in vim for show whitespace EOL I use
highlight whitespaceEOL term=reverse ctermbg=Grey guibg=Grey
match whitespaceEOL /\s\+\(\%#\)\#!$/
but when I use a match for long lines
augroup longLines
autocmd! filetype zsh,sh,python,vim,c,cpp :match ColorColumn /\%>80v.\+/
augroup END
I lost the first match, why ?
:match only matches one pattern at a time.
:2match and :3match exist for exactly this reason.
Alternatively, you can implement this as syntax.
Try:
2match whitespaceEOL /\s\+$/
3match ColorColumn /\%>80v.\+/
Or:
syntax match whitespaceEOL /\s\+$/
syntax match ColorColumn /\%>80v.\+/
Related
I need to remove only characters at the beginning.
GFG2014JP34343
D2013GH43422
HHH2014JP34343
CC2013GH43422
Output:
2014JP34343
2013GH43422
2014JP34343
2013GH43422
I tried REGEXP with different pattern.
We can use a regex replacement here:
SELECT val, REGEXP_REPLACE(val, '^[^[:digit:]]+', '') AS val_out
FROM yourTable;
I want to select all the commas in a string that do not have any white space around. Suppose I have this string:
"He,she, They"
I want to select only the comma between he and she. I tried this in rubular and came up with this regex:
(,[^(,\s)(\s,)])
This selects the comma that I want, but also selects an s which is a character after it.
In your regex (,[^(,\s)(\s,)]) you capture a comma followed by a negated character class that matches not any of the specified characters, which could also be written as (,[^)(,\s]) which will capture for example ,s in a group,
What you could do is use a positive lookahead and a positve lookbehind to check what is on the left and what is on the right is not a \S whitespace character:
(?<=\S),(?=\S)
Regex demo
In Ruby, you may use [[:space:]] to match any (Unicode) whitespace and [^[:space:]] to match any char other than whitespace. Using these character classes inside lookarounds solves the problem:
/(?<=[^[:space:]]),(?=[^[:space:]])/
See the Rubular demo
Here,
(?<=[^[:space:]]) - a positive lookbehind that matches a location that is immediately preceded with a non-whitespace char (if the string start position should also be matched, replace with (?<![[:space:]]))
, - a comma
(?=[^[:space:]]) - a positive lookahead that matches a location that is immediately followed with a non-whitespace char (if the string end position should also be matched, replace with (?![[:space:]])).
Check the regex below and use the code hope it will help you!
re = /[^\s](,)[^\s]/m
str = 'check ,my,domain, qwe,sd'
# Print the match result
str.scan(re) do |match|
puts match.to_s
end
Check LIVE DEMO HERE
I am trying to perform regular expression matching and replacement on the same line in Ruby. I have some libraries that manipulate strings in Ruby and add special formatting characters to it. The formatting can be applied in any order. However, if I would like to change the string formatting, I want to keep some of the original formatting. I'm using regex for that. I have the regular expression matching correctly what I need:
mystring.gsub(/[(\e\[([1-9]|[1,2,4,5,6,7,8]{2}m))|(\e\[[3,9][0-8]m)]*Text/, 'New Text')
However, what I really want is the matching from the first grouping found in:
(\e\[([1-9]|[1,2,4,5,6,7,8]{2}m))
to be appended to New Text and replaced as opposed to just New Text. I'm trying to reference the match in the form of
mystring.gsub(/[(\e\[([1-9]|[1,2,4,5,6,7,8]{2}m))|(\e\[[3,9][0-8]m)]*Text/, '\1' + 'New Text')
but my understanding is that \1 only works when using \d or \k. Is there any way to reference that specific capturing group in my replacement string? Additionally, since I am using an asterik for the [], I know that this grouping could occur more than once. Therefore, I would like to have the last matching occurrence yielded.
My expected input/output with a sample is:
Input: "\e[1mHello there\e[34m\e[40mText\e[0m\e[0m\e[22m"
Output: "\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m"
Input: "\e[1mHello there\e[44m\e[34m\e[40mText\e[0m\e[0m\e[22m"
Output: "\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m"
So the last grouping is found and appended.
You can use the following regex with back-reference \\1 in the replacement:
reg = /(\\e\[(?:[0-9]{1,2}|[3,9][0-8])m)+Text/
mystring = "\\e[1mHello there\\e[34m\\e[40mText\\e[0m\\e[0m\\e[22m"
puts mystring.gsub(reg, '\\1New Text')
mystring = "\\e[1mHello there\\e[44m\\e[34m\\e[40mText\\e[0m\\e[0m\\e[22m"
puts mystring.gsub(reg, '\\1New Text')
Output of the IDEONE demo:
\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m
\e[1mHello there\e[40mNew Text\e[0m\e[0m\e[22m
Mind that your input has backslash \ that needs escaping in a regular string literal. To match it inside the regex, we use double slash, as we are looking for a literal backslash.
I have an string of an email that looks like "<luke#example.com>"
I would like to use regex for deleting "<" and ">", so I wanted something like
"<luke#example.com>".sub /<>/, ""
The problem is quite clear, /<>/ doesn't wrap what I want. I tried with different regex, but I don't know how to choose < AND >, it is there any and operator where I can say: "wrap this and this"?
As written, your regex matches the literal substring "<>" only. You need to use [] to make them a character class so that they're matched individually, and gsub to replace all matches:
"<luke#example.com>".gsub(/[<>]/, "") # => "luke#example.com"
"<luke#example.com>".gsub /[<>]/, ""
http://regex101.com/r/hP3sY2
If you only ever want to strip the < and > from the start and end only, you can use this:
'<luke#example.com>'.sub(/\A<([^<>]+)>\z/, '\1')
You don't need, nor should you use, a regex.
string[1..-2]
is enough.
I want to remove a single * character and any white space from the start of a string.
This is the regex I have /^\*{1}(?:\s+)?/
Here's a Rubular link http://rubular.com/r/r5i4FpQdK2
However Ruby is throwing a warning when I try to use it.
001:0> regex = /^\*{1}(?:\s+)?/
warning: nested repeat operator + and ? was replaced with '*': /^\*{1}(?:\s+)?/
=> /^\*{1}(?:\s+)?/
The actual test still works
002:0> "* foo" =~ regex
=> 0
but I can't figure out what's causing the warning.
Any advice would be appreciated.
Instead of (?:\s+)? use (?:\s*) or just \s*
\s+ allows one or more spaces and the following ? makes it optional, which can be replaced with zero or more space as \s*