Say I have a line that looks like this:
(Type)obj.method());
Is there a motion that will go to the last of a specific character on a line? I'd like to do
^c (motion to ')' character) (insert text) <Esc>
and transform that line to
obj.otherMethod());
I know I can use #f or #t, but I'd rather not count the parens.
As I understand your question, you don't just want to go to the last instance of a character, which you could achieve by going to the end of the line and searching backward. You want to clear text from your current caret position to the last instance of a character, right?
I'd typically use a pattern search to complete a motion to a desired character when there may be n of the same character in between. In your example, you can clear from the current caret position to the last ) by using c/);Enter, since only the last instance of o is followed by ;. You could precede this with a v instead of c, for example, if you wanted to select everything in between.
Using the same example, you could move to the 3rd o with /odEnter.
It may seem tedious at first, but in practice you are probably looking directly at the spot where you want be, so you can already see the additional characters you need and you only need to increase you specificity until you get the match, and you'll have immediate visual feedback as long as you have set incsearch.
Note: If your line did not have a semicolon at the end, you could move to the last paren by using )\n to search for the next ) followed by a line break.
You could use visual mode:
v$F)c
But in this case you're really just inserting some new text and changing the case:
f.aotherEscl~
Related
Let's say I've typed "abcdefg", with the cursor at the end. I want to delete back to the c, so that I only have "abc" left.
Is there a command like d that includes the current character? I know I could do dTcx, but the x feels like a work-around and I suppose there's a better solution.
No. Backward motions always start on the left of the current character for c, y and d which is somehow logical but also unnerving.
The only "clean" solutions I could think of either imply moving to the char after c first and then do a forward delete:
Tcde
or using visual mode:
vTcd
v3hd
But, given your sample and assuming you are entering normal mode just for that correction, the whole thing sounds extremely wasteful to me.
What about staying in insert mode and simply doing ←←←←?
try this:
TcD
this will leave abc for your example... well if the abcdefg is the last word of the line.
if it is not the last word in that line, you may do:
ldTc
or golfing, do it within 3 key-stroke:
3Xx or l4X
See this answer to a similar question : there is a setting to be allowed to go beyond the end of the line
From the doc :
Virtual editing means that the cursor can be positioned where there is
no actual character. This can be halfway into a tab or beyond the end
of the line. Useful for selecting a rectangle in Visual mode and
editing a table.
"onemore" is not the same, it will only allow moving the cursor just
after the last character of the line. This makes some commands more
consistent. Previously the cursor was always past the end of the line
if the line was empty. But it is far from Vi compatible. It may also
break some plugins or Vim scripts. For example because |l| can move
the cursor after the last character. Use with care!
Using the $ command will move to the last character in the line, not
past it. This may actually move the cursor to the left!
The g$ command will move to the end of the screen line.
It doesn't make sense to combine "all" with "onemore", but you will
not get a warning for it.
In short, you could try :set virtualedit=onemore, and see if your environment is stable or not with it.
Use d?c
That will start d mode, search back to 'c' and then delete up to your cursor position.
Edit: nope, that does not include current position...
I may be misunderstanding your request, but does 3hd$ do it?
I would use vFdd in this example. I think it's nicer than the other solutions since the command explicitly shows what to delete. It includes the current character and the specified character when deleting.
v: enter visual mode (mark text)
F: find/goto character backwards
d: the character "d" that will be included for removal.
d: delete command
Since it is visual mode, the cursor can also be moved before executing the actual removal d. This makes the command powerful even for deleting up to a non unique character by first marking a special character close to the character and then adjusting the position.
Let's say I have the following line of code:
something:somethingElse:anotherThing:woahYetAnotherThing
And I want to replace each : with a ; except the first one, such that the line looks like this:
something:somethingElse;anotherThing;woahYetAnotherThing
Is there a way to do this with the :[range]s/[search]/[replace]/[options] command without using the c option to confirm each replace operation?
As far as I can tell, the smallest range that s acts on is a single line. If this is true, then what is the fastest way to do the above task?
