// First remove the leading/trailing whitespace //去掉开始和结束的空白 $str = trim($str);
// Now remove any doubled-up whitespace //去掉跟随别的挤在一块的空白 $str = preg_replace("/s(?=s)/", "", $str);
// Finally, replace any non-space whitespace, with a space //最后,去掉非space 的空白,用一个空格代替 $str = preg_replace("/[nrt]/", " ", $str); // Echo out: "This line contains liberal use of whitespace." echo "<pre>{$str}</pre>"; ?> 希望本文所述对大家的PHP程序设计有所帮助。