如何使用PHP preg_replace（）定义完全匹配

Question

I am trying to remove some parameters from a URL using PHP preg_replace(). For example, i need to remove a[]=1 from the bellow URL.

$my_url = 'www.myhost.com/filter.php?a[]=1&a[]=12&a[]=13&a[]=14'

So i am using:

$without_filter = preg_replace("/(&)?a\[\]=1/", '', $my_url);

I want to remove only a[]=1, but it is removing the portion that contains a[]=1 from the others parameters, so am i getting:

www.myhost.com/filter.php?234

Someone can help me to solve this?

Answer 1

What about: /a\[\]=1(&|\b)/

That way it will capture a[]=1 only if it is followed by a & or end of string.