Question

是否有用于在Java中执行“glob”类型匹配的标准（最好是Apache Commons或类似的非病毒）库？当我必须在Perl中执行类似操作时，我只是将所有“.”更改为“\.”，将“*”更改为“.*”和“ ?“to”.“等等，但我想知道是否有人为我做过这项工作。

类似的问题：Create regex from glob expression

Answer 1

还计划在Java 7中实现>的Globbing。。

请参阅FileSystem.getPathMatcher(String)和the "Finding Files" tutorial。

Answer 2

没有任何内置功能，但将类似glob的内容转换为正则表达式非常简单：

public static String createRegexFromGlob(String glob)
{
    String out = "^";
    for(int i = 0; i < glob.length(); ++i)
    {
        final char c = glob.charAt(i);
        switch(c)
        {
        case '*': out += ".*"; break;
        case '?': out += '.'; break;
        case '.': out += "\\."; break;
        case '\\': out += "\\\\"; break;
        default: out += c;
        }
    }
    out += '$';
    return out;
}

这适用于我，但我不确定它是否涵盖了“标准”，如果有的话：）

Paul Tomblin的更新：我找到了一个执行全局转换的perl程序，并将其调整为Java我最终得到：

    private String convertGlobToRegEx(String line)
    {
    LOG.info("got line [" + line + "]");
    line = line.trim();
    int strLen = line.length();
    StringBuilder sb = new StringBuilder(strLen);
    // Remove beginning and ending * globs because they're useless
    if (line.startsWith("*"))
    {
        line = line.substring(1);
        strLen--;
    }
    if (line.endsWith("*"))
    {
        line = line.substring(0, strLen-1);
        strLen--;
    }
    boolean escaping = false;
    int inCurlies = 0;
    for (char currentChar : line.toCharArray())
    {
        switch (currentChar)
        {
        case '*':
            if (escaping)
                sb.append("\\*");
            else
                sb.append(".*");
            escaping = false;
            break;
        case '?':
            if (escaping)
                sb.append("\\?");
            else
                sb.append('.');
            escaping = false;
            break;
        case '.':
        case '(':
        case ')':
        case '+':
        case '|':
        case '^':
        case '$':
        case '@':
        case '%':
            sb.append('\\');
            sb.append(currentChar);
            escaping = false;
            break;
        case '\\':
            if (escaping)
            {
                sb.append("\\\\");
                escaping = false;
            }
            else
                escaping = true;
            break;
        case '{':
            if (escaping)
            {
                sb.append("\\{");
            }
            else
            {
                sb.append('(');
                inCurlies++;
            }
            escaping = false;
            break;
        case '}':
            if (inCurlies > 0 && !escaping)
            {
                sb.append(')');
                inCurlies--;
            }
            else if (escaping)
                sb.append("\\}");
            else
                sb.append("}");
            escaping = false;
            break;
        case ',':
            if (inCurlies > 0 && !escaping)
            {
                sb.append('|');
            }
            else if (escaping)
                sb.append("\\,");
            else
                sb.append(",");
            break;
        default:
            escaping = false;
            sb.append(currentChar);
        }
    }
    return sb.toString();
}

