Question

我想允许两个主要通配符?和*过滤我的数据。

以下是我现在正在做的事情（正如我在很多网站上看到的那样）：

public boolean contains(String data, String filter) {
    if(data == null || data.isEmpty()) {
        return false;
    }
    String regex = filter.replace(".", "[.]")
                         .replace("?", ".")
                         .replace("*", ".*");
    return Pattern.matches(regex, data);
}

但是我们不应该逃避所有其他正则表达式特殊字符，例如|或(等等吗？而且，如果它们前面有?，我们可以保留*和\吗？例如，像：

filter.replaceAll("([$|\\[\\]{}(),.+^-])", "\\\\$1") // 1. escape regex special chars, but ?, * and \
      .replaceAll("([^\\\\]|^)\\?", "$1.")           // 2. replace any ? that isn't preceded by a \ by .
      .replaceAll("([^\\\\]|^)\\*", "$1.*")          // 3. replace any * that isn't preceded by a \ by .*
      .replaceAll("\\\\([^?*]|$)", "\\\\\\\\$1");    // 4. replace any \ that isn't followed by a ? or a * (possibly due to step 2 and 3) by \\

你怎么看？如果您同意，我是否缺少任何其他正则表达式特殊字符？

编辑＃1 （考虑到dan1111和m.buettner的建议后）：

// replace any even number of backslashes by a *
regex = regex.replaceAll("(?<!\\\\)(\\\\\\\\)+(?!\\\\)", "*");
// reduce redundant wildcards that aren't preceded by a \
regex = regex.replaceAll("(?<!\\\\)[?]*[*][*?]+", "*");
// escape regexps special chars, but \, ? and *
regex = regex.replaceAll("([|\\[\\]{}(),.^$+-])", "\\\\$1");
// replace ? that aren't preceded by a \ by .
regex = regex.replaceAll("(?<!\\\\)[?]", ".");
// replace * that aren't preceded by a \ by .*
regex = regex.replaceAll("(?<!\\\\)[*]", ".*");

这个怎么样？

编辑＃2 （考虑到dan1111的建议后）：

// replace any even number of backslashes by a *
regex = regex.replaceAll("(?<!\\\\)(\\\\\\\\)+(?!\\\\)", "*");
// reduce redundant wildcards that aren't preceded by a \
regex = regex.replaceAll("(?<!\\\\)[?]*[*][*?]+", "*");
// escape regexps special chars (if not already escaped by user), but \, ? and *
regex = regex.replaceAll("(?<!\\\\)([|\\[\\]{}(),.^$+-])", "\\\\$1");
// replace ? that aren't preceded by a \ by .
regex = regex.replaceAll("(?<!\\\\)[?]", ".");
// replace * that aren't preceded by a \ by .*
regex = regex.replaceAll("(?<!\\\\)[*]", ".*");

目标即将到来？

Answer 1

替换字符串中不需要4个反斜杠来写出一个反斜杠。两个反斜杠就足够了。

您可以使用否定的lookbehind来避免替换字符串中的([^\\\\]|^)和$1：

filter.replaceAll("([$|\\[\\]{}(),.+^-])", "\\$1") // 1. escape regex special chars, but ?, * and \
      .replaceAll("(?<!\\\\)[?]", ".")           // 2. replace any ? that isn't preceded by a \ by .
      .replaceAll("(?<!\\\\)[*]", ".*")          // 3. replace any * that isn't preceded by a \ by .*

我真的没有看到你需要的最后一步。不会逃脱逃避元字符的反斜杠（反过来，实际上不会逃避它们）。我忽略了这样一个事实，你的替换呼叫会写出4个反斜杠而不是只有两个。但是说你的原始输入有th|is。然后，您的第一次替换将成为th\|is。然后，最后一次替换会使th\\|is匹配th - 反斜杠或 is。

您需要区分字符串在代码中的编写方式（未编译，反斜杠的两倍）以及编译后的内容（仅包含一半反斜杠）。

您可能还想考虑限制可能*的数量。像.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*!这样的正则表达式（在输入中找不到!）可能需要很长时间才能运行。该问题称为catastrophic backtracking。

Answer 2

最后我采用的解决方案（使用Apache Commons Lang库）：

public static boolean isFiltered(String data, String filter) {
    // no filter: return true
    if (StringUtils.isBlank(filter)) {
        return true;
    }
    // a filter but no data: return false
    else if (StringUtils.isBlank(data)) {
        return false;
    }
    // a filter and a data:
    else {
        // case insensitive
        data = data.toLowerCase();
        filter = filter.toLowerCase();
        // .matches() auto-anchors, so add [*] (i.e. "containing")
        String regex = "*" + filter + "*";
        // replace any pair of backslashes by [*]
        regex = regex.replaceAll("(?<!\\\\)(\\\\\\\\)+(?!\\\\)", "*");
        // minimize unescaped redundant wildcards
        regex = regex.replaceAll("(?<!\\\\)[?]*[*][*?]+", "*");
        // escape unescaped regexps special chars, but [\], [?] and [*]
        regex = regex.replaceAll("(?<!\\\\)([|\\[\\]{}(),.^$+-])", "\\\\$1");
        // replace unescaped [?] by [.]
        regex = regex.replaceAll("(?<!\\\\)[?]", ".");
        // replace unescaped [*] by [.*]
        regex = regex.replaceAll("(?<!\\\\)[*]", ".*");
        // return whether data matches regex or not
        return data.matches(regex);
    }
}

非常感谢@ dan1111和@ m.buettner的宝贵帮助;)

Answer 3

试试这个更简单的版本：

String regex = Pattern.quote(filter).replace("*", "\\E.*\\Q").replace("?", "\\E.\\Q");

引用整个过滤器\Q和\E，然后停止*和?上的引用，将其替换为等效的模式（{{1} }和.*）

我用

测试了它

输出：

String simplePattern = "ab*g\\Ei\\.lmn?p";
String data = "abcdefg\\Ei\\.lmnop";
String quotedPattern = Pattern.quote(simplePattern);
System.out.println(quotedPattern);
String regex = quotedPattern.replace("*", "\\E.*\\Q").replace("?", "\\E.\\Q");
System.out.println(regex);
System.out.println(data.matches(regex));

请注意，这是基于Oracle的\Qab*g\E\\E\Qi\.lmn?p\E \Qab\E.*\Qg\E\\E\Qi\.lmn\E.\Qp\E true实现，我不知道是否还有其他有效的实现。

从通配符到正则表达式

3 个答案: