Regular Expression to Catch multi-line quote

Question

...,"My quote goes on
to multiple lines
like this",...

How would I catch this in a regular expression? I want to do this in a substitution to end up with

....,"My quote goes on to multiple lines like this",...

I tried

"(?<!\")\r\n(?!\")"

This was in an attempt to find a newline that does NOT end with a quote, and the next line does not start with a quote either.

The following substitution was done in R using that regular expression with no luck...

newDF = gsub( "(?<!\")\r\n(?!\")", " ", newDF, perl = TRUE)

Answer 1

You can match a quoted substring and then use gsubfn to replace linebreaks inside the quoted substrings only:

library(gsubfn)
s = "...,\"My quote goes on\r\nto multiple lines\r\nlike this\",..."
gsubfn("\"[^\"]+\"", function(x) gsub("(?:\r?\n)+", " ", x), s)
[1] "...,\"My quote goes on to multiple lines like this\",..."

The "[^"]+" pattern matches all quoted substrings, and then (?:\\r?\\n)+ matches 1 or more sequences of an optional CR ( \\r? ) followed with 1 LF (that are replaced with a space).

Alternatively, you can achieve a similar result with a PCRE regex like

gsub("(?:\r?\n)+(?!(?:[^\"]|\"[^\"]*\")*$)", " ", s, perl=T)
[1] "...,\"My quote goes on to multiple lines like this\",..."

See the regex demo . The (?!(?:[^\\"]|\\"[^\\"]*\\")*$) lookahead makes sure there are no even quotes up to the string end.

Answer 2

> x <- "My quote goes on
+ to multiple lines
+ like this"

> gsub("\\n", " ", x)
[1] "My quote goes on to multiple lines like this"

Don't forget to double the backslashes.