PHP的:从字符串中删除括号/内容?

问题描述：

如果我有这样的字符串:

If I have a string like this:

$str = "blah blah blah (a) (b) blah blah blah";

如何进行正则表达式，以便输出为:

How can I regex so that the output is:

$str = "blah blah blah blah blah blah";

它需要能够在字符串中支持任意数量的括号对.

It needs to be able to support any number of bracket pairs inside a string.

答

这应该可以解决问题:

$str = trim(preg_replace('/\s*\([^)]*\)/', '', $str));

请注意，与其他建议不同，此答案也消除了括号内的空白.

Note, this answer removes whitespace around the bracket too, unlike the other suggestions.

修剪是为了防止字符串以方括号开头，在这种情况下，不会删除其后的空白.

The trim is in case the string starts with a bracketed section, in which case the whitespace following it isn't removed.