Atom 正则表达式：丢弃块周围的多行文本
Atom regexp: discarding multiline text around blocks
假设我有这个文本：
blah blah Bob Loblaw Law blah
keep1 { i want this } blop
blah blob keep2 { and
this too } blaw blat
etc...
我想以
结束
keep1 { i want this }
keep2 { and
this too }
或者也许：
keep1 { i want this }
keep2 { and this too }
我还没有想出如何让 Atom 的正则表达式 find/replace 机制丢弃特定匹配字符串 外部 多行的所有内容。提示？
更新：
在我尝试过的许多事情中，这让我最接近：
[\S\s]+?(keep\d\s+\{[\S\s]+?\})
这导致：
keep1 { i want this }
keep2 { and
this too }
blaw blat
etc...
这可能已经足够好了——我可以编辑尾随的分片——但如果知道如何 trim 那些也会很有用。
使用
[\s\S]*?(keep\d\s+\{[^{}]*\})|(?:(?!keep\d\s+\{[^{}]*\})[\s\S])+$
参见proof。
解释
--------------------------------------------------------------------------------
[\s\S]*? any character of: whitespace (\n, \r, \t,
\f, and " "), non-whitespace (all but \n,
\r, \t, \f, and " ") (0 or more times
(matching the least amount possible))
--------------------------------------------------------------------------------
( group and capture to :
--------------------------------------------------------------------------------
keep 'keep'
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
\s+ whitespace (\n, \r, \t, \f, and " ") (1
or more times (matching the most amount
possible))
--------------------------------------------------------------------------------
\{ '{'
--------------------------------------------------------------------------------
[^{}]* any character except: '{', '}' (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
\} '}'
--------------------------------------------------------------------------------
) end of
--------------------------------------------------------------------------------
| OR
--------------------------------------------------------------------------------
(?: group, but do not capture (1 or more times
(matching the most amount possible)):
--------------------------------------------------------------------------------
(?! look ahead to see if there is not:
--------------------------------------------------------------------------------
keep 'keep'
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
\s+ whitespace (\n, \r, \t, \f, and " ")
(1 or more times (matching the most
amount possible))
--------------------------------------------------------------------------------
\{ '{'
--------------------------------------------------------------------------------
[^{}]* any character except: '{', '}' (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
\} '}'
--------------------------------------------------------------------------------
) end of look-ahead
--------------------------------------------------------------------------------
[\s\S] any character of: whitespace (\n, \r,
\t, \f, and " "), non-whitespace (all
but \n, \r, \t, \f, and " ")
--------------------------------------------------------------------------------
)+ end of grouping
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string
您可以在 Atom 中使用这个简单的正则表达式替换来完成此任务：
\b(keep\d+\s*{[^}]*})|.+?
替换为：</code></p>
<p><a href="https://regex101.com/r/oTpV9g/1" rel="nofollow noreferrer">RegEx Demo</a></p>
<p><strong>正则表达式详细信息：</strong></p>
<ul>
<li><code>\b: 字边界
(keep\d+\s*{[^}]*})：在捕获组 #1 中匹配以 keep 开头的字符串，后跟 1+ 个数字，后跟 0+ 个空格，后跟 {...} 内的任何文本越界也是如此。这假设 { 和 } 是平衡的并且没有转义 { 和 }.|：或.+?: 延迟匹配 1+ 任何东西PS：如果你想删除前导换行符，那么使用：
\n?\b(keep\d+\s*{[^}]*})|.+?
Atom 编辑器演示
替换前：
替换后：
假设我有这个文本：
blah blah Bob Loblaw Law blah
keep1 { i want this } blop
blah blob keep2 { and
this too } blaw blat
etc...
我想以
结束keep1 { i want this }
keep2 { and
this too }
或者也许：
keep1 { i want this }
keep2 { and this too }
我还没有想出如何让 Atom 的正则表达式 find/replace 机制丢弃特定匹配字符串 外部 多行的所有内容。提示？
更新：
在我尝试过的许多事情中，这让我最接近：
[\S\s]+?(keep\d\s+\{[\S\s]+?\})
这导致：
keep1 { i want this }
keep2 { and
this too }
blaw blat
etc...
这可能已经足够好了——我可以编辑尾随的分片——但如果知道如何 trim 那些也会很有用。
使用
[\s\S]*?(keep\d\s+\{[^{}]*\})|(?:(?!keep\d\s+\{[^{}]*\})[\s\S])+$
参见proof。
解释
--------------------------------------------------------------------------------
[\s\S]*? any character of: whitespace (\n, \r, \t,
\f, and " "), non-whitespace (all but \n,
\r, \t, \f, and " ") (0 or more times
(matching the least amount possible))
--------------------------------------------------------------------------------
( group and capture to :
--------------------------------------------------------------------------------
keep 'keep'
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
\s+ whitespace (\n, \r, \t, \f, and " ") (1
or more times (matching the most amount
possible))
--------------------------------------------------------------------------------
\{ '{'
--------------------------------------------------------------------------------
[^{}]* any character except: '{', '}' (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
\} '}'
--------------------------------------------------------------------------------
) end of
--------------------------------------------------------------------------------
| OR
--------------------------------------------------------------------------------
(?: group, but do not capture (1 or more times
(matching the most amount possible)):
--------------------------------------------------------------------------------
(?! look ahead to see if there is not:
--------------------------------------------------------------------------------
keep 'keep'
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
\s+ whitespace (\n, \r, \t, \f, and " ")
(1 or more times (matching the most
amount possible))
--------------------------------------------------------------------------------
\{ '{'
--------------------------------------------------------------------------------
[^{}]* any character except: '{', '}' (0 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
\} '}'
--------------------------------------------------------------------------------
) end of look-ahead
--------------------------------------------------------------------------------
[\s\S] any character of: whitespace (\n, \r,
\t, \f, and " "), non-whitespace (all
but \n, \r, \t, \f, and " ")
--------------------------------------------------------------------------------
)+ end of grouping
--------------------------------------------------------------------------------
$ before an optional \n, and the end of the
string
您可以在 Atom 中使用这个简单的正则表达式替换来完成此任务：
\b(keep\d+\s*{[^}]*})|.+?
替换为：</code></p>
<p><a href="https://regex101.com/r/oTpV9g/1" rel="nofollow noreferrer">RegEx Demo</a></p>
<p><strong>正则表达式详细信息：</strong></p>
<ul>
<li><code>\b: 字边界
(keep\d+\s*{[^}]*})：在捕获组 #1 中匹配以 keep 开头的字符串，后跟 1+ 个数字，后跟 0+ 个空格，后跟 {...} 内的任何文本越界也是如此。这假设 { 和 } 是平衡的并且没有转义 { 和 }.|：或.+?: 延迟匹配 1+ 任何东西PS：如果你想删除前导换行符，那么使用：
\n?\b(keep\d+\s*{[^}]*})|.+?
Atom 编辑器演示
替换前：
替换后：