我在这里和其他网站上阅读了很多帖子,以便使用如下preg_replace
创建半工作的脏话过滤器:
function swearwords($word){
$word = preg_replace('/'bwwww.badword.com'b/', '', $word); //do not remove 'b //to add more copy and put word down
$word = preg_replace('/'bwww.badword.com'b/', '', $word); //do not remove 'b //to add more copy and put word down
$word = preg_replace('/'bbadword'b/', '', $word); //do not remove 'b //to add more copy and put word down
//$word = preg_replace('/'bimg.badword``'b/', '', $word); //do not remove 'b //to add more copy and put word down
return $word;
}
这是有效的,但我想允许 img.badword 的实例
所以在"badword
"前面有.img的地方,我想允许使用
以下是"负面回溯"断言的解决方案:
...
$word = preg_replace('/'b(?<!img'.)badword'b/', '', $word);
...
您也可以尝试以下多合一替代品:
function swearwords($word){
$word = preg_replace('/'b(w{3,4}'.)*(?<!img'.)badword('.com)*'b/', '', $word);
return $word;
}
http://php.net/manual/en/regexp.reference.assertions.php