jetbrick.util.WildcharUtils Maven / Gradle / Ivy
/**
* Copyright 2013-2016 Guoqiang Chen, Shanghai, China. All rights reserved.
*
* Author: Guoqiang Chen
* Email: [email protected]
* WebURL: https://github.com/subchen
*
* Licensed under the Apache License, Version 2.0 (the "License");
* you may not use this file except in compliance with the License.
* You may obtain a copy of the License at
*
* http://www.apache.org/licenses/LICENSE-2.0
*
* Unless required by applicable law or agreed to in writing, software
* distributed under the License is distributed on an "AS IS" BASIS,
* WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
* See the License for the specific language governing permissions and
* limitations under the License.
*/
package jetbrick.util;
/**
* Checks whether a string matches a given wildcard pattern.
* Possible patterns allow to match single characters ('?') or any count of
* characters ('*'). Wildcard characters can be escaped (by an '\').
*
* This method uses recursive matching, as in linux or windows. regexp works the same.
* This method is very fast, comparing to similar implementations.
*/
public class WildcharUtils {
/**
* Checks whether a string matches a given wildcard pattern.
*
* @param string input string
* @param pattern pattern to match
* @return true if string matches the pattern, otherwise false
*/
public static boolean match(String string, String pattern) {
if (string.equals(pattern)) { // speed-up
return true;
}
return match(string, pattern, 0, 0);
}
/**
* Matches string to at least one pattern.
* Returns index of matched pattern, or -1 otherwise.
* @see #match(String, String)
*/
public static int matchOne(String src, String[] patterns) {
for (int i = 0; i < patterns.length; i++) {
if (match(src, patterns[i]) == true) {
return i;
}
}
return -1;
}
/**
* Internal matching recursive function.
*/
private static boolean match(String string, String pattern, int sNdx, int pNdx) {
int pLen = pattern.length();
if (pLen == 1) {
if (pattern.charAt(0) == '*') { // speed-up
return true;
}
}
int sLen = string.length();
boolean nextIsNotWildcard = false;
while (true) {
// check if end of string and/or pattern occurred
if ((sNdx >= sLen) == true) { // end of string still may have pending '*' in pattern
while (pNdx < pLen && pattern.charAt(pNdx) == '*') {
pNdx++;
}
return pNdx >= pLen;
}
if (pNdx >= pLen) { // end of pattern, but not end of the string
return false;
}
char p = pattern.charAt(pNdx); // pattern char
// perform logic
if (nextIsNotWildcard == false) {
if (p == '\\') {
pNdx++;
nextIsNotWildcard = true;
continue;
}
if (p == '?') {
sNdx++;
pNdx++;
continue;
}
if (p == '*') {
char pNext = 0; // next pattern char
if (pNdx + 1 < pLen) {
pNext = pattern.charAt(pNdx + 1);
}
if (pNext == '*') { // double '*' have the same effect as one '*'
pNdx++;
continue;
}
int i;
pNdx++;
// find recursively if there is any substring from the end of the
// line that matches the rest of the pattern !!!
for (i = string.length(); i >= sNdx; i--) {
if (match(string, pattern, i, pNdx) == true) {
return true;
}
}
return false;
}
} else {
nextIsNotWildcard = false;
}
// check if pattern char and string char are equals
if (p != string.charAt(sNdx)) {
return false;
}
// everything matches for now, continue
sNdx++;
pNdx++;
}
}
}