Consider the following Pattern :-
aba
And the foll. source string :-
abababbbaba
01234567890 //Index Positions
Using Pattern and Matcher classes from java.util.regex package, finds this pattern only two times since regex does not consider already consumed characters.
What if I want to reuse a part of already consumed characters. That is, I want 3 matches here, one at position 0, one at 2 (which is ignored previously), and one at 8.
How do I do it??
I think you can use the indexOf () for something like that.
String str = "abababbbaba";
String substr = "aba";
int location = 0;
while ((location = str.indexOf(substr, location)) >= 0)
{
System.out.println(location);
location++;
}
Prints:
0, 2 and 8
You can use a look ahead for that. Now what you have is the first position in group(1)
and the second match in group(2)
. Both making each String of length 3 in the sentence you are searching in.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Question8968432 {
public static void main(String args[]) {
final String needle = "aba";
final String sentence = "abababbbaba";
final Matcher m = Pattern.compile("(.)(?=(..))").matcher(sentence);
while (m.find()) {
final String match = m.group(1) + m.group(2);
final String hint = String.format("%s[%s]%s",
sentence.substring(0, m.start()), match,
sentence.substring(m.start() + match.length()));
if (match.equals(needle)) {
System.out.printf("Found %s starting at %d: %s\n",
match, m.start(), hint);
}
}
}
}
Output:
Found aba starting at 0: [aba]babbbaba
Found aba starting at 2: ab[aba]bbbaba
Found aba starting at 8: abababbb[aba]
You can skip the final String hint
part, this is just to show you what it matches and where.
如果您可以更改正则表达式,则可以简单地使用以下命令:
a(?=ba)
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