[英]regex replacement with a increasing number
我想正则表达式/替换以下集合:
[something] [nothing interesting here] [boring.....]
通过
%0 %1 %2
换句话说,任何用[]
构建的表达式都将成为%
后跟一个递增的数字...
使用Regex可以马上做到吗?
C#中的regex可以做到这一点,因为Regex.Replace
可以将委托作为参数。
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'
您可以使用Regex.Replace
,它有一个方便的重载,需要进行回调 :
string s = "[something] [nothing interesting here] [boring.....]";
int counter = 0;
s = Regex.Replace(s, @"\[[^\]]+\]", match => "%" + (counter++));
由于您所描述的内容具有程序成分,因此不直接存在。 我认为Perl可以通过它的qx运算符允许它(我认为),但是一般来说,您需要遍历字符串,这应该非常简单。
answer = ''
found = 0
while str matches \[[^\[\]]\]:
answer = answer + '%' + (found++) + ' '
PHP和Perl都支持“回调”替换,使您可以将一些代码挂钩来生成替换。 这是在PHP中使用preg_replace_callback进行操作的方法
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
您可以使用全局var和常规函数回调来完成此操作,但是将其包装在类中则更加整洁。
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