java.util.regex.PatternSyntaxException: Dangling meta character '*' near index 0 *welcome* ^ at java.util.regex.Pattern.error(Pattern.java:1528) at java.util.regex.Pattern.sequence(Pattern.java:1645) at java.util.regex.Pattern.expr(Pattern.java:1545) at java.util.regex.Pattern.compile(Pattern.java:1279) at java.util.regex.Pattern.<init>(Pattern.java:1035) at java.util.regex.Pattern.compile(Pattern.java:779) at com.spns.history.TestPattern.main(TestPattern.java:21) Exception in thread "main"
Java Developer, Thailand
posted 11 years ago
The '*' character by itself does not mean anything in regular expressions. If it follows a character, character class, or group, then it means 'zero or more' of the thing it follows.
If your intention with "*welcome" is "zero or more of any character, welcome" then the expression you want is: ".*welcome"; The '.' character will match any character, and the '*' works as stated above.
Reasonable people adapt themselves to the world. Unreasonable people attempt to adapt the world to themselves. All progress, therefore, depends on unreasonable people.