正则表达式：选择以指定字符开头的行的第一个单词

I'm looking for regex pattern like this:

我正在寻找像这样的正则表达式模式:

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

I want to get all strings which begin with '@'. I tried explode in php but can't get rid of the rest string after space. I tried my regex like this :

我想得到所有以'@'开头的字符串。我尝试在php中爆炸,但无法摆脱空间后的其余字符串。我尝试了这样的正则表达式:

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

But still can't. Could someone help me? Thanks.

但还是不行。有人能帮帮我吗?谢谢。

2 个解决方案

#1

Try this one:

试试这个:

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

It selects:

every line starting with @

每行以@开头

all the following characters up to whitespace [^\s]+

所有以下字符直到空格[^ \ s] +

#2

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.

@之前的\ b字边界需要一个字符char(字母,数字或_)。您必须尝试使用\ B而不是非字边界。 [a-zA-Z0-9] *与_不匹配,你应该使用\ w +,一个或多个字母/数字/下划线。

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).

要获得多个匹配,您需要使用preg_match_all,而不是preg_match和带有/ g修饰符的正则表达式(PHP正则表达式不支持)。

Use

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

See PHP demo yielding

请参阅PHP演示产生

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

#1