preg_replace去掉标点符号，空格和占有形式的单词？

Question

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

如果'<tab>'是实际标签\t空间而不是字面'<tab>'

，我希望这样做

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

删除白人的步伐并用一个空格替换它们，标点符号，保留数字或撇号之后的任何东西，而不是's'。

非常感谢你！

Answer 1

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

首先删除's，然后用一个空格替换多个空格。最终它会删除前导和尾随空格。