如何讓正則表達式忽略括號內的所有內容？

Question

考慮以下字符串：

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

我用下面的正則表達式來刪除所有標點符號：

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

此輸出：

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

但我需要的正則表達式來總是忽略任何介於{和}之間。因此，所需的輸出將是：

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

我怎樣才能讓正則表達式忽略之間是什麼{和}？

Answer 1

試試這個

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

看到它here on Regexr

表示匹配不包含在否定字符類的東西，但前提是前面沒有右括號不前的開放，這樣做由負向預測(?![^{]*})。

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

Answer 2

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

另請參閱Split string by delimiter, but not if it is escaped。