I have some ruby code that goes like
some_hash = {
hello: world,
goodbye: moon
}
some_hash.each do |key, value|
# process
end
I want to know how to get searchpair to work to match the '|' character. Every time I execute echo searchpair('|', '', '|', 'W') inside the '|' characters, vim returns 0.
As #Amadan said, searchpair does not work when start and end are equal. The main goal of this was to create operator-pending mappings to match between and around two |. I ended up creating a generic solution that allows me to match on equal patterns on the same line. You can find the code here.
Related
I am parsing a tcl file line line by line and searching for lines with open curly braces so that I can merge them with the next line and read them.
I am struggling to get a single regex to do this. My concern is lines with with a closing } which can be skipped.
Example:
MATCH: test_command -switch1 {
NO MATCH: single_command
NO MATCH: test_tcl -switch2 {arg1 }
Please help with the regex to get the result. I tried this:
% set a "test_command -swithc1 {bye }"
test_command -swithc1 {bye }
% regexp "{" $a match
1
#0 is expected
This is not my intention. I want match only for lines with open curly brace
% set b "test_command -swithc1 {hi"
test_command -swithc1 {hi
% regexp "{" $a match
1
#1 was expected
I'm looking for a regex that will give 0 for the $a and 1 for $b
You really shouldn't be using a regular expression for that; there's a Tcl command specifically for this sort of thing: info complete. Here's how to use it:
set accumulator ""
while {![eof $inputChannel]} {
# Note well: you *must* add the newline
append accumulator [gets $inputChannel] "\n"
if {[info complete $accumulator]} {
handleCompleteChunk $accumulator
set accumulator ""
}
}
This handles various types of bracket matching and the intricacies of backslash sequences, but just to check whether the “line” is complete. (It's also the core of how Tcl's REPL works, except that uses the Tcl C API equivalents.)
You could try a couple "lookarounds", one to say "I see a left bracket" and one to say "I don't see a right bracket":
(?!.*\})(?=.*\{)
https://regex101.com/r/p8bbsF/1/
I need to find all pairs of strings that have the same pattern.
For example:
another string, that is not interesting
la-di-da-di __pattern__ -di-la-di-la
la-di-da-da-di-la __pattern__ -la-da-li-la
and yet another usual string
So I want to delete strings with __pattern__ inside.
I don't know how to do it just with builtin commands and now I have the function, that doesn't work properly:
function! DelDup(pattern)
echom a:pattern
redir => l:count
execute "normal! :%s/a:pattern//n\<cr>"
redir END
echo l:count
endfunction
Here I try to run ":%s/a:pattern//n" to find the count of occurrences of pattern in the text.
And at the same time I try to put it into the variable "l:count".
Then I tried to echo the count I got, but nothing happens when I try to do it.
So the last my problem in function writing is that I can't write the command execution result to variable.
If you have another solution -- please describe it to me.
Update:
Excuse me for bad description. I want to delete only strings, that has pattern-twins in text.
I'm not sure if I understand your question correctly, but I'm assuming you want to remove all lines where there are at least 2 matches. If that's the case you can use the following command:
:g/\(__pattern__.*\)\{2,}/d
How this works is that it deletes all the lines where there is a match (:g/../d).
The pattern is made up of a group (\(..\)) which needs to be matched at least 2 times (\{2,}). And the pattern has a .* at the end so it matches everything between the matches of the pattern.
There are many ways to count occurrences of a pattern, and I'm quite sure there exist a Q/A on the subject. Let's do it yet another way and chain with the next step. (Yes this is completely obfuscated, but it permits to obtain programmatically the information without the need to parse the localized result of :substitute after redirection.)
" declare a list that contain all matches
let matches = []
" replace each occurrence of the "pattern" with:
" the result of the expression "\=" that can be
" interpreted as the last ([-1]) element of the
" list "matches" returned by the function (add)
" that adds the current match (submatch(0)) to the
" list
:%s/thepattern/\=add(matches, submatch(0))[-1]/gn
" The big caveat of this command is that it modifies
" the current buffer.
" We need something like the following to leave it unmodified:
:g/thepattern/call substitute(getline('.'), 'thepattern', '\=add(counter, submatch(0))[-1]', 'g')
" Note however that this flavour won't work with multi-lines patterns
" Now you can test the number of matches or do anything fancy with it
if len(matches) > 1
" replaces matches with nothing
:%s/thepattern//g
endif
Only if you want to define this as a function you'll need to play with:
exe 'normal :%s/'.escape(a:pattern, '/\').'/replacement..../flags....'
