java通配符匹配实现类，java通配符匹配,java通配符匹配的实现

java通配符匹配实现类，java通配符匹配,java通配符匹配的实现

java通配符匹配的实现类，可以计算字符串是否匹配某个通配符，可以计算是否匹配一组通配符中的一个。

// Copyright (c) 2003-2009, Jodd Team (jodd.org). All Rights Reserved./** * 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 '\'). * <p> * 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 Wildcard {  /**   * Checks whether a string matches a given wildcard pattern.   *   * @param string  input string   * @param pattern pattern to match   * @return      <code>true</code> if string matches the pattern, otherwise <code>false</code>   */  public static boolean match(String string, String pattern) {    return match(string, pattern, 0, 0);  }  /**   * Checks if two strings are equals or if they {@link #match(String, String)}.   * Useful for cases when matching a lot of equal strings and speed is important.   */  public static boolean equalsOrMatch(String string, String pattern) {    if (string.equals(pattern) == true) {      return true;    }    return match(string, pattern, 0, 0);  }  /**   * Internal matching recursive function.   */  private static boolean match(String string, String pattern, int stringStartNdx, int patternStartNdx) {    int pNdx = patternStartNdx;    int sNdx = stringStartNdx;    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++;    }  }  // ---------------------------------------------------------------- utilities  /**   * Matches string to at least one pattern.   * Returns index of matched pattern, or <code>-1</code> 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;  }}