In vi (from cygwin), when I do searching:
:%s/something
It just replaces the something with empty string like
:%s/something// .
I've googled for a while but nothing really mentions this. Is there anything I should add to the .vimrc or .exrc to make this work?
Thanks!
In vi and vim, when you search for a pattern, you can search it again by simply typing /. It is understood that the previous pattern has to be used when no pattern is specified for searching.
(Though, you can press n for finding next occurence)
Same way, when you give a source (pattern) and leave the replacement in substitute command, it assumes that the replacement is empty and hence the given pattern is replaced with no characters (in other words, the pattern is removed)
In your case, you should understand that % stand for whole file(buffer) and s for substitute. To search, you can simply use /, followed by a pattern. To substitute , you will use :s. You need not confuse searching and substituting. Hence, no need for such settings in ~/.exrc. Also, remember that / is enough to search the whole buffer and % isnt necessary with /. / searches the entire buffer implicitly.
You may also want to look at :g/Pattern/. Learn more about it by searching :help global or :help :g in command line.
The format of a substitution in vim is as follows:
:[range]s[ubstitute]/{pattern}/{string}/[flags] [count]
In your case you have omitted the string from the substitution command and here what vim documentation stated about it:
If the {string} is omitted the substitute is done as if it's empty.
Thus the matched pattern is deleted. The separator after {pattern}
can also be left out then. Example: >
:%s/TESTING This deletes "TESTING" from all lines, but only one per line.
For compatibility with Vi these two exceptions are allowed:
"/{string}/" and "\?{string}?" do the same as "//{string}/r".
"\&{string}&" does the same as "//{string}/".
E146
Instead of the '/' which surrounds the pattern and replacement string, you can
use any other single-byte character, but not an alphanumeric
character, '\', '"' or '|'. This is useful if you want to include a
'/' in the search pattern or replacement string. Example: >
:s+/+//+
In other words :%s/something and :%s;something or :%s,something have all the same behavior because the / ; and , in the last examples are considered only as SIMPLE SEPARATOR
Related
How to search for ** in a file using vim like **kwargs , when I use a /** it highlights the entire file.
* is a meta-character that needs to be escaped to become a regular character:
/\*\*
Vim refers to this as "magic": if the pattern is "magic", you need to escape meta-characters, if it is "nomagic", you don't (this is a bit over-simplified, see :help magic).
Therefore, you can force Vim to treat the pattern as "nomagic" with \M or even "very nomagic", with \V:
/\V**
Which doesn't help much, here, but well…
As for why /** highlights the whole buffer, a hint can be found under :help /star:
Exception: When "*" is used at the start of the pattern or just after
"^" it matches the star character.
The first * matches (at least some of) the stars in your buffer,
The second * matches zero or more of the preceding atom, as many as possible, so, essentially, every character.
You have to escape the * symbol, try
/\*\*
On the text below I added extra spaces to be easier to read
#Don't use this one / \* \*
I have a bunch of t_ prefixed fonction names in a header file and i would like to remove this prefix from some of them.
I have ended by typing :
:s/\(\s+\)t_\([^(]+\)(/\1\2(/c
But vim complains with Pattern not found
What is wrong in my pattern ?
You forgot to put a backslash before + to give it its quantifier meaning. You can also probably simplify it using \< to match the start of a word instead of capturing spaces.
I suggest breaking the problem down, first the search:
/\v\s+\zst_\ze.*\(
Now the substitution:
:s///\c
We can reuse a search pattern simply typing an empty search on the substitution
OBS: I suppose your functions have () at the end of the line, so .*().
Another thing; \zs and \ze to match only t_, for more see: :h \zs.
If you are using neovim you can see what happens befor hitting enter, just put these lines on your init.vim:
if has("nvim")
set inccommand=nosplit
endif
When i need to replace a string with a new string in vim.
First i would use search-mode to check that the search pattern is correct.
/search pattern
Then use the 's' command to do substitution.
:%s/search pattern/new string/
The search pattern need to type twice. If it is too complex, it would be boring.
Is there a method to avoid this?
You can simply omit the pattern in the substitution command, e.g.
:%s//new string/
This is documented in :help last-pattern (emphasis mine):
The last used pattern and offset are remembered. They can be used to
repeat the search, possibly in another direction or with another
count. Note that two patterns are remembered: One for 'normal' search
commands and one for the substitute command ":s". Each time an empty
pattern is given, the previously used pattern is used.
Also (in addition to Marco Baldelli's correct answer), the last search pattern is stored in the special register /. You can insert this in the command-line via Ctrl + R, followed by /. (This also works in insert mode, and also with other registers.) It's helpful when you want to tweak your search pattern before substituting.
The search command in vim allows you to place the cursor relative to the search results. For example, /hello/b+2 places the cursor on the first l.
How do I do that with the substitute command?
s/hello/b+2/_/
does not work.
I need this to replace not the entire search string, but a portion of it only (specifically, to blank out all but the first character of a word).
You generally have two options: similar to other regex engines zero-width matches (though with different syntax):
:s/\(he\)\#<=llo/_/
or vim-specific “set the start of the match here”:
:s/he\zsllo/_/
. Also, there is a workaround which will look similar in almost every other regex engine:
:s/\(he\)llo/\1_/
: this captures text that should be unchanged and makes replacement include it.
I'm trying to search and replace $data['user'] for $data['sessionUser'].
However, no matter what search string I use, I always get a "pattern not found" as the result of it.
So, what would be the correct search string? Do I need to escape any of these characters?
:%s/$data['user']/$data['sessionUser']/g
:%s/\$data\[\'user\'\]/$data['sessionUser']/g
I did not test this, but I guess it should work.
Here's a list of all special search characters you need to escape in Vim: `^$.*[~)+/
There's nothing wrong with with the answers given, but you can do this:
:%s/$data\['\zsuser\ze']/sessionUser/g
\zs and \ze can be used to delimit the part of the match that is affected by the replacement.
You don't need to escape the $ since it's the at the start of the pattern and can't match an EOL here. And you don't need to escape the ] since it doesn't have a matching starting [. However there's certainly no harm in escaping these characters if you can't remember all the rules. See :help pattern.txt for the full details, but don't try to digest it all in one go!
If you want to get fancy, you can do:
:%s/$data\['\zsuser\ze']/session\u&/g
& refers to the entire matched text (delimited by \zs and \ze if present), so it becomes 'user' in this case. The \u when used in a replacement string makes the next character upper-case. I hope this helps.
Search and replace in vim is almost identical to sed, so use the same escapes as you would with that:
:%s/\$data\['user'\]/$data['session']/g
Note that you only really need to escape special characters in the search part (the part between the first set of //s). The only character you need to escape in the replace part is the escape character \ itself (which you're not using here).
The [ char has a meaning in regex. It stands for character ranges. The $ char has a meaning too. It stands for end-line anchor. So you have to escape a lot of things. I suggest you to try a little plugin like this or this one and use a visual search.