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.
if I use a scanner to read system input, how do i store the input in one string?
So far I have something like this.
Scanner user_input = new Scanner(System.in);
System.out.print("Enter your name: ");
String name = user_input.next();
If in the terminal I type, "Bob Saget", name = "Bob". I want name = "Bob Saget"
Can anyone give me detailed instructions, if they involve eliminating white space or using tokenizers or something?
Use nextLine instead of next in order to read the entire input line into your name variable :
String name = user_input.nextLine();
Use user_input.nextLine() method , it reads till ENTER key is pressed
Use the nextLine(); method. The next(); method only reads the first token, the input until the first space (separator).
nextLine();
reads the entire line.
When I run my program instead of reading the string and storing it in tempAddress my program simply prints the next line before I enter input. Using next works for the first two because I am only using one word but the third one encompasses multiple words so need something else and through my research I found nextLine() was the answer but I am not able to get it to work as others have, thanks in advance.
System.out.println("Enter Employee First Name: ");
String tempFirstName = input.next();
employeesArray[i].setFirstName(tempFirstName);
System.out.println("Enter Employee Last Name: ");
String tempLastName = input.next();
employeesArray[i].setLastName(tempLastName);
System.out.println("Enter Employee Address: ");
String tempAddress = input.nextLine();
employeesArray[i].setAddress(tempAddress);
System.out.println("Enter Employee Title: ");
String tempTitle = input.next();
employeesArray[i].setTitle(tempTitle);
Basically Scanner tokenizes the input by default using whitespace. Using next() method of scanner returns the first token before the space and the pointer stays there. Using nextLine() returns the whole line and then moves the pointer to the next line.
The reason your nextLine() was not behaving fine was because, your previous input for employee last name using next() cause the pointer to stay in the line hence, when you reach the point to take the employee address using nextLine(), the pointer returns remainder of the previous input next() which was obviously empty (when supplied one word as input to next()). Assume you entered two words separated by space for last name, the next() will store the first word in last name field and pointer waits after first token before second token and as soon as you reach nextLine() pointer returns the second token and moves to new line.
The solution is to execute nextLine() after reading the input for last name to make sure that your pointer is in new line waiting for input for address.
I updated my code by inserting a input.nextLine() there to make sure that scanner input is consumed and pointer is moved to the next line.
System.out.println("Enter Employee First Name: ");
String tempFirstName = input.next();
employeesArray[i].setFirstName(tempFirstName);
System.out.println("Enter Employee Last Name: ");
String tempLastName = input.next();
employeesArray[i].setLastName(tempLastName);
//feed this to move the scanner to next line
input.nextLine();
System.out.println("Enter Employee Address: ");
String tempAddress = input.nextLine();
employeesArray[i].setAddress(tempAddress);
System.out.println("Enter Employee Title: ");
String tempTitle = input.next();
employeesArray[i].setTitle(tempTitle);
When you have input.next(), it reads the input, but not the newline character, it leaves it in the input stream. input.nextLine() ends with a newline character. So when input.nextLine() is executed, it stops without taking any input because it already got the newline (\n) char from the input stream.
Solution: read the newline before you execute inupt.nextLine():
System.out.println("Enter Employee Address: ");
input.next();//read the newline char - and don't store it, we don't need it.
String tempAddress = input.nextLine();
see also: https://stackoverflow.com/a/13102066/3013996
I am trying to write a string input to a text file using the Scanner object.
The string input is a film name. If the file name has two words, though, the scanner object only takes the first word.
I need it to take both words. Here is my code:-
Scanner new_dvd_info = new Scanner(System.in);
System.out.println("Enter name of new film");`
String film_name = new_dvd_info.next();
Can anybody shed any light please?
Replace new_dvd_info.next() with new_dvd_info.nextLine() to grab the entire line.
Documentation of Scanner.next() method says
Finds and returns the next complete token from this scanner.
A complete token is preceded and followed by input that matches
the delimiter pattern. This method may block while waiting for input
to scan, even if a previous invocation of {#link #hasNext} returned
<code>true</code>.
So it would just pick up until it finds " " as delimiter in your case. You could use next line method on scanner to get whole string new_dvd_info.nextLine() alternatively you could just loop over like:
while(scanner.hasNext) {
//append to string using scanner.next();
}
The problem here is that you are using new_dvd_info.next() which returns the first complete token. If any delimiter such as space is encountered it considers the next word as a separate token.
Scanner sc=new Scanner(System.in);
String s=sc.next();
System.out.println(s);
In the above code if you give the name of the movie as Age of Ultron it will return you just the tokenAge as there is a delimiter after the token age.
In case you want the complete String separated by delimiter you should use
Scanner sc=new Scanner(System.in);
String s=sc.nextLine();
System.out.println(s);
This will give you the desired output i.e. Age of Ultron
In my current program one method asks the user to enter the description of a product as a String input. However, when I later attempt to print out this information, only the first word of the String shows. What could be the cause of this? My method is as follows:
void setDescription(Product aProduct) {
Scanner input = new Scanner(System.in);
System.out.print("Describe the product: ");
String productDescription = input.next();
aProduct.description = productDescription;
}
So if the user input is "Sparkling soda with orange flavor", the System.out.print will only yield "Sparkling".
Any help will be greatly appreciated!
Replace next() with nextLine():
String productDescription = input.nextLine();
Use input.nextLine(); instead of input.next();
The javadocs for Scanner answer your question
A Scanner breaks its input into tokens using a delimiter pattern,
which by default matches whitespace.
You might change the default whitespace pattern the Scanner is using by doing something like
Scanner s = new Scanner();
s.useDelimiter("\n");
input.next() takes in the first whitsepace-delimited word of the input string. So by design it does what you've described. Try input.nextLine().
Javadoc to the rescue :
A Scanner breaks its input into tokens using a delimiter pattern,
which by default matches whitespace
nextLine is probably the method you should use.