Regex - find if beginning of line contains a string. If so replace with backspace-CodePudding

I am attempting to write a regular expression to help locate some carriage return issues. I generated a string at the beginning of each line that I know should start that line. So I can assume that if the line does not contain that string there was a carriage return.

I was able to do it like so:

\n[^SOMEWORD]

the \n also gets the previous lines space so in the Replace With section I would like to replace that with a backspace so that line will be moved back up but retain that data.

SOMEWORD,111111,22222222,null,null,"1333333
4444",55555,66,null,7777777,8888 99999,null,null,null,0,null,false,false,1212121,null,null,null,false,null,null,333333,null

SOMEWORD,111111,22222222,null,null,"1333333
4444",55555,66,null,7777777,8888 99999,null,null,null,0,null,false,false,1212121,null,null,null,false,null,null,333333,null

would look like:

SOMEWORD,111111,22222222,null,null,"1333333, 4444",55555,66,null,7777777,8888 99999,null,null,null,0,null,false,false,1212121,null,null,null,false,null,null,333333,null
SOMEWORD,111111,22222222,null,null,"1333333, 4444",55555,66,null,7777777,8888 99999,null,null,null,0,null,false,false,1212121,null,null,null,false,null,null,333333,null

CodePudding user response：

You can search using this regex:

(?m)\n(?!SOMEWORD,|$)

And replace with ", " string.

RegEx Demo

RegEx Details:

(?m): Enable multiline mode
\n: Match a line break
(?!SOMEWORD,|$): Assert that there is no SOMEWORD, or an empty line ahead