Regex：高级删除单行PHP注释

我想删除一些PHP代码中的所有单行注释，比如使用Visual Basic的代码:

<?php
some code
// "(\/\/[^;)]*)(?=\?\>)"         <-- NOT REMOVED
/* a first comment */
/* a second comment with some // inside
// a single line ; comment inside         <-- NOT REMOVED
// a comment with a ; website http://www.google.com         <-- NOT REMOVED
a third comment
*/
/* a fourth comment with some // inside */
/* a fifth comment with some ftp://ftp.google;com//onefolder inside */
some code
if {// a comment         <-- NOT REMOVED
    doit();
} /* another comment */
some more code// a comment         <-- NOT REMOVED
some more code2// a comment         <-- NOT REMOVED
?>
<?php  // a comment with a website http://www.google.com ?>
<?php and some more code // and a comment ?>
<?php var = 'http://www.google;com'; // and a comment ?>
<?php var = 'https://www.google;com'; // and a comment ?>
<?php var = 'ftp://www.google;com'; // and a comment ?>
<?php var = 'ftp://www.google;com'; // a comment with a website http://www.google.com ?>
<?php var = 'ftp://www.google;com'; // a comment ?         <-- NOT REMOVED
?>

我目前使用的是"(\/\/[^;)]*)(?=\?\>)"模式，这在一定程度上是可行的，但我仍然有剩余的行(参见<-- NOT REMOVED标记).我没有成功地告诉正则表达式删除，直到它找到一个?>序列，在行尾之前是if it is there…

你能帮我改进这个正则表达式吗？

因此，它给出了这个(或根据剩余的\n和注释清理后留下的倍数空格，我可以在下一次清理中删除):

<?php
some code  

/* a first comment */
/* a second comment with some // inside <-- END CAN BE REMOVED, NO PROBLEM
  

a third comment
*/
/* a fourth comment with some // inside */ <-- CAN BE REMOVED IF THE ENDING */ IS KEPT
/* a fifth comment with some ftp://ftp.google;com//onefolder inside */
some code
if {
    doit();
} /* another comment */
some more code
some more code2
?>
<?php ?>
<?php and some more code ?>
<?php var = 'http://www.google;com'; ?>
<?php var = 'https://www.google;com'; ?>
<?php var = 'ftp://www.google;com'; ?>
<?php var = 'ftp://www.google;com'; ?>
<?php var = 'ftp://www.google;com'; 
?>

注意:如果这会导致一个编译器可读的代码，那么我在多次传递中这样做没有问题.