替换记事本++中的字符但排除单引号内的字符（2nd）

Question

对已经回答了第一个类似问题的所有用户（尤其是 Avinash Raj）表示抱歉 - 我只是忘记了第二种字符串。（而且（那是可悲的事情）-我无法将第一个类似问题的解决方案调整为第二类字符串...）

我有两个不同的字符串：

SELECT column_name FROM table_name WHERE column_name IN ('A' , 'st9u' ,'Meyer', ....);
WHERE    a.object_type IN (' 'TABLE'', ''MATEerialIZED VIE3W'   ')

我想从上到下替换notepad++中的所有字符，但不包括替换字符单引号。

condition: It exists no solid structure before/behind/between the single quotation marks part!

(That means - I can not use the keyword "IN" or signs like "," or "(" or ")" or ";" for this regex ...!)

The once thing is, that two structures for single quotation marks are possible: 'Word|Number' or ''Word|Number'' (but, as shown in 2nd example, with different number of spaces between every single quotation mark!).

目标字符串（单引号内的字符必须保持不变）：

select column_name from table_name where column_name in ('A' , 'st9u' ,'Meyer', ....);
where    a.object_type in (' 'TABLE'', ''MATerialIZED VIE3W'   ')

如何在记事本++中排除单引号部分（替换）？

Answer 1

可以组合一个recursive part for getting '... '.... abc '' and use \K for resetting after. This is the part that needs to be skipped. And using kind of the trick，匹配管道中剩余的单词：

'\s*(?0)?[^']*'\K|(\w+)

(?0)? 这里的模式可以选择从头开始粘贴
\s 是 shorthand 用于 空白字符 [ \t\r\n\f]
\w是字字符的缩写 [A-Za-z0-9_]

并替换为 \L 以降低在第一个捕获组中捕获的词或 \U 以提高。
使用您的示例数据在 NP++ 6.7.9.2 中为我工作。 See regex101 用于测试模式。

替换记事本++中的字符但排除单引号内的字符（2nd）

replace characters in notepad++ BUT exclude characters inside single quotation marks(2nd)

regex

notepad++

regex-negation

regex-lookarounds