I would like to use vim's substitute function (:%s) to search and replace a certain pattern of code. For example if I have code similar to the following:
if(!foo)
I would like to replace it with:
if(foo == NULL)
However, foo is just an example. The variable name can be anything.
This is what I came up with for my vim command:
:%s/if(!.*)/if(.* == NULL)/gc
It searches the statements correctly, but it tries to replace it with ".*" instead of the variable that's there (i.e "foo"). Is there a way to do what I am asking with vim?
If not, is there any other editor/tools I can use to help me with modifications like these?
Thanks in advance!
You need to use capture grouping and backreferencing in order to achieve that:
Pattern String sub. flags
|---------| |------------| |-|
:%s/if(!\(.*\))/if(\1 == NULL)/gc
|---| |--|
| ^
|________|
The matched string in pattern will be exactly repeated in string substitution
:help /\(
\(\) A pattern enclosed by escaped parentheses. /\(/\(\) /\)
E.g., "\(^a\)" matches 'a' at the start of a line.
E51 E54 E55 E872 E873
\1 Matches the same string that was matched by /\1 E65
the first sub-expression in \( and \). {not in Vi}
Example: "\([a-z]\).\1" matches "ata", "ehe", "tot", etc.
\2 Like "\1", but uses second sub-expression, /\2
... /\3
\9 Like "\1", but uses ninth sub-expression. /\9
Note: The numbering of groups is done based on which "\(" comes first
in the pattern (going left to right), NOT based on what is matched
first.
You can use
:%s/if(!\(.*\))/if(\1 == NULL)/gc
By putting .* in \( \) you make numbered captured group, which means that the regex will capture what is in .*
When the replace starts then by using \1 you will print the captured group.
A macro is easy in this case, just do the following:
qa .............. starts macro 'a'
f! .............. jumps to next '!'
x ............... erase that
e ............... jump to the end of word
a ............... starts append mode (insert)
== NULL ........ literal == NULL
<ESC> ........... stop insert mode
q ............... stops macro 'a'
:%norm #a ........ apply marco 'a' in the whole file
:g/^if(!/ norm #a apply macro 'a' in the lines starting with if...
Try the following:
%s/if(!\(.\{-}\))/if(\1 == NULL)/gc
The quantifier .\{-} matches a non-empty word, as few as possible (more strict than .*).
The paranthesis \( and \) are used to divide the searched expression into subexpressions, so that you can use those subgroups in the substitute string.
Finally, \1 allows the user to use the first matched subexpression, in our case it is whatever is caught inside the paranthesis.
I hope this is more clear, more information can be found here. And thanks for the comment that suggests improving the answer.
I have a variable from which I have to grep the which in middle of %% adn the word which starts with $$. I used split it works... but for only some scenarios.
Example:
#!/usr/bin/perl
my $lastline ="%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)";
my #lastline_temp = split(/%/,$lastline);
print #lastline_temp;
my #var=split("\\$\\$",$lastline_temp[2]);
print #var;
I get the o/p as expected. But can i get the same using Grep command. I mean I dont want to use the array[2] or array[1]. So that I can replace the values easily.
I don't really see how you can get the output you expect. Because you put your data in "busy" quotes (interpolating, double, ...), it comes out being stored as:
'%Filters_LN_RESS_DIR%ARCOptionsPegaCHF_Vega$01212_GV_DATE_LDN)'
See Quote and Quote-like Operators and perhaps read Interpolation in Perl
Notice that the backslashes are gone. A backslash in interpolating quotes simply means "treat the next character as literal", so you get literal 'A', literal 'O', literal 'P', ....
That '0' is the value of $( (aka $REAL_GROUP_ID) which you unwittingly asked it to interpolate. So there is no sequence '$$' to split on.
Can you get the same using a grep command? It depends on what "the same" is. You save the results in arrays, the purpose of grep is to exclude things from the arrays. You will neither have the arrays, nor the output of the arrays if you use a non-trivial grep: grep {; 1 } #data.
Actually you can get the exact same result with this regular expression, assuming that the single string in #vars is the "result".
m/%([^%]*)$/
Of course, that's no more than
substr( $lastline, rindex( $lastline, '%' ) + 1 );
which can run 8-10 times faster.
First, be very careful in your use of quotes, I'm not sure if you don't mean
'%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)'
instead of
"%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)"
which might be a different string. For example, if evaluated, "$$" means the variable $PROCESS_ID.