I'm fairly new to vim myself; I think you're right about range being lines-only (not 100% certain), but for this specific example you might try replacing all of the instances with a global flag, and then putting back the first one by omitting the global -- something like :s/:/;/g|s/;/:/.
Note: if the line contains a ; before the first : then this will not work.
Here you go...
:%s/\(:.*\):/\1;/|&|&|&|&
This is a simple regex substitute that takes care of one single not-the-first :.
The & command repeats the last substitute.
The | syntax separates multiple commands on one line. So, each substitute is repeated as many times as there are |& things.
Here is how you could use a single keystroke to do what you want (by mapping capital Q):
map Q :s/:/;/g\|:s/;/:<Enter>j
Every time you press Q the current line will be modified and the cursor will move to the next line.
In other words, you could just keep hitting Q multiple times to edit each successive line.
Explanation:
This will operate globally on the current line:
:s/:/;/g
This will switch the first semi-colon back to a colon:
:s/;/:
The answer by #AlliedEnvy combines these into one statement.
My map command assigns #AlliedEnvy's answer to the capital Q character.
Another approach (what I would probably do if I only had to do this once):
f:;r;;.
Then you can repeatedly press ;. until you reach the end of the line.
(Your choice to replace a semi-colon makes this somewhat comfusing)
Explanation:
f: - go to the first colon
; - go to the next colon (repeat in-line search)
r; - replace the current character with a semi-colon
; - repeat the last in-line search (again)
. - repeat the last command (replace current character with a semi-colon)
Long story short:
fx - moves to the next occurrence of x on the current line
; repeats the last inline search
While the other answers work well for this particular case, here's a more general solution:
Create a visual selection starting from the second element to the end of the line. Then, limit the substitution to the visual area by including \%V:
:'<,'>s/\%V:/;/g
Alternatively, you can use the vis.vim plugin
:'<,'>B s/:/;/g
Using vim I would like to replace all characters up to a certain one with another character, say a blank space - without affecting the layout/number of characters in the line. Here's an example:
Before:
real(kind=R12), intent(out) :: my_var
After replacing , intent(out) with blanks (i.e. starting from ,, and going up to )):
real(kind=R12) :: my_var
I know about r to replace one character, and about nr to replace n characters, but I would like to know whether I can accomplish my task without first having to count the characters I want to replace.
Thanks a lot for your replies!
Visual mode is probably the shortest way here:
vt:r
v enter visual mode
t: select till :
r (note space after r) replace selected region with spaces.
In command mode type 'df?' to delete up to that (?) character. Then 'i' to go back to insert.
For example if the following sentence is in your view:
The wizard quickly jinxed the gnomes before they vaporized.
and you enter dfs
You will be left with:
before they vaporized.
I know about r to replace one character
Did you know that R will keep you in that replace mode? So you could hit R and then hold Space until you've replaced everything you want.
However, I'd still go with Thor's answer. Visual mode allows you to use the efficient text navigation methods in vim without having to count out characters.
But if you disagree, there's always EasyMotion.
You can use regular expression here (use (.*?) to reference all values up to a token).
For instance:
The regex: (.*?)foo will get rid of everything up to foo.
I have the following characters being repeated at the end of every line:
^[[00m
How can I remove them from each line using the Vim editor?
When I give the command :%s/^[[00m//g, it doesn't work.
You could use :%s/.\{6}$// to literally delete 6 characters off the end of each line.
The : starts ex mode which lets you execute a command. % is a range that specifies that this command should operate on the whole file. The s stands for substitute and is followed by a pattern and replace string in the format s/pattern/replacement/. Our pattern in this case is .\{6}$ which means match any character (.) exactly 6 times (\{6}) followed by the end of the line ($) and replace it with our replacement string, which is nothing. Therefore, as I said above, this matches the last 6 characters of every line and replaces them with nothing.
I would use the global command.