我正在编辑这个答案而不是自己创作，因为这个答案让我走上正轨。

Answer 3

感谢大家的贡献。我写了比以往任何一个答案更全面的转换：

/**
 * Converts a standard POSIX Shell globbing pattern into a regular expression
 * pattern. The result can be used with the standard {@link java.util.regex} API to
 * recognize strings which match the glob pattern.
 * <p/>
 * See also, the POSIX Shell language:
 * http://pubs.opengroup.org/onlinepubs/009695399/utilities/xcu_chap02.html#tag_02_13_01
 * 
 * @param pattern A glob pattern.
 * @return A regex pattern to recognize the given glob pattern.
 */
public static final String convertGlobToRegex(String pattern) {
    StringBuilder sb = new StringBuilder(pattern.length());
    int inGroup = 0;
    int inClass = 0;
    int firstIndexInClass = -1;
    char[] arr = pattern.toCharArray();
    for (int i = 0; i < arr.length; i++) {
        char ch = arr[i];
        switch (ch) {
            case '\\':
                if (++i >= arr.length) {
                    sb.append('\\');
                } else {
                    char next = arr[i];
                    switch (next) {
                        case ',':
                            // escape not needed
                            break;
                        case 'Q':
                        case 'E':
                            // extra escape needed
                            sb.append('\\');
                        default:
                            sb.append('\\');
                    }
                    sb.append(next);
                }
                break;
            case '*':
                if (inClass == 0)
                    sb.append(".*");
                else
                    sb.append('*');
                break;
            case '?':
                if (inClass == 0)
                    sb.append('.');
                else
                    sb.append('?');
                break;
            case '[':
                inClass++;
                firstIndexInClass = i+1;
                sb.append('[');
                break;
            case ']':
                inClass--;
                sb.append(']');
                break;
            case '.':
            case '(':
            case ')':
            case '+':
            case '|':
            case '^':
            case '$':
            case '@':
            case '%':
                if (inClass == 0 || (firstIndexInClass == i && ch == '^'))
                    sb.append('\\');
                sb.append(ch);
                break;
            case '!':
                if (firstIndexInClass == i)
                    sb.append('^');
                else
                    sb.append('!');
                break;
            case '{':
                inGroup++;
                sb.append('(');
                break;
            case '}':
                inGroup--;
                sb.append(')');
                break;
            case ',':
                if (inGroup > 0)
                    sb.append('|');
                else
                    sb.append(',');
                break;
            default:
                sb.append(ch);
        }
    }
    return sb.toString();
}

该单元测试证明它有效：

/**
 * @author Neil Traft
 */
public class StringUtils_ConvertGlobToRegex_Test {

    @Test
    public void star_becomes_dot_star() throws Exception {
        assertEquals("gl.*b", StringUtils.convertGlobToRegex("gl*b"));
    }

    @Test
    public void escaped_star_is_unchanged() throws Exception {
        assertEquals("gl\\*b", StringUtils.convertGlobToRegex("gl\\*b"));
    }

    @Test
    public void question_mark_becomes_dot() throws Exception {
        assertEquals("gl.b", StringUtils.convertGlobToRegex("gl?b"));
    }

    @Test
    public void escaped_question_mark_is_unchanged() throws Exception {
        assertEquals("gl\\?b", StringUtils.convertGlobToRegex("gl\\?b"));
    }

    @Test
    public void character_classes_dont_need_conversion() throws Exception {
        assertEquals("gl[-o]b", StringUtils.convertGlobToRegex("gl[-o]b"));
    }

    @Test
    public void escaped_classes_are_unchanged() throws Exception {
        assertEquals("gl\\[-o\\]b", StringUtils.convertGlobToRegex("gl\\[-o\\]b"));
    }

    @Test
    public void negation_in_character_classes() throws Exception {
        assertEquals("gl[^a-n!p-z]b", StringUtils.convertGlobToRegex("gl[!a-n!p-z]b"));
    }

    @Test
    public void nested_negation_in_character_classes() throws Exception {
        assertEquals("gl[[^a-n]!p-z]b", StringUtils.convertGlobToRegex("gl[[!a-n]!p-z]b"));
    }

    @Test
    public void escape_carat_if_it_is_the_first_char_in_a_character_class() throws Exception {
        assertEquals("gl[\\^o]b", StringUtils.convertGlobToRegex("gl[^o]b"));
    }

    @Test
    public void metachars_are_escaped() throws Exception {
        assertEquals("gl..*\\.\\(\\)\\+\\|\\^\\$\\@\\%b", StringUtils.convertGlobToRegex("gl?*.()+|^$@%b"));
    }

    @Test
    public void metachars_in_character_classes_dont_need_escaping() throws Exception {
        assertEquals("gl[?*.()+|^$@%]b", StringUtils.convertGlobToRegex("gl[?*.()+|^$@%]b"));
    }

    @Test
    public void escaped_backslash_is_unchanged() throws Exception {
        assertEquals("gl\\\\b", StringUtils.convertGlobToRegex("gl\\\\b"));
    }