After trying to solve riddles (not sure about that), and quoting your string
my $lastline =
'%Filters_LN_RESS_DIR%\ARC\Options\Pega\CHF_Vega\$$(1212_GV_DATE_LDN)'
differently, I'd use:
my ($w1, $w2) = $lastline =~ m{ % # the % char at the start
([^%]+) # CAPTURE everything until next %
[^(]+ # scan to the first brace
\( # hit the brace
([^)]+) # CAPTURE everything up to closing brace
}x;
print "$w1\n$w2";
to extract your words. Result:
Filters_LN_RESS_DIR
1212_GV_DATE_LDN
But what do you mean by replace the values easily. Which values?
Addendum
Now lets extract the "words" delimited by '\'. Using a simple split:
my #words = split /\\/, # use substr to start split after the first '\\'
substr $lastline, index($lastline,'\\');
you'll get the words between the backslashes if you drop the last entry (which is the $$(..) string):
pop #words; # remove the last element '$$(..)'
print join "\n", #words; # print the other elements
Result:
ARC
Options
Pega
CHF_Vega
Does this work better with grep? Seems to:
my #words = grep /^[^\$%]+$/, split /\\/, $lastline;
and
print join "\n", #words;
also results in:
ARC
Options
Pega
CHF_Vega
Maybe that is what you are after? What do you want to do with these?
Regards
rbo
By outside, I want solutions that does not use Vim's scripting hacks but try to reuse certain basic *ix tools. Inside Vim stuff asks for solutions to get the column-increment with inside stuff such as scripting.
1 1
1 2
1 3
1 ---> 4
1 5
1 6
. .
. .
Vim has a script that does column-vise incrementing, VisIncr. It has gathered about 50/50 ups and down, perhaps tasting a bit reinventing-the-wheel. How do you column-increment stuff in Vim without using such script? Then the other question is, how do you column-increment stuff without/outside Vim?
Most elegant, reusable and preferably-small wins the race!
I don't see a need for a script, a simple macro would do
"a yyp^Ayy
then play it, or map to play it.
Of course, there is always the possibility that I misunderstood the question entirely...
The optimal choice of a technique highly depends on the actual circumstances
of the transformation. There are at least two points variations affecting
implementation:
Whether the lines to operate on are the only ones in a file? If not,
is the range of lines defined by context (i.e. it separated by blank
lines, like a paragraph) or is it arbitrary and should be specified by
user?
Are those lines already contain numbers that should be changed or is
it necessary to insert new ones leaving the text on the lines in tact?
Since there is no information to answer these questions, below we will try to
construct a flexible solution.
A general solution is a substitution operating on the beginnings of the lines
in the range specified by the user. Visual mode is probably the simplest way
of selecting an arbitrary range of lines, so we assume here that boundaries of
the range are defined by the visual selection.
:'<,'>s/^\d\+/\=line(".")-line("''")+1/
If it is necessary to number every line in a buffer, the command can be
simplified as follows.
:%s/^\d\+/\=line('.')/
In any case, if the number should be merely inserted at the beginnings of the
lines (without modifying the ones that already exist), one can change the
pattern from ^\d\+ to ^, and optionally add a separator:
:'<,'>s/^\d\+/\=(line(".")-line("''")+1).' '/
or
:%s/^/\=line('.').' '/
respectively.
For a solution based on command-line tools, one can consider using stream
editors like Sed or text extraction and reporting tools like AWK.
To number each of the lines in a file using Sed, run the commands
$ sed = filename | sed 'N;s/\n/ /'
In order to do the same in AWK, use the command
$ awk '{print NR " " $0}' filename
which could be easily modfied to limit numbering to a particular range of lines
satisfying a certain condition. For example, the following command numbers the
lines two through eight.
$ awk '{print (2<=NR && NR<=8 ? ++n " " : "") $0}' filename
Having an interest in how commands similar to those from the script linked in
the question statement are implemented, one can use the following command as
a reference.
vnoremap <leader>i :call EnumVisualBlock()<cr>
function! EnumVisualBlock() range
if visualmode() != "\<c-v>"
return
endif
let [l, r] = [virtcol("'<"), virtcol("'>")]
let [l, r] = [min([l, r]), max([l, r])]
let start = matchstr(getline("'<"), '^\d\+', col("'<")-1)
let off = start - line("'<")
let w = max(map([start, line("'>") + off], 'len("".v:val)'))
exe "'<,'>" 's/\%'.l.'v.*\%<'.(r+1).'v./'.
\ '\=printf("%'.w.'d",line(".")+off).repeat(" ",r-l+1-w)'
endfunction
If you want change 1 1 1 1 ... to 1 2 3 4 .... (Those numbers should be on different lines.)
:let i=1 | g/1/s//\=i/g | let i+=1
If some of 1 1 1 1 ... are in the same line:
:let g:i = 0
:func! Inc()
: let g:i+=1
: return g:i
:endfun
:%s/1/\=Inc()/g