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问题:
I am using the Scanner methods nextInt() and nextLine() for reading input.
It looks like this:
System.out.println("Enter numerical value");
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string");
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:
Enter numerical value
3 // This is my input
Enter 1st string // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string // ...and this line is executed and waits for my input
I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.
回答1:
That's because the Scanner.nextInt method does not read the newline character in your input created by hitting "Enter," and so the call to Scanner.nextLine returns after reading that newline.
You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).
Workaround:
Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline
int option = input.nextInt();
input.nextLine(); // Consume newline left-over
String str1 = input.nextLine();
Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.
int option = 0;
try {
option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
e.printStackTrace();
}
String str1 = input.nextLine();
回答2:
The problem is with the input.nextInt() method - it only reads the int value. So when you continue reading with input.nextLine() you receive the "\n" Enter key. So to skip this you have to add the input.nextLine(). Hope this should be clear now.
Try it like that:
System.out.print("Insert a number: ");
int number = input.nextInt();
input.nextLine(); // This line you have to add (It consumes the \n character)
System.out.print("Text1: ");
String text1 = input.nextLine();
System.out.print("Text2: ");
String text2 = input.nextLine();
回答3:
It's because when you enter a number then press Enter, input.nextInt() consumes only the number, not the "end of line". When input.nextLine() executes, it consumes the "end of line" still in the buffer from the first input.
Instead, use input.nextLine() immediately after input.nextInt()
回答4:
There seem to be many questions about this issue with java.util.Scanner. I think a more readable/idiomatic solution would be to call scanner.skip("[\r\n]+") to drop any newline characters after calling nextInt().
EDIT: as @PatrickParker noted below, this will cause an infinite loop if user inputs any whitespace after the number. See their answer for a better pattern to use with skip: https://stackoverflow.com/a/42471816/143585
回答5:
It does that because input.nextInt(); doesn't capture the newline. you could do like the others proposed by adding an input.nextLine(); underneath.
Alternatively you can do it C# style and parse a nextLine to an integer like so:
int number = Integer.parseInt(input.nextLine());
Doing this works just as well, and it saves you a line of code.
回答6:
Things you need to know:
text which represents few lines also contains non-printable characters between lines (we call them line separators) like
- carriage return (CR - in String literals represented as
"\r")
- line feed (LF - in String literals represented as
"\n")
when you are reading data from the console, it allows the user to type his response and when he is done he needs to somehow confirm that fact. To do so, the user is required to press "enter"/"return" key on the keyboard.
What is important is that this key beside ensuring placing user data to standard input (represented by System.in which is read by Scanner) also sends OS dependant line separators (like for Windows \r\n) after it.
So when you are asking the user for value like age, and user types 42 and presses enter, standard input will contain "42\r\n".
Problem
Scanner#nextInt (and other Scanner#nextType methods) doesn't allow Scanner to consume these line separators. It will read them from System.in (how else Scanner would know that there are no more digits from the user which represent age value than facing whitespace?) which will remove them from standard input, but it will also cache those line separators internally. What we need to remember, is that all of the Scanner methods are always scanning starting from the cached text.
Now Scanner#nextLine() simply collects and returns all characters until it finds line separators (or end of stream). But since line separators after reading the number from the console are found immediately in Scanner's cache, it returns empty String, meaning that Scanner was not able to find any character before those line separators (or end of stream).
BTW nextLine also consumes those line separators.
Solution
So when you want to ask for number and then for entire line while avoiding that empty string as result of nextLine, either
- consume line separator left by
nextInt from Scanners cache by
- calling
nextLine,
- or by calling
skip("\\R") to let Scanner skip part matched by \R which represents line separator (more info about \R: https://stackoverflow.com/a/31060125)
- don't use
nextInt (nor next, or any nextTYPE methods) at all. Instead read entire data line-by-line using nextLine and parse numbers from each line (assuming one line contains only one number) to proper type like int via Integer.parseInt.
BTW: Scanner#nextType methods can skip delimiters (by default all whitespaces like tabs, line separators) including those cached by scanner, until they will find next non-delimiter value (token). Thanks to that for input like "42\r\n\r\n321\r\n\r\n\r\nfoobar" code
int num1 = sc.nextInt();
int num2 = sc.nextInt();
String name = sc.next();
will be able to properly assign num1=42 num2=321 name=foobar.
回答7:
Instead of input.nextLine() use input.next(), that should solve the problem.
Modified code:
public static Scanner input = new Scanner(System.in);
public static void main(String[] args)
{
System.out.print("Insert a number: ");
int number = input.nextInt();
System.out.print("Text1: ");
String text1 = input.next();
System.out.print("Text2: ");
String text2 = input.next();
}
回答8:
If you want to read both strings and ints, a solution is to use two Scanners:
Scanner stringScanner = new Scanner(System.in);
Scanner intScanner = new Scanner(System.in);
intScanner.nextInt();
String s = stringScanner.nextLine(); // unaffected by previous nextInt()
System.out.println(s);
intScanner.close();
stringScanner.close();
回答9:
In order to avoid the issue, use nextLine(); immediately after nextInt(); as it helps in clearing out the buffer. When you press ENTER the nextInt(); does not capture the new line and hence, skips the Scanner code later.
Scanner scanner = new Scanner(System.in);
int option = scanner.nextInt();
scanner.nextLine(); //clearing the buffer
回答10:
If you want to scan input fast without getting confused into Scanner class nextLine() method , Use Custom Input Scanner for it .