    @Test
    public void slashQ_and_slashE_are_escaped() throws Exception {
        assertEquals("\\\\Qglob\\\\E", StringUtils.convertGlobToRegex("\\Qglob\\E"));
    }

    @Test
    public void braces_are_turned_into_groups() throws Exception {
        assertEquals("(glob|regex)", StringUtils.convertGlobToRegex("{glob,regex}"));
    }

    @Test
    public void escaped_braces_are_unchanged() throws Exception {
        assertEquals("\\{glob\\}", StringUtils.convertGlobToRegex("\\{glob\\}"));
    }

    @Test
    public void commas_dont_need_escaping() throws Exception {
        assertEquals("(glob,regex),", StringUtils.convertGlobToRegex("{glob\\,regex},"));
    }

}

Answer 4

有几个库可以进行类似Glob的模式匹配，它们比列出的更现代：

Theres Ants Directory Scanner 和斯普林斯AntPathMatcher

我推荐其他解决方案，因为 Ant Style Globbing已经成为Java世界中的标准glob语法（Hudson，Spring，Ant和I think Maven）。

Answer 5

我最近不得不这样做并使用\Q和\E来逃避glob模式：

private static Pattern getPatternFromGlob(String glob) {
  return Pattern.compile(
    "^" + Pattern.quote(glob)
            .replace("*", "\\E.*\\Q")
            .replace("?", "\\E.\\Q") 
    + "$");
}

Answer 6

这是一个简单的Glob实现，可以处理*和？在模式中

public class GlobMatch {
    private String text;
    private String pattern;

    public boolean match(String text, String pattern) {
        this.text = text;
        this.pattern = pattern;

        return matchCharacter(0, 0);
    }

    private boolean matchCharacter(int patternIndex, int textIndex) {
        if (patternIndex >= pattern.length()) {
            return false;
        }

        switch(pattern.charAt(patternIndex)) {
            case '?':
                // Match any character
                if (textIndex >= text.length()) {
                    return false;
                }
                break;

            case '*':
                // * at the end of the pattern will match anything
                if (patternIndex + 1 >= pattern.length() || textIndex >= text.length()) {
                    return true;
                }

                // Probe forward to see if we can get a match
                while (textIndex < text.length()) {
                    if (matchCharacter(patternIndex + 1, textIndex)) {
                        return true;
                    }
                    textIndex++;
                }

                return false;

            default:
                if (textIndex >= text.length()) {
                    return false;
                }

                String textChar = text.substring(textIndex, textIndex + 1);
                String patternChar = pattern.substring(patternIndex, patternIndex + 1);

                // Note the match is case insensitive
                if (textChar.compareToIgnoreCase(patternChar) != 0) {
                    return false;
                }
        }

        // End of pattern and text?
        if (patternIndex + 1 >= pattern.length() && textIndex + 1 >= text.length()) {
            return true;
        }

        // Go on to match the next character in the pattern
        return matchCharacter(patternIndex + 1, textIndex + 1);
    }
}

Answer 7

与Tony Edgecombe的answer类似，这是一个简短的简单的globber，如果有人需要，则不使用正则表达式来支持*和?。

public static boolean matches(String text, String glob) {
    String rest = null;
    int pos = glob.indexOf('*');
    if (pos != -1) {
        rest = glob.substring(pos + 1);
        glob = glob.substring(0, pos);
    }

    if (glob.length() > text.length())
        return false;

    // handle the part up to the first *
    for (int i = 0; i < glob.length(); i++)
        if (glob.charAt(i) != '?' 
                && !glob.substring(i, i + 1).equalsIgnoreCase(text.substring(i, i + 1)))
            return false;

    // recurse for the part after the first *, if any
    if (rest == null) {
        return glob.length() == text.length();
    } else {
        for (int i = glob.length(); i <= text.length(); i++) {
            if (matches(text.substring(i), rest))
                return true;
        }
        return false;
    }
}