Try this:
:g/$/norm $xxxxxx
or even:
:g/$/norm $5Xx
I think the key to this problem is to keep it generic and not specific to the characters you are trying to delete. That way the technique you learn will be applicable to many other situations.
Assuming this is an ANSI escape sequence, the ^[ stands for a single <Esc> character. You have to enter it by pressing Ctrl + V (or Ctrl + Q) on many Windows Vim installations), followed by Esc. Notice how this is then highlighted in a slightly different color, too.
It's easy enough to replace the last six characters of every line being agnostic to what those characters are, but it leaves considerable room for error so I wouldn't recommend it. Also, if ^[ is an escape character, you're really looking for five characters.
Escape code
Using ga on the character ^[ you can determine whether it's an escape code, in which case the status bar would display
<^[> 27, Hex 1b, Octal 033
Assuming it is, you can replace everything using
:%s/\%x1b\[00m$//gc
With \%x1b coming from the hex value above. Note also that you have to escape the bracket ([) because it's a reserved character in Vim regex. $ makes sure it occurs at the end of a line, and the /gc flags will make it global and confirm each replacement (you can press a to replace all).
Not escape code
It's a simple matter of escaping then. You can use either of the two below:
:%s/\^\[\[00m$//gc
:%s/\V^[[00m\$//gc
If they are all aligning, you can do a visual-block selection and delete it then.
Otherwise, if you have a sequence unknown how to input, you can visually select it by pressing v, then mark and yank it y (per default into register "). Then you type :%s/<C-R>"//g to delete it.
Note:
<C-R>" puts the content of register " at the cursor position.
If you yanked it into another register, say "ay (yank to register a - the piglatin yank, as I call it) and forgot where you put it, you can look at the contents of your registers with :reg.
<C-R> is Vim speak for Ctrl+R
This seems to work fine when the line is more than 5 chars long:
:perldo $_ = substr $_, 0, -5
but when the line is 5 or less chars long it does nothing.
Maybe there is a easy way in perl to delete the last 5 chars of a string, but I don't really know it:)
Use this to delete:
:%s/^[[00m//gc
I'm learning the power of g and want to delete all lines containing an expression, to the end of the sentence (marked by a period). Like so:
There was a little sheep. The sheep was black. There was another sheep.
(Run command to find all sentences like There was and delete to the next period).
The sheep was black.
I've tried:
:g/There was/d\/\. in an attempt to "delete forward until the next period" but I get a trailing characters error.
:g/There was/df. but get a df. is not an editor command error.
Any thoughts?
The action associated with g must be able to act on the line without needing position information from the pattern match that g implies. In the command you are using, the delete forward command needs a starting position that is not being provided.
The problem is that g only indicates a line match, not a specific character position for it's pattern match. I did the following and it did what I think you want:
:g/There was/s/There was[^.]*[.]//
This found lines that matched the pattern There was, and performed a substitution of the regular expression There was[^.]*[.] with the empty string.
This is equivalent to:
:1,$s/There was[^.]*[.]//g
I'm not sure what the g is getting you in your use case, except the automatic application to the entire file line range (same as 1,$ or %). The g in this latter example has to do with applying the substitution to all patterns on the same line, not with the range of lines affected by the substitution command.
I'd just use a regex:
%s/There was\_.\{-}\.\s\?//ge
Note how \_. allows for cross-line sentences
You can use :norm like this:
:g/There was/norm 0weldf.
This finds lines with "There was" then executes the normal commands 0weldf..
0: go to beginning of line
w: go to next word (in this case, "was")
e: go the end of the word (so cursor is on the 's' of "was")
l: move one character to the right (so we don't delete any of "was")
df.: delete until the next '.', inclusive.
If you want to keep the period use dt. instead of df..
If you don't want to delete from the beginning of the line and instead want to do sentences, the :%s command is probably more appropriate here. (e.g. :%s/\(There was\)[^.]*\./\1/g or %s/\(There was\)[^.]*\./\1./g if you want to keep the period at the end of the sentence.
Use search and replace:
:%s/There was[^.]*\.\s*//g