Code :
class ScanReader {
/**
* @author Nikunj Khokhar
*/
private byte[] buf = new byte[4 * 1024];
private int index;
private BufferedInputStream in;
private int total;
public ScanReader(InputStream inputStream) {
in = new BufferedInputStream(inputStream);
}
private int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) return -1;
}
return buf[index++];
}
public char scanChar(){
int c=scan();
while (isWhiteSpace(c))c=scan();
return (char)c;
}
public int scanInt() throws IOException {
int integer = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = scan();
}
}
return neg * integer;
}
public String scanString() throws IOException {
int c = scan();
while (isWhiteSpace(c)) c = scan();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = scan();
} while (!isWhiteSpace(c));
return res.toString();
}
private boolean isWhiteSpace(int n) {
if (n == ' ' || n == '\n' || n == '\r' || n == '\t' || n == -1) return true;
else return false;
}
public long scanLong() throws IOException {
long integer = 0;
int n = scan();
while (isWhiteSpace(n)) n = scan();
int neg = 1;
if (n == '-') {
neg = -1;
n = scan();
}
while (!isWhiteSpace(n)) {
if (n >= '0' && n <= '9') {
integer *= 10;
integer += n - '0';
n = scan();
}
}
return neg * integer;
}
public void scanLong(long[] A) throws IOException {
for (int i = 0; i < A.length; i++) A[i] = scanLong();
}
public void scanInt(int[] A) throws IOException {
for (int i = 0; i < A.length; i++) A[i] = scanInt();
}
public double scanDouble() throws IOException {
int c = scan();
while (isWhiteSpace(c)) c = scan();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = scan();
}
double res = 0;
while (!isWhiteSpace(c) && c != '.') {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, scanInt());
}
res *= 10;
res += c - '0';
c = scan();
}
if (c == '.') {
c = scan();
double m = 1;
while (!isWhiteSpace(c)) {
if (c == 'e' || c == 'E') {
return res * Math.pow(10, scanInt());
}
m /= 10;
res += (c - '0') * m;
c = scan();
}
}
return res * sgn;
}
}
Advantages :
- Scans Input faster than BufferReader
- Reduces Time Complexity
- Flushes Buffer for every next input
Methods :
- scanChar() - scan single character
- scanInt() - scan Integer value
- scanLong() - scan Long value
- scanString() - scan String value
- scanDouble() - scan Double value
- scanInt(int[] array) - scans complete Array(Integer)
- scanLong(long[] array) - scans complete Array(Long)
Usage :
- Copy the Given Code below your java code.
- Initialise Object for Given Class
ScanReader sc = new ScanReader(System.in);
3. Import necessary Classes :
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
4. Throw IOException from your main method to handle Exception
5. Use Provided Methods.
6. Enjoy
Example :
import java.io.BufferedInputStream;
import java.io.IOException;
import java.io.InputStream;
class Main{
public static void main(String... as) throws IOException{
ScanReader sc = new ScanReader(System.in);
int a=sc.scanInt();
System.out.println(a);
}
}
class ScanReader....
回答11:
sc.nextLine() is better as compared to parsing the input.
Because performance wise it will be good.
回答12:
I guess I'm pretty late to the party..
As previously stated, calling input.nextLine() after getting your int value will solve your problem. The reason why your code didn't work was because there was nothing else to store from your input (where you inputted the int) into string1. I'll just shed a little more light to the entire topic.
Consider nextLine() as the odd one out among the nextFoo() methods in the Scanner class. Let's take a quick example.. Let's say we have two lines of code like the ones below:
int firstNumber = input.nextInt();
int secondNumber = input.nextInt();
If we input the value below (as a single line of input)
54 234
The value of our firstNumber and secondNumber variable become 54 and 234 respectively. The reason why this works this way is because a new line feed (i.e \n) IS NOT automatically generated when the nextInt() method takes in the values. It simply takes the "next int" and moves on. This is the same for the rest of the nextFoo() methods except nextLine().
nextLine() generates a new line feed immediately after taking a value; this is what @RohitJain means by saying the new line feed is "consumed".
Lastly, the next() method simply takes the nearest String without generating a new line; this makes this the preferential method for taking separate Strings within the same single line.
I hope this helps.. Merry coding!
回答13:
Use 2 scanner objects instead of one
Scanner input = new Scanner(System.in);
System.out.println("Enter numerical value");
int option;
Scanner input2 = new Scanner(System.in);
option = input2.nextInt();
回答14:
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int i = scan.nextInt();
scan.nextLine();
double d = scan.nextDouble();
scan.nextLine();
String s = scan.nextLine();
System.out.println("String: " + s);
System.out.println("Double: " + d);
System.out.println("Int: " + i);
}
回答15:
if I expect a non-empty input
public static Function<Scanner,String> scanLine = (scan -> {
String s = scan.nextLine();
return( s.length() == 0 ? scan.nextLine() : s );
});
used in above example:
System.out.println("Enter numerical value");
int option = input.nextInt(); // read numerical value from input
System.out.println("Enter 1st string");
String string1 = scanLine.apply( input ); // read 1st string
System.out.println("Enter 2nd string");
String string2 = scanLine.apply( input ); // read 2nd string
回答16:
As nextXXX() methods don't read newline, except nextLine(). We can skip the newline after reading any non-string value (int in this case) by using scanner.skip() as below:
Scanner sc = new Scanner(System.in);
int x = sc.nextInt();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(x);
double y = sc.nextDouble();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(y);
char z = sc.next().charAt(0);
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(z);
String hello = sc.nextLine();
System.out.println(hello);
float tt = sc.nextFloat();
sc.skip("(\r\n|[\n\r\u2028\u2029\u0085])?");
System.out.println(tt);
回答17:
Why not use a new Scanner for every reading? Like below. With this approach you will not confront your problem.
int i = new Scanner(System.in).nextInt();
