Can someone show me example?
You can get split a String using .split(String s) and put it in a String[]
String editMe;
Scanner user_input = new Scanner( System.in );
editMe = user_input.nextLine();
String[] edit1 = editMe.split(" ");
If you would like to see the values in the System you can use
int i =0;
for(String s:edit1)
{
System.out.println(s);
i++;
}
See more information on the String variable and how to use it here.
Input obtained from scanning the input stream can be split based on the space character.
Scanner scanner = new Scanner(System.in);
String x = scanner.nextLine();
String array[] =x.split(" ");
In this way, the words are stored in array.

What's the difference between next() and nextLine() methods from Scanner class?

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.

How to input a specific number of words in java?

I'm supposed to input an array of strings in java according to the following specification:
Input:A text containing K English words (where K <= 5000), with spaces and punctuation marks
My approach is to use an array of strings, each string containing a word and taking input using java.util.Scanner.next() inside a loop.
My problem is how to stop taking inputs when the user hits enter. Any idea?
Use a Scanner. Read the lines that the user enters. If the user enters an empty line then exit the loop:
public static void main(String[] args) throws Exception {
final Scanner scanner = new Scanner(System.in);
while (true) {
final String read = scanner.nextLine();
if ("".equals(read)) {
break;
}
System.out.println(read);
}
}
You could use two Scanner objects to handle this like so:
Scanner inScan = new Scanner(System.in);
String line = sinScan.nextLine();
while (line.length() > 0) {
Scanner lineScan = new Scanner(line);
while(lineScan.hasNext()) {
String word = lineScan.next();
// process this word.
}
line = inScan.nextLine();
}
// loop should exit when the user enters a blank line.

Printing user input in Java?

Say I had the Scanner method, and I want to print the user input after that.
Take this code for instance:
String randomWords;
Scanner kb = new Scanner(System.in);
System.out.print("Please enter two words separated: ");
randomWords = kb.next();
System.out.println(randomWords);
And if the two separated words entered were
hello world
Only
hello
is printed
Why is this? and how can I print both of the words with the space included?
Thank you.
Use Scanner#nextLine() instead,
This method returns the rest of the current line, excluding any line
separator at the end.
randomWords = kb.nextLine();
Scanner#next() reads the next complete token basing on the delimiter.
Finds and returns the next complete token from this scanner. A
complete token is preceded and followed by input that matches the
delimiter pattern
As default delimiter of the scanner is whitespace, you should explicitly define the delimiter for your scanner using Scanner#useDelimiter(str).
If you use \n next line as delimiter your curretn code would work.
Scanner kb = new Scanner(System.in).useDelimiter("\n");
System.out.print("Please enter two words separated: ");
randomWords = kb.next();
System.out.println(randomWords);
Use Scanner.nextLine() method to read up to the line break (noninclusive).
randomWords = kb.nextLine();
public static void main(String[] args) {
String randomWords;
Scanner kb = new Scanner(System.in);
System.out.print("Please enter two words separated: ");
randomWords = kb.nextLine();
System.out.println(randomWords);
}

Categories