how to replace a current line with what in a buffer using vi

Question

Say i have line stored in buffer k. how do I replace some line with the content of the buffer?

Answer 1

You can use ctrl-v and select what you want to copy and press "y" to "yank" it. Then ctrl-v or shift-v to select the "some lines" that you want to replace and press "p" to paste it.

Answer 2

go anywhere on the line to be replaced. Execute a buffer P (Put above). Use dd to delete the current line.

so "xPdd

Answer 3

The quote key " is what you need. That makes your yank/put register specific. So you have something in register k and you wanted to replace the current line with it you'd type:

^c$<esc>"kp

Answer 4

Go on the line you wish to change, and execute

V"kp

Answer 5

best way i can think on the spot is

"ayy (this yanks / copies the line to the "a buffer

then

dd (delete the line to the standard buffer)

then

"aP which inserts buffer"a before the current line

Answer 6

As others have said, the overall answer is to use dd"kP. I'd like to add that you might want to use :g, so that if you want to replace all lines that match 'foo' with the content of register k, you can do:

:g/foo/normal dd"kP

Note that using p instead of P will cause some problems if the first line of your buffer matches the pattern.