I am working on creating a program for my course, in which I am required to divide the string: This;is;the;first;line;;This;is;the;second;line!;;;;Done!;;.
In the requirements, I need to read a single semicolon as a space, and a double semicolon as a new line. How do I create a regular expression in the useDelimiter() method that allows me to parse through and differentiate between both ; and ;;? Thank you!
Assignment Excerpt:
Instead of hard-coding the string, this time you will read it from the console. Study the useDelimiter() method and use it to set the delimiter for the scanner input. This time allow either colons or semicolons as the delimiters. One might prefer to use the String Tokenizer here, but don’t -- use the Scanner’s useDelimiter() method to set the delimiter in the Scanner and process each token as it comes.
import java.util.Scanner;
public class Hw3p2 {
public static void main(String[] args) {
// Initializes scanner class.
Scanner input = new Scanner(System.in);
// Prompts user for input.
System.out.println("Enter the string you wish to filter & parse: ");
// Reads user input.
String filterString = input.nextLine();
// Initiates new scanner reading the user inputted string.
Scanner a = new Scanner(filterString);
a.useDelimiter(";|;;");
System.out.printf("\n");
// Loop that parses through string while there are more tokens.
while(a.hasNext()) {
System.out.print(a.next());
}
}
}
The Expected output is to be:
You may use this code:
public class Hw3p2 {
public static void main(String[] args) {
// Initializes scanner class.
Scanner input = new Scanner(System.in).useDelimiter(";;");
// Prompts user for input.
System.out.print("Enter the string you wish to filter & parse: ");
// Reads user input.
while(input.hasNext()) {
String filterString = input.next();
//System.err.println("filterString: " + filterString);
// Initiates new scanner reading the user inputted string.
Scanner a = new Scanner(filterString).useDelimiter(";");
// Loop that parses through string while there are more tokens.
while(a.hasNext()) {
System.out.print(a.next() + " ");
}
System.out.println();
a.close();
}
input.close();
}
}
Output:
Enter the string you wish to filter & parse: This;is;the;first;line;;This;is;the;second;line!;;;;Done!;;
This is the first line
This is the second line!
Done!
Note use of outer scanner with delimiter ;; and an inner one with ;.
This code finds the first word "horror", but does not show me the whole line, only the word found.
File f = new File("MyFile.txt");
Scanner scan = new Scanner(f);
while (scan.hasNextLine()) {
String str = scan.next();
if (str.contains("horror")) {
System.out.println(str + " este horror");
}
}
Why is that?
The Scanner class has many methods for reading different types, and each has a corresponding hasNext...() method, for example nextInt() and hasNextInt(). You checked hasNextLine(), but used next() which returns the next word instead of nextLine() which returns the next line.
Change your code from:
String str = scan.next(); // read next word ❌
to:
String str = scan.nextLine(); // read next line ✅
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.
Im having some trouble with scanning user input in one of my first java programs. When I compile and run this, I am immediately prompted for input (i.e the command line stops and blinks). When I enter anything, the first line is printed, asking me to enter an integer. Then the second line is printed and I'm prompted to enter another value.
The output from this program is the first two values that I input. This is hard to explain, but it basically asks for 3 input values and only uses two.
import java.util.Scanner;
public class objects
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
System.out.println("Enter an integer please...");
int input = sc.nextInt();
System.out.println("Enter your name please...");
String name = sc.nextLine();
System.out.println("The read values: " + input + ", " + name);
sc.close();
}
}
Put a System.out.flush() command after your println statements if you're reading from the console directly afterward
just use this:
Scanner sc = new Scanner(System.in);
System.out.print ("Enter your name please... ");
String name = sc.nextLine();
System.out.print ("Enter an integer please... ");
int input = sc.nextInt();
System.out.println ("The read values: " + input + ", " + name);
i just moved the integer below the name and it sorta fixed it. hahaha
When you introduce a number you press enter key, nextInt() uses the number but the enter (\n) remains buffered. After this if you call again nextInt(), Java tries to convert \n into a number giving you a NumberFormatException, but if you invoke nextLine() they read the enter as empty string
Here you have a better explanation and one solution
Can't use Scanner.nextInt() and Scanner.nextLine() together
It seems this is an error to do with my installation of VirtualBox. No matter what I try, the problem persists. Even if i try to only read ONE integer, it will ask me to input two values.
Thanks for everyone who tried to help, I learned a lot just trying to debug this.
Sorry, should've added more
import java.util.Scanner;
public class rpg {
public static void main(String[] args) {
Scanner keyboard = new Scanner(System.in);
String input;
boolean live = true;
if (live == true){
System.out.println("DETECTIVE RPG");
while (true) { // << ENTER CHECK
System.out.print("\n Press Enter...");
input = keyboard.next();
if (input.equals(null))
break;
} // ENTER CHECK >>
System.out.print("Past check.");
keyboard is what I named the Scanner.
This prints "Press Enter" to the console and makes sure the input box is empty. If they press Enter it should break the While loop and move onto the next statement, right? When I press Enter it shows another input box and never gets to the next print statement.
There is no token to return yet from the Scanner when just "Enter" is hit. The next() method will block until it can read a token.
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 hasNext() returned true.
Switch to 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.
It will return the whole line of input. When the line is empty, it will return an empty string, "", not null. Try
input = keyboard.nextLine();
if ("".equals(input))
break;