有越来越多的正则表达式替换

Question

我想正则表达式/替换下面的一组：由

%0 %1 %2 

换句话说

[something] [nothing interesting here] [boring.....] 

，这是建立与[]任何表述将成为%其次是越来越多的数字...

是否有可能立即与正则表达式？

Answer 1

这是可能的正则表达式在C＃中，因为Regex.Replace可以采取委托作为参数。

 Regex theRegex = new Regex(@"\[.*?\]"); 
     string text = "[something] [nothing interesting here] [boring.....]"; 
     int count = 0; 
     text = theRegex.Replace(text, delegate(Match thisMatch) 
     { 
      return "%" + (count++); 
     }); // text is now '%0 %1 %2' 

Answer 2

不是直接的，因为你描述的是一个程序组件。我认为Perl可能会允许这个，尽管它的qx操作符（我认为），但一般来说，你需要遍历字符串，应该很简单。

answer = '' 
found = 0 
while str matches \[[^\[\]]\]: 
    answer = answer + '%' + (found++) + ' ' 

Answer 3

您可以使用Regex.Replace，它有一个handy overload that takes a callback：

string s = "[something] [nothing interesting here] [boring.....]"; 
int counter = 0; 
s = Regex.Replace(s, @"\[[^\]]+\]", match => "%" + (counter++)); 

Answer 4

PHP和Perl都支持“回调”更换，让你上钩一些代码，生成的替代品。这里是你如何与preg_replace_callback

class Placeholders{ 
    private $count; 

    //constructor just sets up our placeholder counter 
    protected function __construct() 
    { 
     $this->count=0; 
    } 

    //this is the callback given to preg_replace_callback 
    protected function _doreplace($matches) 
    { 
     return '%'.$this->count++; 
    } 

    //this wraps it all up in one handy method - it instantiates 
    //an instance of this class to track the replacements, and 
    //passes the instance along with the required method to preg_replace_callback  
    public static function replace($str) 
    { 
     $replacer=new Placeholders; 
     return preg_replace_callback('/\[.*?\]/', array($replacer, '_doreplace'), $str); 
    } 
} 


//here's how we use it 
echo Placeholders::replace("woo [yay] it [works]"); 

//outputs: woo %0 it %1 

你可以用全局变量和常规功能的回调做这个做在PHP中，但在课堂上包裹起来有点整洁。