Answer 8

我不知道“标准”实现，但我知道在BSD许可下发布的sourceforge项目实现了文件的全局匹配。它是在one file中实现的，也许您可以根据自己的要求进行调整。

Answer 9

这可能是一个有点古怪的方法。我已经从NIO2的Files.newDirectoryStream(Path dir, String glob)代码中找到了答案。请注意，将创建每个匹配的新Path对象。到目前为止，我只能在Windows FS上进行测试，但是，我认为它也可以在Unix上运行。

// a file system hack to get a glob matching
PathMatcher matcher = ("*".equals(glob)) ? null
    : FileSystems.getDefault().getPathMatcher("glob:" + glob);

if ("*".equals(glob) || matcher.matches(Paths.get(someName))) {
    // do you stuff here
}

更新在Mac和Linux上均可使用。

Answer 10

很久以前我正在做一个大规模的全局驱动的文本过滤，所以我写了一小段代码（15行代码，没有JDK之外的依赖）。它只处理'*'（对我来说足够了），但可以很容易地扩展为'？'。它比预编译的正则表达式快几倍，不需要任何预编译（基本上每次匹配模式时都是字符串与字符串的比较）。

代码：

  public static boolean miniglob(String[] pattern, String line) {
    if (pattern.length == 0) return line.isEmpty();
    else if (pattern.length == 1) return line.equals(pattern[0]);
    else {
      if (!line.startsWith(pattern[0])) return false;
      int idx = pattern[0].length();
      for (int i = 1; i < pattern.length - 1; ++i) {
        String patternTok = pattern[i];
        int nextIdx = line.indexOf(patternTok, idx);
        if (nextIdx < 0) return false;
        else idx = nextIdx + patternTok.length();
      }
      if (!line.endsWith(pattern[pattern.length - 1])) return false;
      return true;
    }
  }

用法：

  public static void main(String[] args) {
    BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    try {
      // read from stdin space separated text and pattern
      for (String input = in.readLine(); input != null; input = in.readLine()) {
        String[] tokens = input.split(" ");
        String line = tokens[0];
        String[] pattern = tokens[1].split("\\*+", -1 /* want empty trailing token if any */);

        // check matcher performance
        long tm0 = System.currentTimeMillis();
        for (int i = 0; i < 1000000; ++i) {
          miniglob(pattern, line);
        }
        long tm1 = System.currentTimeMillis();
        System.out.println("miniglob took " + (tm1-tm0) + " ms");

        // check regexp performance
        Pattern reptn = Pattern.compile(tokens[1].replace("*", ".*"));
        Matcher mtchr = reptn.matcher(line);
        tm0 = System.currentTimeMillis();
        for (int i = 0; i < 1000000; ++i) {
          mtchr.matches();
        }
        tm1 = System.currentTimeMillis();
        System.out.println("regexp took " + (tm1-tm0) + " ms");

        // check if miniglob worked correctly
        if (miniglob(pattern, line)) {
          System.out.println("+ >" + line);
        }
        else {
          System.out.println("- >" + line);
        }
      }
    } catch (IOException e) {
      // TODO Auto-generated catch block
      e.printStackTrace();
    }
  }

从here

复制/粘贴

Answer 11

Vincent Robert / dimo414的previous solution依赖于Pattern.quote() ... \Q来实现\E，API中未对此进行记录，因此可能其他/未来的Java实现则不是这样。以下解决方案通过转义所有\E而不是使用quote()来消除实现依赖。万一要匹配的字符串包含换行符，它还会激活DOTALL模式（(?s)）。

    public static Pattern globToRegex(String glob)
    {
        return Pattern.compile(
            "(?s)^\\Q" +
            glob.replace("\\E", "\\E\\\\E\\Q")
                .replace("*", "\\E.*\\Q")
                .replace("?", "\\E.\\Q") +
            "\\E$"
        );
    }

Answer 12

顺便说一句，看起来好像你在Perl中做得很难

这可以解决Perl的问题：

my @files = glob("*.html")
# Or, if you prefer:
my @files = <*.html>

是否有“glob”类型模式的java.util.regex等价物？

12 个答案: