This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 26 days ago.
I am having this problem a lot. When I use a Scanner a lot of times, it doesn't get input from user.
Scanner scan = new Scanner(System.in);
System.out.println("1---");
int try1 = scan.nextInt();
System.out.println("2---");
int try2 = scan.nextInt();
System.out.println("3---");
String try3 = scan.nextLine();
System.out.println("4---");
String try4 = scan.nextLine();
When I run this code, result it :
1---
12
2---
321
3---
4---
aa
As you can see, it skipped at 3rd input. Why this is happening? I solve this problem by using new Scanners, sometimes I have 5-6 different Scanners and it looks so complicated.
Another problem is : there is an error "Resource leak: scan is never closed". I am really confused.
The problem is that by using scanner.nextInt() you only read an integer value, but not the whole line and you don't consume the newline character (\n) that is appended to the line when you press Enter.
Then, when you process reading with scanner.nextLine() you consume the newline character (\n) and from the previous row and don't read the line you want to read. In order to force it to read it, you have to add an additional input.nextLine() statement.
System.out.println("2---");
int try2 = scan.nextInt();
System.out.println("3---");
scan.nextLine(); //<-- fake statement, to move the cursor on the next line
String try3 = scan.nextLine();
Not related to the question, you have to close the Scanner, after finishing work, otherwise the compiler complains with a warning:
scan.close();
Use next API rather than nextLine as when you do nextInt you press enter and it generates number + \n and nextInt is only going to take an integer and won't take \n which in turn gets passed to next input i.e. try3 which is why it gets skipped.
Scanner scan = new Scanner(System.in);
System.out.println("1---");
int try1 = scan.nextInt();
System.out.println("2---");
int try2 = scan.nextInt();
System.out.println("3---");
String try3 = scan.next();
System.out.println("4---");
String try4 = scan.next();
Well, for the first question use
scan.next()
insted of using
scan.nextLine();
For the second question I'd recommend using try, assuming you need to close your scan as your compiler is warning: "Resource leak: scan is never closed"
Scanner scanner= null;
try {
scanner= new Scanner(System.in);
}
finally {
if(scanner!=null)
scanner.close();
}
Related
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 1 year ago.
In this program, string variable fd doesn't wait taking input.
Can someone help me with this program.
i can get the input if i use the new scanner object though.
import java.util.Scanner;
public class Strmethod {
public static void main(String[] args)
{
String ch,fd;
int s,e;
Scanner sc=new Scanner(System.in);
System.out.print("Enter A String:");
ch=sc.nextLine();
System.out.println("String is "+ch);
System.out.println("Enter Two Numbers For Substring:");
s=sc.nextInt();
e=sc.nextInt();
System.out.println("Substring:"+ch.substring(s,e));
System.out.println("Enter a Word to search:");
fd=sc.nextLine();
System.out.println(ch.contains(f));
String jn=String.join("/","hello","g","u","y","s");
System.out.println(jn);
System.out.println(ch.startsWith("H"));
System.out.println(ch.startsWith("e"));
System.out.println("Length:"+ch.length());
String newstr=ch.replace("Hello","Hey");
System.out.println("String is "+ch);
}
}
The above picture shows the input in the form of a buffer your program is using.
Explanation:
When you first use the nextLine() function it reads the whole buffer up to \n character. Now the next buffer line starts and you have used nextInt() to read an integer 3 again nextInt() to read 6 now you have pressed enter i.e. \n character that is appended to the last of the buffer.
The current position of the buffer is that already contains the \n character so when you use the nextLine() method it consumes the \n character from the buffer and takes no input from the console.
Solution:
The solution is to use the nextLine() whenever using the last nextInt() or similar functions that don't read the full line.
The final program would be
...
System.out.println("Enter Two Numbers For Substring:");
s=sc.nextInt();
e=sc.nextInt();
sc.nextLine(); // <--- to read the last \n character from the buffer
System.out.println("Substring:" + ch.substring(s,e));
System.out.println("Enter a Word to search:");
fd=sc.nextLine();
...
This is the program
public class bInputMismathcExceptionDemo {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean continueInput = true;
do {
try {
System.out.println("Enter an integer:");
int num = input.nextInt();
System.out.println("the number is " + num);
continueInput = false;
}
catch (InputMismatchException ex) {
System.out.println("Try again. (Incorrect input: an integer is required)");
}
input.nextLine();
}
while (continueInput);
}
}
I know nextInt() only read the integer not the "\n", but why should we need the input.nextLine() to read the "\n"? is it necessary?? because I think even without input.nextLine(), after it goes back to try {}, the input.nextInt() can still read the next integer I type, but in fact it is a infinite loop.
I still don't know the logic behind it, hope someone can help me.
The reason it is necessary here is because of what happens when the input fails.
For example, try removing the input.nextLine() part, run the program again, and when it asks for input, enter abc and press Return
The result will be an infinite loop. Why?
Because nextInt() will try to read the incoming input. It will see that this input is not an integer, and will throw the exception. However, the input is not cleared. It will still be abc in the buffer. So going back to the loop will cause it to try parsing the same abc over and over.
