Regex Find and Replace：删除除以“开始”的单词以外的所有内容#&引用；？

我是regex的新手，找不到一个简单的方法。我想删除所有不以#开头的单词，并在它们之间加一个逗号，例如，如果我有：

Cookie Recipe for n00bs
#cookie #recipe #chocolate To do this you have to etc...
Bla bla bla mumbo jumbo

因此，我想得到：

cookie, recipe, chocolate

---

如果你能帮我，那就太好了，谢谢，祝你今天愉快

您没有告诉我您使用的是哪种编程语言。下面是一个PHP示例，它使用与perl兼容的正则表达式：

$text = <<<EOF
Cookie Recipe for n00bs
#cookie #recipe #chocolate To do this you have to etc...
Bla bla bla mumbo jumbo
EOF;

$pattern = '/((?<=#)\w+)/';
preg_match_all($pattern, $text, $matches);

echo implode(', ', $matches[0]);

---

$text=试试这个：

$re = "/#\\w+/";
$str = "#cookie #recipe #chocolate To do this you have to etc...";
$str .= "#cookie #recipe #chocolate To do this you have to etc...";

preg_match_all($re, $str, $matches);
$r=@implode(", ",@$matches[0]);  // for adding comma(,) and space( ).
var_dump( $matches,$r);

array (size=1)
  0 => 
    array (size=6)
      0 => string '#cookie' (length=7)
      1 => string '#recipe' (length=7)
      2 => string '#chocolate' (length=10)
      3 => string '#cookie' (length=7)
      4 => string '#recipe' (length=7)
      5 => string '#chocolate' (length=10)

string '#cookie, #recipe, #chocolate, #cookie, #recipe, #chocolate' (length=58)

输出：

$re = "/#\\w+/";
$str = "#cookie #recipe #chocolate To do this you have to etc...";
$str .= "#cookie #recipe #chocolate To do this you have to etc...";

preg_match_all($re, $str, $matches);
$r=@implode(", ",@$matches[0]);  // for adding comma(,) and space( ).
var_dump( $matches,$r);

array (size=1)
  0 => 
    array (size=6)
      0 => string '#cookie' (length=7)
      1 => string '#recipe' (length=7)
      2 => string '#chocolate' (length=10)
      3 => string '#cookie' (length=7)
      4 => string '#recipe' (length=7)
      5 => string '#chocolate' (length=10)

string '#cookie, #recipe, #chocolate, #cookie, #recipe, #chocolate' (length=58)

与其删除所有前面没有
#
的单词，为什么不尝试匹配前面有
#
的单词？您使用什么语言/工具？我尝试在查找和替换上下文中使用它。我正在创建一个键盘Maestro宏，该宏按照查找和替换规则过滤剪贴板。我不知道如何只复制匹配的结果。