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Asked: 2026-06-17T19:41:40+00:00

I’ve got a CSV file with some 600 records where I need to replace

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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.

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1 Answer

  1. Editorial Team

    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]+([^"]*+")

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