((\d{1,2})/(\d{1,2})/(\d{2,4}))
Is there a way to retrieve a list of all the capture groups with the Pattern
object. I debugged the object and all it says is how many groups there are (5).
I need to retrieve a list of the following capture groups.
Example of output:
0 ((\d{1,2})/(\d{1,2})/(\d{2,4}))
1 (\d{2})/(\d{2})/(\d{4})
2 \d{2}
3 \d{2}
4 \d{4}
Update:
I am not necessarily asking if a regular expression exists, but that would be most favorable. So far I have created a rudimentary parser (I do not check for most out-of-bounds conditions) that only matches inner-most groups. I would like to know if there is a way to hold reference to already-visited parenthesis. I would probably have to implement a tree structure?
import java.util.ArrayList;
import java.util.List;
import java.util.regex.Pattern;
import java.util.regex.PatternSyntaxException;
public class App {
public final char S = '(';
public final char E = ')';
public final char X = '\\';
String errorMessage = "Malformed expression: ";
/**
* Actual Output:
* Groups: [(//), (\d{1,2}), (\d{1,2}), (\d{2,4})]
* Expected Output:
* Groups: [\\b((\\d{1,2})/(\\d{1,2})/(\\d{2,4}))\\b, ((\\d{1,2})/(\\d{1,2})/(\\d{2,4})), (\d{1,2}), (\d{1,2}), (\d{2,4})]
*/
public App() {
String expression = "\\b((\\d{1,2})/(\\d{1,2})/(\\d{2,4}))\\b";
String output = "";
if (isValidExpression(expression)) {
List<String> groups = findGroups(expression);
output = "Groups: " + groups;
} else {
output = errorMessage;
}
System.out.println(output);
}
public List<String> findGroups(String expression) {
List<String> groups = new ArrayList<>();
int[] pos;
int start;
int end;
String sub;
boolean done = false;
while (expression.length() > 0 && !done) {
pos = scanString(expression);
start = pos[0];
end = pos[1];
if (start == -1 || end == -1) {
done = true;
continue;
}
sub = expression.substring(start, end);
expression = splice(expression, start, end);
groups.add(0, sub);
}
return groups;
}
public int[] scanString(String str) {
int[] range = new int[] { -1, -1 };
int min = 0;
int max = str.length() - 1;
int start = min;
int end = max;
char curr;
while (start <= max) {
curr = str.charAt(start);
if (curr == S) {
range[0] = start;
}
start++;
}
end = range[0];
while (end > -1 && end <= max) {
curr = str.charAt(end);
if (curr == E) {
range[1] = end + 1;
break;
}
end++;
}
return range;
}
public String splice(String str, int start, int end) {
if (str == null || str.length() < 1)
return "";
if (start < 0 || end > str.length()) {
System.err.println("Positions out of bounds.");
return str;
}
if (start >= end) {
System.err.println("Start must not exceed end.");
return str;
}
String first = str.substring(0, start);
String last = str.substring(end, str.length());
return first + last;
}
public boolean isValidExpression(String expression) {
try {
Pattern.compile(expression);
} catch (PatternSyntaxException e) {
errorMessage += e.getMessage();
return false;
}
return true;
}
public static void main(String[] args) {
new App();
}
}
Here is my solution ... I simply provided a regex of the regex as @SotiriosDelimanolis commented out.
public static void printGroups() {
String sp = "((\\(\\\\d\\{1,2\\}\\))\\/(\\(\\\\d\\{1,2\\}\\))\\/(\\(\\\\d\\{2,4\\}\\)))";
Pattern p = Pattern.compile(sp);
Matcher m = p.matcher("(\\d{1,2})/(\\d{1,2})/(\\d{2,4})");
if (m.matches())
for (int i = 0; i <= m.groupCount(); i++)
System.out.println(m.group(i));
}
Pay attention that you cannot remove the if
-statement because in order to use the group
method you should call the matches
method first (I didn't know it!). See this link as a reference about it.
Hope this is what you were asking for ...
来源:https://stackoverflow.com/questions/19790210/java-pattern-print-capturing-groups