Scanner is skipping nextLine after using next or nextFoo

Question

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

User · Answer

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

User · Answer

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

User · Answer

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  gt   total                index   0              total   in read buf               if  total  lt   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  gt    0   amp  amp  n  lt    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  gt    0   amp  amp  n  lt    9                     integer    10                  integer    n -  0                   n   scan                                    return neg   integer             public void scanLong long   A  throws IOException           for  int i   0  i  lt  A length  i    A i    scanLong               public void scanInt int   A  throws IOException           for  int i   0  i  lt  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   amp  amp  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

User · Answer

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

User · Answer

TL DR Use scanner skip  quot   R quot    since skip uses regex where  R represents line separators  before each scanner newLine   call  which is executed after   scanner next   scanner next TYPE    method  like scanner nextInt     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  quot  r quot    line feed  LF - in String literals represented as  quot  n quot    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  quot enter quot   quot return quot  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  quot 42 r n quot   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 IMO more readable way would be by calling skip  quot   R quot   or skip  quot  r n  r  n quot   to let Scanner skip part matched by 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  quot 42 r n r n321 r n r n r nfoobar quot  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

User · Answer

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

User · Answer

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

User · Answer

To resolve this problem just make a  scan nextLine    where scan is an instance of the Scanner object  For example  I am using a simple HackerRank Problem for the explanation  package com company  import java util Scanner   public class hackerrank   public static void main String   args        Scanner scan   new Scanner System in       int i   scan nextInt        double d   scan nextDouble        scan nextLine       This line shall stop the skipping the nextLine        String s   scan nextLine        scan close              Write your code here       System out println  quot String   quot    s       System out println  quot Double   quot    d       System out println  quot Int   quot    i

User · Answer

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

User · Answer

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

User · Answer

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  quot  n quot  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  quot Insert a number   quot    int number   input nextInt    input nextLine       This line you have to add  It consumes the  n character  System out print  quot Text1   quot    String text1   input nextLine    System out print  quot Text2   quot    String text2   input nextLine

User · Answer

sc nextLine   is better as compared to parsing the input  Because performance wise it will be good

User · Answer

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

User · Answer

if I expect a non-empty input  public static Function lt Scanner String gt  scanLine    scan - gt        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

User · Answer

In one of my usecase  I had the scenario of reading a string value preceded by a couple of integer values  I had to use a  for   while loop  to read the values  And none of the above suggestions worked in this case    Using input next   instead of input nextLine   fixed the issue  Hope this might be helpful for those dealing with similar scenario

User · Answer

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

User · Answer

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

User · Answer

Use this code it will fix your problem   System out println  Enter numerical value        int option  option   input nextInt       Read numerical value from input input nextLine    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

User · Answer

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

User · Answer

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

User · Answer

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

[java] Scanner is skipping nextLine() after using next() or nextFoo()?

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