Notepad ++中的RegEx查找少于或多于n个管道的行

Question

I have a large pipe-delimited text file that should have one 3-column record per line. Many of the records are split up by line breaks within a column.

I need to do a find/replace to get three, and only three, pipes per line/record.

Here's an example (I added the line breaks ( \\r\\n ) to demonstrate where they are and what needs to be replaced):

12-1234|The quick brown fox jumped over the lazy dog.|Every line should look similar to this one|\r\n
56-7890A|This record is split\r\n
\r\n
on to multiple lines|More text|\r\n
09-1234AS|\r\n
||\r\n
\r\n
56-1234|Some text|Some more text\r\n
|\r\n
76-5432ABC|A record will always start with two digits, a dash and four digits|There may or may not be up to three letters after the four digits|\r\n

The caveat is that I need to retain those mid-record line breaks for the target system. They need to be replaced with \\.br\\ . So the final result of the above should look like this:

12-1234|The quick brown fox jumped over the lazy dog.|Every line should look similar to this one|\r\n
56-7890A|This record is split\.br\\.br\on multiple lines|More text|\r\n
09-1234AS|\.br\||\.br\\r\n
56-1234|Some text|Some more text\.br\|\r\n
76-5432ABC|A record will always start with two digits, a dash and four digits|There may or may not be up to three letters after the four digits|\r\n

As you can see the mid-record line breaks have all been replaced with \\.br\\ and the end-of-line line breaks have been retained to keep each three-column/pipe record on its own line. Note the last record's text, explaining how each line/record begins. I included that in case that would help in building a regex to properly identify the beginning of a record.

I'm not sure if this can be done in one find/replace step or if it needs to be (or just should be) split up into a couple of steps.

I had the thought to first search for |\\r\\n , since all records end with a pipe and a CRLF , and replace those with dummy text !@#$ . Then search for the remaining line breaks with \\r\\n , which will be mid-column line breaks and replace those with \\.br\\ , then replace the dummy text with the original line breaks that I want to keep |\\r\\n .

That worked for all but records that looked like the third record in the first example, which has several line breaks after a pipe within the record. In such a large file as I am working with it wasn't until much later that I found that the above process I was using didn't properly catch those instances.

Answer 1

You can use

(?:\G(?!^(?<!.))|^\d{2}-\d+[A-Z]*\|[^|]*?(?:\|[^|]*?)?)\K\R+

Replace with \\\\.br\\\\ . See the regex demo . Details :

• (?:\\G(?!^(?<!.))|^\\d{2}-\\d+[AZ]*\\|[^|]*?(?:\\|[^|]*?)?) - either the end of the previous match ( \\G(?!^(?<!.)) ) or ( | ) start of a line, two digits, 0 , one or more digits, zero or more letters, a | , then any zero or more chars other than | , as few as possible, and then an optional sequence of | and any zero or more chars other than |, as few as possible (see ^\\d{2}-\\d+[AZ]*\\|[^|]*?(?:\\|[^|]*?)? )
• \\K - omit the text matched
• \\R+ - one or more line breaks.

See the Notepad++ demo:

If you need to remove empty lines after this, use Edit > Line Operations > Remove Empty Lines .