java.lang.Object
java.lang.Throwable
java.lang.Exception
java.lang.RuntimeException
java.lang.IllegalArgumentException
java.util.regex.PatternSyntaxException
- All Implemented Interfaces:
- Serializable
- See Also:
- Top Examples, Source Code
public String getDescription()
- Geek's Notes:
- Description Add your codes or notes Search More Java Examples
[1927]PatternSyntaxException
By Anonymous on 2007/08/15 11:29:57 Rate
A PatternSyntaxException is an unchecked exception that indicates a syntax error in a regular expression pattern.
public int getIndex()
- Geek's Notes:
- Description Add your codes or notes Search More Java Examples
public String getMessage()
- See Also:
- Throwable
- Geek's Notes:
- Description Add your codes or notes Search More Java Examples
[1928]PatternSyntaxException.getMessage()
By Anonymous on 2007/08/15 11:30:35 Rate
Returns a multi-line string containing the description of the syntax error and its index, the erroneous regular-expression pattern, and a visual indication of the error index within the pattern.
public String getPattern()
- Geek's Notes:
- Description Add your codes or notes Search More Java Examples
public PatternSyntaxException(String desc,
String regex,
int index)- Geek's Notes:
- Description Add your codes or notes Search More Java Examples
[1929]How to use PatternSyntaxException
By Anonymous on 2007/08/15 11:31:41 Rate
import java.io.Console;
import java.util.regex.Pattern;
import java.util.regex.Matcher;
import java.util.regex.PatternSyntaxException;
public class RegexTestHarness2 {
public static void main ( String [ ] args ) {
Pattern pattern = null;
Matcher matcher = null;
Console console = System.console ( ) ;
if ( console == null ) {
System.err.println ( "No console." ) ;
System.exit ( 1 ) ;
}
while ( true ) {
try {
pattern =
Pattern.compile ( console.readLine ( "%nEnter your regex: " ) ) ;
matcher =
pattern.matcher ( console.readLine ( "Enter input string to search: " ) ) ;
}
catch ( PatternSyntaxException pse ) {
console.format ( "There is a problem with the regular expression!%n" ) ;
console.format ( "The pattern in question is: %s%n",pse.getPattern ( ) ) ;
console.format ( "The description is: %s%n",pse.getDescription ( ) ) ;
console.format ( "The message is: %s%n",pse.getMessage ( ) ) ;
console.format ( "The index is: %s%n",pse.getIndex ( ) ) ;
System.exit ( 0 ) ;
}
boolean found = false;
while ( matcher.find ( ) ) {
console.format ( "I found the text \"%s\" starting at " +
"index %d and ending at index %d.%n",
matcher.group ( ) , matcher.start ( ) , matcher.end ( ) ) ;
found = true;
}
if ( !found ) {
console.format ( "No match found.%n" ) ;
}
}
}
}
To run this test, enter ?i ) foo as the regular expression. This mistake is a common scenario in which the programmer has forgotten the opening parenthesis in the embedded flag expression ( ?i ) . Doing so will produce the following results:
Enter your regex: ?i )
There is a problem with the regular expression!
The pattern in question is: ?i )
The description is: Dangling meta character '?'
The message is: Dangling meta character '?' near index 0
?i )
^
The index is: 0
From this output, we can see that the syntax error is a dangling metacharacter ( the question mark ) at index 0. A missing opening parenthesis is the culprit. 
Java API By Example, From Geeks To Geeks.