String delimiter in string split method

Question

I have following data   1  1  Abdul-Jabbar  Karim  1996  1974   I want to delimit the tokens   Here the delimiter is        My delimiter setter is   public void setDelimiter String delimiter        char   c   delimiter toCharArray        this delimiter                 c 0           c 1              System out println  Delimiter string is      this delimiter       However    String   tokens   line split delimiter     is not giving the required result

User · Answer

Double quotes are interpreted as literals in regex; they are not special characters. You are trying to match a literal "||".

Just use Pattern.quote(delimiter):

As requested, here's a line of code (same as Sanjay's)

final String[] tokens = line.split(Pattern.quote(delimiter));

If that doesn't work, you're not passing in the correct delimiter.

User · Answer

StringTokenizer st   new StringTokenizer  1  1  Abdul-Jabbar  Karim  1996  1974                       while st hasMoreTokens          System out println st nextElement         Answer will print     1 1 Abdul-Jabbar Karim 1996 1974

User · Answer

String   splitArray   subjectString split              You use a function   public String   stringSplit String string        String   splitArray   string split                return splitArray

User · Answer

There is no need to set the delimiter by breaking it up in pieces like you have done   Here is a complete program you can compile and run   import java util Arrays  public class SplitExample       public static final String PLAYER    1  1  Abdul-Jabbar  Karim  1996  1974       public static void main String   args            String   data   PLAYER split                    System out println Arrays toString data              If you want to use split with a pattern  you can use Pattern compile or Pattern quote   To see compile and quote in action  here is an example using all three approaches   import java util Arrays  import java util regex Pattern  public class SplitExample       public static final String PLAYER    1  1  Abdul-Jabbar  Karim  1996  1974       public static void main String   args            String   data   PLAYER split                    System out println Arrays toString data             Pattern pattern   Pattern compile                    data   pattern split PLAYER           System out println Arrays toString data             pattern   Pattern compile Pattern quote                 data   pattern split PLAYER           System out println Arrays toString data              The use of patterns is recommended if you are going to split often using the same pattern   BTW the output is    1  1  Abdul-Jabbar  Karim  1996  1974   1  1  Abdul-Jabbar  Karim  1996  1974   1  1  Abdul-Jabbar  Karim  1996  1974

User · Answer

The problem is because you are adding quotes to your delimiter  It should be removed  and it will work fine   public void setDelimiter String delimiter        char   c   delimiter toCharArray        this delimiter          c 0           c 1       System out println  Delimiter string is      this delimiter

User · Answer

String   strArray  str split Pattern quote           where   str    1  1  Abdul-Jabbar  Karim  1996  1974   Pattern quote       will ignore the special character   split function will split the string for every occurrence of     strArray will have the array of string that is delimited by

User · Answer

Use the Pattern quote   method for escaping     Try   final String   tokens   myString split Pattern quote           This is required because   is an alternation character and hence gains a special meaning when passed to split call  basically the argument to split is a regular expression in string form

User · Answer

There is something wrong in your setDelimiter   function   You don t want to double quote the delimiters  do you   public void setDelimiter String delimiter        char   c   delimiter toCharArray        this delimiter          c 0           c 1       System out println  Delimiter string is      this delimiter       However  as other users have said  it s better to use the Pattern quote   method to escape your delimiter if your requirements permit

User · Answer

Pipe     is a special character in regex  to escape it  you need to prefix it with backslash      But in java  backslash is also an escape character  so again you need to escape it with another backslash  So your regex should be        e g   String   tokens   myString split

User · Answer

Split uses regex  and the pipe char   has special meaning in regex  so you need to escape it  There are a few ways to do this  but here s the simplest   String   tokens   line split

[java] String delimiter in string.split method

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