Question

我正在尝试替换记事本++中引号之间的字符串中的所有逗号。

我几乎与regex to remove comma between double quotes notepad++到达那里，但并不完全。目前它只是取代了第一个逗号。

总结这篇文章，它使用了查找内容：("[^",]+),([^"]+")替换：\1\2（我更改为\1~\2）

正则表达式中是否有一种方法可以捕获引号之间逗号的所有实例？

编辑：添加一些代表性字符串：

1,G 32,170696,01/06/2015,Jun-17,"M12 X 1,50 - 4H GO SRG",P U7,,,,SRG ,"G 32_170696_06-2017_M12 X 1,50 - 4H GO SRG_P U7.pdf"
3,13247,163090,01/11/2015,Nov-17,"PG 0,251 to 0,500 inch",P U7,,,,,"13247_163090_11-2017_PPG 0,251 to 0,500 inch_P U7.pdf"
9,PI 1496,182411,01/04/2015,Apr-17,"6,000 - 6,018mm GO-NOGO  PPG",,,,,PPG,"PI 1496_182411_04-2017_6,000 - 6,018mm GO-NOGO  PPG.pdf"

Answer 1

您可以使用此模式一次性完成：

(?:\G(?!^)|([^"]*(?:"[^,"]*"[^"]*)*"))[^",]*\K,([^",]*+(?:"(?1)|$))?

使用此替换：~\2

demo

细节：

(?:
    \G(?!^) # contiguous to a previous match
  | # OR
    ([^"]*(?:"[^,"]*"[^"]*)*") # capture group 1: first match
                               # reach the first quoted part with commas
)
[^",]* \K ,  #"#
( # capture group 2: succeeds after the last comma
    [^",]*+  #"#
    (?:
        " (?1) #"# reach the next quoted part with commas
               # (using the capture group 1 subpattern)
      | # OR
        $ # end of the string: (this set a default behavior: when
          # the closing quote is missing)
    )
)?

在notepad ++中用regex替换引号中的所有逗号

1 个答案: