I want search one invisible character(control character, such as 0x05) using less command in Linux.
I find the character is displayed as ^E in less. I can not search it by directly using the characters "^E" . Neither ctrl+vctrl+e in VI or just search by /\0x05.
CtrlE means something special at the beginning of the search field regardless of how it is entered. Type another non-"special" character first, then press the sequence, then remove the first character.
Related
I'm trying to convert multiple instances of Unicode codes to their corresponding characters.
I have some text with this format:
U+00A9
And I want to generate the following next to it:
©
I have tried to select the code in visual mode and use the selection range '<,'> in command mode as input for i_CTRL_V but I don't know how to use special keys on a command.
I haven't found anything useful in the manual with :help command-mode . I could solve this problem using other tools but I want to improve my vim knowledge. Any hint is appreciated.
Edit:
As #m_mlvx has pointed out my goal is to visually select, then run some command that looks up the Unicode and does the substitution. Manually input a substitution like :s/U+00A9/U+00A9 ©/g is not what I'm interested in as it would require manually typing each of the special characters on every substitution.
Any hint is appreciated.
Here are a whole lot of them…
:help i_ctrl-v is about insert mode and ranges matter in command-line mode so :help command-mode is totally irrelevant.
When they work on text, Ex commands only work on lines, not arbitrary text. This makes ranges like '<,'> irrelevant in this case.
After carefully reading :help i_ctrl-v_digit, linked from :help i_ctrl-v, we can conclude that it is supposed to be used:
with a lowercase u,
without the +,
without worrying about the case of the value.
So both of these should be correct:
<C-v>u00a9
<C-v>u00A9
But your input is U+00A9 so, even if you somehow manage to "capture" that U+00A9, you won't be able to use it as-is: it must be sanitized first. I would go with a substitution but, depending on how you want to use that value in the end, there are probably dozens of methods:
substitute('U+00A9', '\(\a\)+\(.*\)', '\L\1\2', '')
Explanation:
\(\a\) captures an alphabetic character.
+ matches a literal +.
\(.*\) captures the rest.
\L lowercases everything that comes after it.
\1\2 reuses the two capture groups above.
From there, we can imagine a substitution-based method. Assuming "And I want to generate the following next to it" means that you want to obtain:
U+00A9©
you could do:
v<motion>
y
:call feedkeys("'>a\<C-v>" . substitute(#", '\(\a\)+\(.*\)', '\L\1\2', '') . "\<Esc>")<CR>
Explanation:
v<motion> visually selects the text covered by <motion>.
y yanks it to the "unnamed register" #".
:help feedkeys() is used as low-level way to send a complex series of characters to Vim's input queue. It allows us to build the macro programatically before executing it.
'> moves the cursor to the end of the visual selection.
a starts insert mode after the cursor.
<C-v> + the output of the substitution inserts the appropriate character.
That snippet begs for being turned into a mapping, though.
In case you would like to just convert unicodes to corresponding characters, you could use such nr2char function:
:%s/U+\(\x\{4\}\)/\=nr2char('0x'.submatch(1))/g
Brief explanation
U+\(\x\{4\}\) - search for a specific pattern (U+ and four hexadecimal characters which are stored in group 1)
\= - substitute with result of expression
'0x'.submatch(1) - append 0x to our group (U+00A9 -> 0x00A9)
In case you would like to have unicode character next to text you need to modify slightly right side (use submatch(0) to get full match and . to append)
In case someone wonders how to compose the substitution command:
'<,'>s/\<[uU]+\(\x\+\)\>/\=submatch(0)..' '..nr2char(str2nr(submatch(1), 16), 1)/g
The regex is:
word start
Letter "U" or "u"
Literal "plus"
One or more hex digits (put into "capture group")
word end
Then substituted by (:h sub-replace-expression) concatenation of:
the whole matched string
single space
character by UTF-8 hex code taken from "capture group"
This is to be executed in Visual/command mode and works over selected line range.
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.
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 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 have a file that was converted from EBCDIC to ASCII. Where there used to be new lines there are now characters that show up as <85> (a symbol representing a single character, not the four characters it appears to be) and the whole file is on one line. I want to search for them and replace them all with new lines again, but I don't know how.
I tried putting the cursor over one and using * to search for the next occurrence, hoping that it might show up in my / search history. That didn't work, it just searched for the word that followed the <85> character.
I searched Google, but didn't see anything obvious.
My goal is to build a search and replace string like:
:%s/<85>/\n/g
Which currently just gives me:
E486: Pattern not found: <85>
I found "Find & Replace non-printable characters in vim" searching Google. It seems like you should be able to do:
:%s/\%x85/\r/gc
Omit the c to do the replacement without prompting, try with c first to make sure it is doing what you want it to do.
In Vim, typing :h \%x gives more details. In addition to \%x, you can use \%d, \%o, \%u and \%U for decimal, octal, up to four and up to eight hexadecimal characters.
For special character searching, win1252 for example, for the case of <80>,<90>,<9d>...
type:
/\%u80, \/%u90, /\%u9d ...
from the editor.
Similarly for octal, decimal, hex, type: /\%oYourCode, /\%dYourCode, /\%xYourCode.
try this: :%s/<85>/^M/g
note: press Ctrl-V together then M
or if you don't mind using another tool,
awk '{gsub("<85>","\n")}1' file