preg_replace去掉標點符號，空格和佔有形式的單詞？

Question

$text = 
"<tab><return><nbsp>O'Neil really 
likes his pudding's, he really really does!!!1.5" 

我想有這種可能的話，其中'<tab>'是一個實際的標籤\t空間不是字面上'<tab>'

$text = "O'Neil really likes his pudding he really really does 1.5" 

取出白色的步伐，並用一個空格，標點符號，保持數字或替換它們撇號之後的任何東西都不是's'。

非常感謝！

Answer 1

$string = trim(preg_replace(array('~[:;!?]|[.,](?![0-9])|\'s~', '~\s+~'), array('', ' '), $string)); 

首先擺脫's，然後用一個空格替換多個空格。最終它將刪除前導空白和尾隨空白。