Php regex，找到关键字时获取行后的第一个字符

我一直在与正则表达式构建器和教程混搭，但我无法理解这一点。我有一个关键字，我可以找到，我需要找到同一行或下一行的时间戳

我的文本看起来像这样

16:51:35.536 This is a bunch of completely random text although it has a keyword. This keyword is unique
16:51:35.806 This is another line which has all sorts of crap i dont really care about.

it has a keyword (.*) ([0-9){2}:[0-9){2}:[0-9){2}) 

现在我要捕捉的是获取时间戳的前8个字符，即16:51:35

所以我试着做这样的事情

16:51:35.536 This is a bunch of completely random text although it has a keyword. This keyword is unique
16:51:35.806 This is another line which has all sorts of crap i dont really care about.

it has a keyword (.*) ([0-9){2}:[0-9){2}:[0-9){2}) 

但不管出于什么原因，即使我在末尾添加/m或/s或/s，它也会卡在该行上

这有助于进入下一行：

([\s\S]*)

所以我试过了

it has a keyword ([\s\S]*)(?=[0-9){2}:[0-9){2}:[0-9){2}) but that didnt work either

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我很确定这比我做的要简单，任何想法

如果你想要一行的前八个字符带有关键字字符串“foobar”，那么你可以使用：

^(.{8})[^\n]*foobar.*?$

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您可以看到，这项工作很容易修改，以获得下一行的前八个字符，或者对前八个字符进行限制。

首先，您的意思是

[0-9）

而不是

[0-9]

？