在记事本++中寻求正则表达式以仅在两个引号[“]之间搜索和替换CRLF

Question

I've got a CSV file with some 600 records where I need to replace some [CRLF] with a [space] but only when the [CRLF] is positioned between two ["] (quotation marks). When the second ["] is encountered then it should skip the rest of the line and go to the next line in the text.

I don't really have a starting point. Hope someone comes up with a suggestion.

Example:

John und Carol,,Smith,,,J.S.,,,,,,,,,,,,,+11 22 333 4444,,,,,"streetx 21[CRLF]
New York City[CRLF]
USA",streetx 21,,,,New York City,,,USA,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,Normal,,My Contacts,[CRLF]

In this case the two [CRLF] after the first ["] need to be replaced with a space [ ]. When the second ["] is encountered, skip the end of the line and go to next line.

Then again, now on the next line, after the first ["] is encountered replace all [CRLF] until the second ["] is encountered. The [CRLF]s vary in numbers. In the CSV-file the amount of commas [,] before (23) and after (65) the 2 quotation marks ["] is constant.

So maybe a comma counter could be used. I don't know.

Thanks for feedback.

Answer 1

This will work using one regex only (tested in Notepad++):

Enter this regex in the Find what field:

((?:^|\\r\\n)[^"]*+"[^\\r\\n"]*+)\\r\\n([^"]*+")

Enter this string in the Replace with field:

$1 $2

Make sure the Wrap around check box (and Regular expression radio button) are selected.

Do a Replace All as many times as required (until the "0 occurrences were replaced" dialog pops up).

Explanation:

(
  (?:^|\r\n)     Begin at start of file or before the CRLF before the start of a record
  [^"]*+         Consume all chars up to the opening "
  "              Consume the opening "
  [^\r\n"]*+     Consume all chars up to either the first CRLF or the closing "
)                Save as capturing group 1 (= everything in record before the target CRLF)
\r\n             Consume the target CRLF without capturing it
(
  [^"]*+         Consume all chars up to the closing "
  "              Consume the closing "
)                Save as capturing group 2 (= the rest of the string after the target CRLF)

Note: The *+ is a possessive quantifier. Use them appropriately to speed up execution.

Update:

This more general version of the regex will work with any line break sequence ( \\r\\n , \\r or \\n ):

((?:^|[\\r\\n]+)[^"]*+"[^\\r\\n"]*+)[\\r\\n]+([^"]*+")

Answer 2

In this case the source data is generated by the export function in GMail for your contacts. After the modification outlined below (without RegEx) the result can be used to tidy up your contacts database and re-import it to GMail or to MS Outlook. Yes, I am standing on the shoulders of @alan and @robinCTS. Thank you both.

Instructions in 5 steps:

use Notepad++ / find replace / extended search mode / wrap around = on

-1- replace all [CRLF] with a unique set characters or a string (I used [ ~~ ])

find: \\r\\n and replace with: ~~ The file contents are now on one line only.

-2- Now we need to separate the header line. For this move to where the first record starts exactly before the 88th. comma (including the word after the 87th. comma [,]) and enter the [CRLF] manually by hitting the return key. There are two lines now: header and records.

-3- now find all [ ,~~ ] and replace with [ ,\\r\\n ] The result is one record per line.

-4- remove the remaining [~~] find: ~~ and replace with: [ ] a space. The file is now clean of unwanted [CRLF]s.

-5- Save the file and use it as intended.

Answer 3

Maybe do it in three steps (assuming you have 88 fields in the CSV, because you said there are 23 commas before, and 65 after each second " )

Step 1 : replace all CR/LF with some character not anywhere in the file, like ~

Search: \\r\\n Replace: ~

Step 2 : replace all ~ after every 88th 'comma group' (or however many fields in CSV) with \\r\\n -- to reinsert the required CSV linebreaks:

Search: ((?:[^,]*?,){88})~ Replace: $1\\r\\n

Step 3 : replace all remaining ~ with space

Search ~ Replace: <space>