java - 从通配符到正则表达式

Question

我想允许两个主要的通配符 ? 和 * 来过滤我的数据。

这是我现在的做法(正如我在许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);
}

但我们不应该转义所有其他正则表达式特殊字符，例如 | 或 ( 等吗？而且，也许我们可以保留 ？ 和 * 是否以 \ 开头？例如，类似于:

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，我是否遗漏了任何其他正则表达式特殊字符？

---

编辑 #1(在考虑了 dan1111 和 m.buettner 的建议之后):

// 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(在考虑了 dan1111 的建议之后):

// 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("(?<!\\\\)[*]", ".*");

目标在望？

Answer 1

您不需要在替换字符串中使用 4 个反斜杠来写出一个。两个反斜杠就足够了。

并且您可以通过使用负向后视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 .*

我真的不明白你需要最后一步做什么。这不会逃避转义元字符的反斜杠(反过来，实际上并没有转义它们)。我忽略了一个事实，即您的替换调用会写出 4 个反斜杠而不是两个。但是假设您的原始输入有 th|is。然后您的第一个替换将成为 th\|is。然后最后一个替换将使 th\\|is 匹配 th-反斜杠 或 is。

您需要区分字符串在代码中的外观(未编译，反斜杠数量翻倍)和编译后的外观(仅包含一半的反斜杠)。

您可能还想考虑限制可能的 * 的数量。正则表达式，如 .*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*.*!(其中 ! 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 .