undo with %s// replacement jump

Question

undo with %s// replacement jump

kyx0r opened this issue 3 years ago · comments

Currently if one does any successful %s// replacement and then uses u to undo it, the screen always jumps to line 0 of the open file. I don't think this makes sense and is sort of annoying. What it should do instead is jump to the first change that has been done by %s// in lbuf and of course undo the result, OR it should just stay in the same place, no jump. @aligrudi

Kyryl · Answer 1 · Thu Feb 25 2021 21:02:22 GMT+0800 (China Standard Time)

The easiest solution maybe to just write the current postion into * jump register right after the substitute command.

Kyryl · Answer 2 · Sat Feb 27 2021 06:42:13 GMT+0800 (China Standard Time)

I fixed it in my fork, here are the changes if it was hard to understand initially.
kyx0r/nextvi@0c7b420

Ali Gholami Rudi · Answer 3 · Wed Mar 03 2021 17:33:34 GMT+0800 (China Standard Time)

Kyryl <notifications@github.com> wrote:

I fixed it in my fork, here are the changes if it was hard to understand initially. kyx0r/nextvi@0c7b420

Thanks. Applied. Ali

Kyryl · Answer 4 · Thu Mar 04 2021 04:18:37 GMT+0800 (China Standard Time)

Thank you Ali. In case you did not notice, my commit on the fork had a typo where it should of been xrows, not xcols. I have fixed it next commit. I prefer the undo-redo to stay in place if the affected area is bigger than terminal rows, but that's just what I am doing, kinda settling on a middle ground between 2 possible solutions.