Using nextLine() will clear the buffer, so that the next input you read after an error is going to be the fresh input that's after the bad line you have entered.
but why should we need the input.nextLine() to read the "\n"? is it necessary??
Yes (actually it's very common to do that), otherwise how will you consume the remaining \n? If you don't want to use nextLine to consume the left \n, use a different scanner object (I don't recommend this):
Scanner input1 = new Scanner(System.in);
Scanner input2 = new Scanner(System.in);
input1.nextInt();
input2.nextLine();
or use nextLine to read the integer value and convert it to int later so you won't have to consume the new line character later.
Also you can use:
input.nextInt();
input.skip("\\W*").nextLine();
or
input.skip("\n").nextLine();
if you need whitespaces before line
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Scanner issue when using nextLine after nextXXX [duplicate]
Closed 7 years ago.
I tried to get inputs via scanner and in the past, I use enter to get to the next set of inputs.
For ex.
Input 1 <enter>
Input 2 <enter>
However this time, it only accepts in the same line , taking spaces as delimiter.
Scanner in = new Scanner(System.in);
int a,b,n,t;
String input_line;
String inputs[]= new String[3];
t = in.nextInt();
in.reset(); //Tried resetting Scanner to see if this works
input_line = in.nextLine();
inputs = input_line.split(" ");
for(String s:inputs)
System.out.println(s);
For instance, I expect to take the variable t in first line and then move on to the second line for input_line scanning. But if I hit enter after entering t, the program ends.
What am I missing here?
(Merging with another question was suggested but , let me explain, the Scanner does not skip any inputs).
Without any testing I would think you would need something like this
Scanner in = new Scanner(System.in);
int a,b,n,t;
String input_line;
String[] input_numbers = new String[3];
t = in.nextInt();
in.nextLine();
input_line = in.nextLine();
while(!input_line.equals("")){
input_numbers = input_line.split(" ");
// do what you want with numbers here for instance parse to make each string variable into int or create new scanner to do so
input_line = in.nextLine();
}
}
This is the program
public class bInputMismathcExceptionDemo {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
boolean continueInput = true;
do {
try {
System.out.println("Enter an integer:");
int num = input.nextInt();
System.out.println("the number is " + num);
continueInput = false;
}
catch (InputMismatchException ex) {
System.out.println("Try again. (Incorrect input: an integer is required)");
}
input.nextLine();
}
while (continueInput);
}
}
I know nextInt() only read the integer not the "\n", but why should we need the input.nextLine() to read the "\n"? is it necessary?? because I think even without input.nextLine(), after it goes back to try {}, the input.nextInt() can still read the next integer I type, but in fact it is a infinite loop.
I still don't know the logic behind it, hope someone can help me.
The reason it is necessary here is because of what happens when the input fails.
For example, try removing the input.nextLine() part, run the program again, and when it asks for input, enter abc and press Return
The result will be an infinite loop. Why?
Because nextInt() will try to read the incoming input. It will see that this input is not an integer, and will throw the exception. However, the input is not cleared. It will still be abc in the buffer. So going back to the loop will cause it to try parsing the same abc over and over.
Using nextLine() will clear the buffer, so that the next input you read after an error is going to be the fresh input that's after the bad line you have entered.
but why should we need the input.nextLine() to read the "\n"? is it necessary??
Yes (actually it's very common to do that), otherwise how will you consume the remaining \n? If you don't want to use nextLine to consume the left \n, use a different scanner object (I don't recommend this):
Scanner input1 = new Scanner(System.in);
Scanner input2 = new Scanner(System.in);
input1.nextInt();
input2.nextLine();
or use nextLine to read the integer value and convert it to int later so you won't have to consume the new line character later.
Also you can use:
input.nextInt();
input.skip("\\W*").nextLine();
or
input.skip("\n").nextLine();
if you need whitespaces before line
This question already has answers here:
Scanner is skipping nextLine() after using next() or nextFoo()?
(24 answers)
Closed 6 years ago.
I am using Java's Scanner to read user input. If I use nextLine only once, it works OK. With two nextLine first one doesnt wait for user to enter the string(second does).
Output:
X: Y: (wait for input)
My code
System.out.print("X: ");
x = scanner.nextLine();
System.out.print("Y: ");
y = scanner.nextLine();
Any ideas why could this happen? Thanks
It's possible that you are calling a method like nextInt() before. Thus a program like this:
Scanner scanner = new Scanner(System.in);
int pos = scanner.nextInt();
System.out.print("X: ");
String x = scanner.nextLine();
System.out.print("Y: ");
String y = scanner.nextLine();
demonstatres the behavior you're seeing.
The problem is that nextInt() does not consume the '\n', so the next call to nextLine() consumes it and then it's waiting to read the input for y.
You need to consume the '\n' before calling nextLine().
System.out.print("X: ");
scanner.nextLine(); //throw away the \n not consumed by nextInt()
x = scanner.nextLine();
System.out.print("Y: ");
y = scanner.nextLine();
(actually a better way would be to call directly nextLine() after nextInt()).