Insert a pattern before another one

Question

I have a line looking like this:

pattern1 any text pattern2 any text pattern2 any text pattern2 any text

And I want it to become:

pattern1 any text
pattern1 pattern2 any text
pattern1 pattern2 any text
pattern1 pattern2 any text

and I'll be applying this to thousands of lines

This command: %s/\(pattern1\).\{-}\zs.\(pattern2\)\|\(pattern2\)/\r\1 \2 \3/g

Is only producing:

pattern1 any text
pattern1 pattern2  any text
  pattern2 any text
  pattern2 any text

Because of the \| in the search string I think.

Any pointers?

Answer 1

That answer works for plain text, but not for patterns - you can't have patterns in the replace string unless you use an evaluated expression \=.

I've figured this out:

:%s/pattern2/\="\r" . matchstr(getline('.'), 'pattern1') . submatch(0)/g

So in my actual file, I'd use

pattern1 = .\{-}:  (everything up to the first colon)
pattern2 = \d\+ \a\+: (a date in the format dd mmm:)

resulting in

:%s/\d\+ \a\+:/\="\r" . matchstr(getline('.'), '.\{-}:') . submatch(0)/g

Answer 2

Why not just replace pattern2 by \n pattern1 pattern2?

:%s/pattern2/\rpattern1 pattern2/g

or without repeating pattern2:

:%s/\(pattern2\)/\rpattern1 \1/g

Both change

pattern1 any text pattern2 any text pattern2 any text pattern2 any text

to

pattern1 any text 
pattern1 pattern2 any text 
pattern1 pattern2 any text 
pattern1 pattern2 any text