I know that in command mode, ct{char} removes all text from current cursor position till next matching {char}, which can then be replaced by continuing to type.
For example, suppose the text is:
abcdefgh
and cursor is on b, then typing ctg will remove bcdef, which can be replaced by continuing to type.
But this works only for a single matching character. Is it possible to do this by matching a character sequence, for example, using gh instead of g, in above example?
Note: I know that the s/// could be used, but this is a little faster, and more convenient.
Yes. Use the search command:
c/gh
The search is a motion for the previous command. So it will delete characters until if finds a gh string.
Related
Please let me know, How I can remove the last word from each line using vim commands?
Before :
abcdef 123
xyz 1256
qwert 2
asdf 159
after :
abcdef
xyz
qwert
asdf
Similarly please let me know how to remove the second word from each line using vim command?
Use the following regex (in ex mode):
%s/\s*\w\+\s*$//
This tells it to find optional whitespace, followed by one or more word characters, followed by optional whitespace, followed by end of line—then replace it with nothing.
The question's been answered already, but here's what I'd more likely end up doing:
Record a macro:
qq to record a macro into register "q"
$ to go to the end of the line
daw to delete a word
q to stop recording
Then select the rest of the lines:
j to go down a line
vG to select to the end of the file
And apply the macro:
:norm #q
Some similar alternatives:
:%norm $daw
qq$dawjq (note the added j) then 999#q to replay the macro many times. (Macro execution stops at the first "error" -- in this case, you'd probably hit the bottom of the file, j would not work, and the macro would stop.)
The key for this is the :substitute command; it is very powerful (and often used in vi / Vim).
You need to come up with a regular expression pattern that matches what you want to delete. For the last word, that's whitespace (\s), one or more times \+ (or any number (*), depending on how you want to treat single-word lines), followed by word characters (\w\+), anchored to the end of the line ($). Note that word has a special meaning in Vim; you may want to use a different atom (e.g. \S). Voila:
:%s/\s\+\w\+$//
For the second word, you can make use of the special \zs and \ze atoms that assert for matches, but do not actually match: Anchored at the start (^), match a word, then start the match for a second one:
:%s/^\w\+\s\+\zs\w\+\s\+//
Soon, you'll also want to reorder things, not just remove them. For that, you need to know capturing groups: \(...\). The text matched by those can then be referred to in the replacement part. For example, to swap the first and second words:
:%s/^\(\w\+\s\+\)\(\w\+\s\+\)/\2\1/
For details, have a look at the help, especially :help :substitute and :help pattern.
To remove the second word from the start of a line, use the following:
:%s/^\(\s*\w\+\s\+\)\w\+\s*/\1/
Update
To treat special characters as part of the word, you have to use the \S (which matches all non-whitespace characters) instead of \w (which matches only word characters [0-9A-Za-z_]). Then, the command would be:
:%s/^\(\s*\S\+\s\+\)\S\+\s*/\1/
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$F)c
But in this case you're really just inserting some new text and changing the case:
f.aotherEscl~
I have the following characters being repeated at the end of every line:
^[[00m
How can I remove them from each line using the Vim editor?
When I give the command :%s/^[[00m//g, it doesn't work.
You could use :%s/.\{6}$// 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
I’m writing a function that edits a certain environment in LaTeX.
The environment basically looks like this:
\begin{quicktikz}
...some stuff...
\end{quicktikz}
or like this:
\begin*{quicktikz}
...some stuff...
\end{quicktikz}
I want to write a function that toggles between the two, when called from within the environment. Since my Vim knowledge ain’t all that, I’m coming up with a simple solution:
Get cursor position with let save_cursor=getpos(".").
Backward search for \begin{quicktikz} using: ?\\begin{quicktikz}\|\\begin\*{quicktikz}.
Search for the { and move left using: normal 0f{h.
Check if the item under cursor equals *:
if it does, do normal x;
if it doesn’t, do normal a*<esc>.
Restore cursor position using call setpos('.',save_cursor).
I know how to do all of this except for step 3. How can I check if the char under the cursor equals to * or not?
If you know a better way of doing this, sharing this would be welcome.
I think the easiest way to retrieve the char under cursor is:
getline(".")[col(".")-1]
Alternatively, you can do it with strpart()
strpart(getline("."), col(".")-1, 1)
The first expression first calls the function getline() passing "." as
argument which means the line where the cursor is positioned will be returned.
Then we use the so called expr8 or expr-[] (see the help) to retrieve a
single character. The number passed comes from another function, col()
which returns the current cursor column. As indexes start in 0, one is subtracted.
You can use it like
if getline(".")[col(".")-1] == '*'
...
Let me propose an alternative implementation of the technique you describe.
:?\\begin\>\zs\*\=?s//\='*'[submatch(0)!='']/|norm!``
The above command consists of two separate commands chained with | (see
:help :bar) in one line. The first one is a substitution (see :help :s)
performed for each line in the specified range,
?\\begin\>\zs\*\=?
According to the range syntax (see :help :range), this range specifies the
only line, that is the previous line where the \\begin\>\zs\*\= pattern
matches the word begin preceded with a backslash and followed by by optional
star character.1 The \zs atom between parts of the pattern
matching \begin and *, sets the start of the match there. So, the match
of the whole pattern is either empty or contains single star character. This
is not necessary for specifying a line in the range, it is useful for reusing
the same pattern later in the :substitute command, where only that star
character or its empty place should be replaced. For details about the
pattern's syntax see :help /\>, :help /\=, :help /\zs.
The substitution itself,
s//\='*'[submatch(0)!='']/
replaces the first occurrence of the last search pattern (which is set by the
backward search in the range) with a string to which the expression
'*'[submatch(0)!=''] evaluates (see :help sub-replace-\=). As the pattern
matches only an empty string or a star character, the subexpression
submatch(0)!='' evaluates to zero if there is no star after \begin, or to
one otherwise. Zero subscript from the string '*' results in a substring
containing the first character of that one-character string. Index one is
equal to the length of the string, therefore subscript results in an empty
string. Thus, when there is a star after \begin, it gets replaced with an
empty string, when a star is not present, zero-width interval just after
\begin is substituted with *.
The second command,
:norm!``
takes advantage of the fact that :substitute command stores the current
cursor position before it actually starts replacement. The `` movement
command jumps back to the position before the latest jump (which occurs in the
aforementioned substitution command) restoring position of the
cursor.2
1 Be careful with search, since in ranges, as usual, it wraps
around the end of file, when the wrapscan option is enabled (it is turned on
by default).
2 Do not confuse `` with the '' command which moves the
cursor to the first non-blank character in the line of the location before the
latest jump.