如何让正则表达式忽略括号之间的所有内容？

Question

考虑以下字符串：

I have been driving to {Palm.!.Beach:100} and it . was . great!!

我使用以下正则表达式删除所有标点符号：

$string preg_replace('/[^a-zA-Z ]+/', '', $string);

输出：

I have been driving to PalmBeach and it  was  great!!

但我需要正则表达式始终忽略{和}之间的任何内容。所以期望的输出是：

I have been driving to {Palm.!.Beach:100} and it  was  great

如何让正则表达式忽略{和}之间的内容？

Answer 1

试试这个

[^a-zA-Z {}]+(?![^{]*})

见here on Regexr

表示匹配否定字符类中未包含的任何内容，但仅当前面没有前面没有右括号时，这是由否定前瞻(?![^{]*})完成的。

$string preg_replace('/[^a-zA-Z {}]+(?![^{]*})/', '', $string);

Answer 2

$str = preg_replace('(\{[^}]+\}(*SKIP)(*FAIL)|[^a-zA-Z ]+)', '', $str);

另见Split string by delimiter, but not if it is escaped。