正则表达式：选择以指定字符开头的行的第一个单词

我正在寻找这样的正则表达式:

<html>
<body>
@info
<input>.........</input>
@ok_test somthin here
</body>
</html>

我想要得到所有以“@”开头的字符串。 我尝试了 in php，但无法摆脱休息字符串后的空间。 我这样试了试我的正则表达式:

\b@[a-zA-Z0-9]*

但是仍然不能。 有人可以帮我吗？ 谢谢。

Try this one:

/(^\@)[^\s]+/gm

It selects:

  • every line starting with @
  • all the following characters up to whitespace [^\s]+

The \b word boundary before @ requires a word char (a letter, or digit, or a _) before it. You must have tried with \B instead, a non-word boundary. The [a-zA-Z0-9]* is not matching _, you should have use \w+, one or more letters/digits/underscores.

To get multiple matches, you need to use preg_match_all, not preg_match with a regex having /g modifier (that PHP regex does not support).

Use

$input = "@info and more here
Text here: @ok_test somthin here";
preg_match_all('~\B@\w+~', $input, $matches);
print_r($matches[0]);

See PHP demo yielding

Array
(
    [0] => @info
    [1] => @ok_test
)