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Regex to replace character between square brackets, but not between parenthesis


By : DSA
Date : August 01 2020, 06:00 AM
hop of those help? I have a list of Markdown links that look like this: , You may use
code :
(?:\G(?!\A)|\[)[^][-]*\K-(?=[^][]*])


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Regex: match text square brackets and what is contained within square brackets then search and replace


By : Stephen Kalmakis
Date : March 29 2020, 07:55 AM
this one helps. I am trying to match (a) and replace (b) the following occurrences: , How about this?
code :
:s/\<array\[[^\]]\+\]/atoi(\0)/

Regex to replace right square brackets inside outer square brackets


By : Merit Platform
Date : March 29 2020, 07:55 AM
I wish this helpful for you No looping is necessary. We want to match every right-hand bracket but the last one and the one that immediately precedes the dot (.) separator. We simply require that our right-hand bracket is followed by a single character and specify that it must be a non-dot character. I'm not sure about the proper .NET way to say this, but you want to match on ]([^\.]) and replace with ]]$1 where $1 is a backreference to the stuff in parentheses in the match pattern. If you expect whitespace at the end of your input(s), I'd suggest cleaning that first, as this would replace the last ] if it were followed by a space.
You can test here: http://www.regular-expressions.info/javascriptexample.html
code :
~ $ echo "                              
[dim[e]nsion].[this] [member] name]
[dim[en]sion n[a]me ].[this [member] name]
[[dimension] name].[this member [name]]
" > input.txt
~ $ cat input.txt

[dim[e]nsion].[this] [member] name]
[dim[en]sion n[a]me ].[this [member] name]
[[dimension] name].[this member [name]]

~ $ sed 's/]\([^\\.]\)/]]\1/g' input.txt

[dim[e]]nsion].[this]] [member]] name]
[dim[en]]sion n[a]]me ].[this [member]] name]
[[dimension]] name].[this member [name]]]

~ $ 

Regex matching square brackets followed by parenthesis where the square brackets can also contain other square brackets


By : Rohan Khanna
Date : March 29 2020, 07:55 AM
this will help I have some text like this, it's written in a custom markdown style format. For example: , Here you go.
code :
preg_match_all("~
    \[(              # open outer square brackets and capturing group
    (?:              # open subpattern for optional inner square brackets
        [^[\]]*      # non-square-bracket characters
        \[           # open inner square bracket
        [^[\]]*      # non-square-bracket characters
        ]            # close inner square bracket
    )*               # end subpattern and repeat it 0 or more times
    [^[\]]*          # non-square-bracket characters
    )]               # end capturing group and outer square brackets
    (?:              # open subpattern for optional parentheses
        \((          # open parentheses and capturing group
        [a-z]+       # letters
        )\)          # close capturing group and parentheses
    )?               # end subpattern and make it optional
    ~isx",
    $input,
    $matches);
"~\[((?:[^[\]]*\[[^[\]]*])*[^[\]]*)](?:\(([a-z]+)\))?~isx"

Regex Unmatched parenthesis with double Square brackets


By : Daryl Bing
Date : March 29 2020, 07:55 AM
may help you . If you only want to match 16 digits, no more, no less
and the number must start with one of your combinations,
code :
 ^ 
 (?= \d{16} $ )
 (                             # (1 start)
      6011
   |  65
   |  64 [4-9] 
   |  622 
 )                             # (1 end)
 ( \d{1,14} )                  # (2) or even just a \d* will do here
 $

Regex to replace every character with another character EXCEPT for what's in square brackets


By : saitou
Date : March 29 2020, 07:55 AM
this one helps. Hi I need help with a regex.. if it's possible I'm not sure..
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