In vim I can use for instance 10j to go 10 lines down. And I can use . to repeat the last deletion.
Now, in bash script I have many commented lines like this:
# ... # ... # ... # ... # ... # ... # ... # ... # ... etc... Say, there are 52 such lines. Is there a way to combine moving 52j and repeating the deletion of the # via x and delete 52 lines at once?
I do approve @mMontu comment suggesting to use a comment plugin (vim-commentary is an option but NERDCommenter has my preference over vim-commentary).
But you could do it in other ways:
First if all your # are aligned on the same column you can block select them with ctrl+v52j and then delete your selection with x.
If they are not aligned you could use a macro:
qq^xjq qq Record a macro in the register q ^ Go to beginning of line x Suppress the first character j Go one line down q Stop recording You can then repeat your macro as many time as necessary with 52@q
And with NERDCommenter:
For more details:
I like the approaches statox suggested. Here's another one:
52:norm x This only works if the # is the first character on each line. Otherwise, I would do
52:s/# These two work very similarly. Essentially, what <count>: is doing, is setting up a range so that the next ex command is applied to the next <count> lines. norm x means
Press 'x' as if it was in normal mode
which would obviously remove the first character on that line.
s/# is just shorthand for :substitute/#// which will remove the first # character on each line.
Assumptions : commented lines are continuous and commenting character (# in your case) is at beginning of all those lines.
References: What's a quick way to comment/uncomment lines in Vim?
Namaste π
:scommand^v54jx(ctrl+v then 54jx), i.e. make a visual block on current column on all 54 lanes and delete it. Similarly, you can revert this with^v54I#<ESC>(or evengvI#<ESC>[gv = reopen previous visual mode])