如何删除 java 中包含特定字符的单词？

Question

有没有办法使用正则表达式删除包含“@”字符的整个单词？

例如，应删除以下单词：kʌn@tənɪu 和 b@ilu

我试过：

line = br.readLine();
line = line.replaceAll("\b[@]+\b","");
System.out.println(line);

...只删除“@”符号，不删除单词本身。

Answer 1

line = br.readLine();
line = line.replaceAll("\b.+?[@]+.+?\b","");
System.out.println(line);

编辑：.*? 不好。我测试了最新的编辑，它对我有用。

    String test =  "  x+1@ three@two @@ o@ne @ two";
    test = test.replaceAll("\b.+?[@]+.+?\b","");
    System.out.println(test);

给予

two

Answer 2

使用正则表达式

[^@\s]*@\S*

[^@\s]* 查找零个或多个不是“@”或空格的字符

@ 寻找一个 '@'

\s* 查找零个或多个非空白字符

所以代码行是：

line = line.replaceAll("[^@\s]*@\s*","");

Answer 3

您可以使用如下正则表达式：

(\s+)([^\s]*?@[^\s]*?)(\s+)

如果您想用它周围的两个空格替换它，您可以将其替换为 </code>。 如果你只想用一个空格替换它（这样看起来更好），你可以用 <code> 或 </code>. 替换它 <a href="https://regex101.com/r/dI8qN3/2" rel="nofollow noreferrer">Here's a preview of the regex in action</a> 在 regex101.com 上，它是这样工作的： <blockquote> <code>\s matches any whitespace character
\s+ maches any whitespace character one or more times.

[^\s] matches any non-whitespace character
[^\s]*? matches any non-whitespace character any amount of times, and tries to find the smallest match.

@ matches the character @ literally.

[^\s] matches any non-whitespace character
[^\s]*? matches any non-whitespace character any amount of times, and tries to find the smallest match.

\s matches any whitespace character
\s+ maches any whitespace character one or more times.

如何删除 java 中包含特定字符的单词？

How do I delete words containing a certain charater in java?

java

regex

string

symbols

filereader