获取子字符串的Reqular Expression，如果不存在则获取其他字符串

I have a reqular expression for getting sub-string from a string, if a particular string doesn't exist then try to get another sub-string.

我有一个reqular表达式,用于从字符串中获取子字符串,如果某个特定字符串不存在,则尝试获取另一个子字符串。

Reqular expression i am trying is this:

我正在尝试的Reqular表达式是这样的:

\@\s*\((.*?)\)|\((.*?)\)

But this does not work, and always get the second option sub-string instead of the first option.

但这不起作用,并始终获得第二个选项子字符串而不是第一个选项。

Example string is

示例字符串是

some text (2nd Sub-string) @ (First sub-string)

一些文字(第二个子串)@(第一个子串)

And it give me this result:

它给了我这个结果:

Array
(
   [0] =>  (2nd Sub-string)
   [1] => 
   [2] => 2nd Sub-string
)

1 个解决方案

#1

Why don't you simply get both strings (for which your regexp works correctly) and check their existence programmatically? Something like:

为什么不简单地获取两个字符串(正则表达式正常工作)并以编程方式检查它们的存在?就像是:

$num = preg_match_all(
    "/\@\s*\((.*?)\)|\((.*?)\)/", 
    "some text (2nd Sub-string) @ (First sub-string)", 
    $matches, PREG_SET_ORDER
    );
var_dump($num, $matches);
if($num < 2)
    {
    // no second match, read first
    }
if(!array_key_exists(2, $matches[1]))
    {
    // another way to put it
    }

HTH.