如何匹配任何单词但忽略以多个空格开头的单词？

Question

我想要实现的是匹配文本中的所有单词，但忽略那些以 4 个空格开头的行中（新行之前）的单词。

---

示例

查找单词的文本文件：

Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do 
eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut 
enim ad minim veniam, quis nostrud exercitation ullamco laboris 
nisi ut aliquip ex ea commodo consequat.

    This must NOT be matched. Because it has 4 whitespaces at the beginning.

Lorem ipsum dolor sit amet. Ut enim ad minim veniam.

因此，下一行中的单词不应该被认为与模式匹配：

This must NOT be matched. Because it has 4 whitespaces at the beginning.

代码

这是我的正则表达式，它可以找到所有单词：

\\b[A-Za-z]+\\b

我知道在 Java 的 RegEx 语法中有 except 是 ^ 符号，但我只知道如何在更简单的表达式中使用它。

Answer 1

也许遵循 sn-p 可能是您想要实现的目标的基础。

String[] lines = {"Lorem ipsum dolor sit amet, consectetur adipiscing elit, sed do",
    "eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut",
    "enim ad minim veniam, quis nostrud exercitation ullamco laboris",
    "nisi ut aliquip ex ea commodo consequat.",
    "",
    "    This must NOT be matched. Because it has 4 whitespaces at the beginning.",
    "",
    "Lorem ipsum dolor sit amet. Ut enim ad minim veniam."};
for (String line : lines) {
    if (!line.startsWith("    ")) {
        String[] words = line.split("[\\p{IsPunctuation}\\p{IsWhite_Space}]+");
        System.out.println("words = " + Arrays.toString(words));
    }
}

输出

words = [Lorem, ipsum, dolor, sit, amet, consectetur, adipiscing, elit, sed, do]
words = [eiusmod, tempor, incididunt, ut, labore, et, dolore, magna, aliqua, Ut]
words = [enim, ad, minim, veniam, quis, nostrud, exercitation, ullamco, laboris]
words = [nisi, ut, aliquip, ex, ea, commodo, consequat]
words = []
words = []
words = [Lorem, ipsum, dolor, sit, amet, Ut, enim, ad, minim, veniam]

PS：正则表达式是从this answer借来的

Answer 2

以下应该做到这一点

(?<!\s{4})\\b[A-Za-z]+\\b

它以negative lookbehind 开头，因此它不会匹配前面带有\s{4} 的任何内容。