java - 将正则表达式匹配回同一行的开头？

Question

给定:

-- 输入--

Keep this.
And keep this.

And keep this too.
Chomp this chomp:
Anything beyond here gets chomped.

-- 输出(预期)--

Keep this.
And keep this.

And keep this too.

我如何为每个分组匹配一个正则表达式，以便一旦找到“chomp:”，从该行开头以及之后的所有内容都会被压缩(删除)？

String text = "Keep this.\nAnd keep this.\n\nAnd keep this too.\n"
        + "This could be anything here chomp:\nAnything beyond here gets chomped.";
Pattern CHOMP= Pattern.compile("^((.*)chomp:(.*))$",  Pattern.MULTILINE | Pattern.DOTALL);
Matcher m = CHOMP.matcher(text);
if (m.find()) {
    int count = m.groupCount();
    //         
    // How can I match a group here to either delete or keep for expected output?
    //
    // text = <match a group to assign or replace non-desired text>;
    System.out.println(text);  // Should output contents from above -- output (expected) --
}

Answer 1

这是一种方法，a demonstration on ideaone .

我稍微简化了模式；然而，我的代码中最大的变化是它在没有 DOTALL 选项的情况下运行 - 使用 DOTALL . 将错误地匹配 多行。

^(.*)chomp:(.*)

模式应该匹配一次(这似乎是意图)，用“chomp:”之前/之后的文本填充组 1 和 2 以及其余数据将被“消耗”，因为它只是未处理。要在正则表达式匹配(而不是匹配)之前获取数据，我使用以下结构:

StringBuffer sb = new StringBuffer();
matcher.appendReplacement(sb, "");

(虽然这可以用子字符串替换，但我想，这个成语 mirrors other patterns 。)

---

如果您希望进行面向行的处理(适合大流)，那么正确的做法是依次处理每一行。我自己可能会使用拆分或扫描器方法，但我希望将这个答案保留在最初提出的原始整体正则表达式方法中。

例如:

Scanner s = new Scanner(input);
while (s.hasNextLine()) {
    // process next line and "break" if it matches the end-line condition
}

---

ideone 的片段:

String text = "Keep this.\nAnd keep this.\n\nAnd keep this too.\n"
        + "Chomp this chomp:\nAnything beyond here gets chomped.";
Pattern CHOMP= Pattern.compile("^(.*)chomp:(.*)",  Pattern.MULTILINE);
Matcher m = CHOMP.matcher(text);
if (m.find()) {
    System.out.println("  LINE:" + m.group(0));
    System.out.println("BEFORE:" + m.group(1));
    System.out.println(" AFTER:" + m.group(2));
    System.out.println(">>>");
    StringBuffer sb = new StringBuffer();
    m.appendReplacement(sb, "");
    System.out.print(sb);
    System.out.println("<<<");
}