Is it possible to replace the dot with a underscore but just inside curly braces using only regex ?
eg. a.b.c={{c.d.f}}
after the replace it should look like
a.b.c={{c_d_f}}
The curly braces are always balanced and there will always be two open curly braces and two closed ones.
You can use this lookahead regex for search:
\.(?=[^{}]*\})
Lookahead (?=[^{}]*\}) asserts that there is a } ahead after 0 or more non { and non } characters.
and replace by _
RegEx Demo
You can use \G like (assuming . inside only {{ and no nesting)
(?:(\{\{)|\G(?!\A))([^.}]*)[.]
and replace with
\1\2_
Regex Demo
If u assume there is one character between dots use:
Search:({{.).(.).(.}})
Replace with:\1_\2_\3
If one or more:
Search:({{.+).(.+).(.+}})
With same replace
Related
How can I replace the words like:
something(name_one,name_two).
into
something(name_two,name_one).
I've tried something like ^(.+),(.+)$ and replace with \2,\1, but not success and it turns out to be like this:
name_two).,something(name_one
You have to exclude from the capturing groups what you don't want to be swapped.
Replace
something\((.+),(.+)\)
with
something(\2,\1)
(see demo)
Or
Replace
\((.+),(.+)\)
with
(\2,\1)
(see demo)
Please notice that \( and \) are escaped parentheses, so they will match literally an open parenthesis and a closed one.
Find what: \(\K(.*),(.*)(?=\))
Replace with: $2,$1
Suppose a string on the following format:
Use \hyperlink{aaa}{apple {pear} banana} and \hyperlink{bbb}{banana {pear} {apple}}.
I want to extract:
\hyperlink{aaa}{apple {pear} banana}
\hyperlink{bbb}{banana {pear} {apple}}
What regex could be used for such an extraction?
I got stuck with this:
\\hyperlink{\S+}{.+}
Here how you can do it with a recursive regex
\\hyperlink\{[^}]+?\}(\{(?>[^{}]+|(?1))+\})(?=\s|$)
Regex Demo
Recursive regex
If there is no arbitrary nesting, you can use a pattern with negated }{ like
\\hyperlink{[^}{]*}{[^}{]*(?:{[^}{]*}[^}{]*)*}
Similar this answer but unrolled. See the demo at regex101. To {extract} use groups (demo).
Depending on your environment / regex flavor it can be necessary to escape the opening { by a backslash for the braces that are not inside a character class to match them literally.
Further note that \S+ can consume } and .+ can match more than desired if unaware.
I have tried
[^\(]\(.*\)
but in a string like
Tamara(PER) Jorquiera
The pattern returns
a(PER)
How can I get only the text inside parenthesis, assuming the open and close parenthesis occur once?
[^\(]\(.*\)
It matches any character that is not ( (a in the example).. and a literal ( then everything till )
You can use the following:
\([^)]*\) // this will match only the text between parentheses
// including parantheses
If you want only the text use lookahead and lookbehinds:
(?<=\()[^)]*(?=\))
See DEMO
Use following regex :
\(([^)]*?)\)
Demo
Note that you don't need [^\(] out side of the parenthesis just use
[^)]*? within a capture group.
[^)]*? is a non-greedy regex that match anything except closing parenthesis.
I'm so close to understanding regex. I'm a bit stumped, I thought i understood lazy and greedy.
Here is my current regex: <g_n><!\[CDATA\[([^]]+)(?=]]><\/g_n>)
My current regex makes:
<g_n><![CDATA[xxxxxxxxxx]]></g_n>
match to:
<g_n><![CDATA[xxxxxxxxxx
But I want to make it match like this:
xxxxxxxxxx
You want
<g_n><!\[CDATA\[(.*?)]]></g_n>
then if you want to replace it use
\1
in the replacement box
Your matching the whole string, the brackets around the .*? match all of that and put it in the \1 variable
So the match will be all of the string with \1 referring to what you want
To change the xxxxx
Regex :
(<g_n><![CDATA[)(?:.*?)(]]></g_n>)
Replacement
\1WHAT YOU WANT TO CHANGE TO\2
It looks like you need to add escape slashes to the two closing square brackets, as they are literals from the string you're parsing.
<g_n><!\[CDATA\[.*+?\]\]><\/g_n>
^ ^
Any square brackets not being escaped by backslashes will be treated as regex operational brackets, which in this case won't catch the input string.
EDIT, I think the +? is redundant.
\[.*\]\]> ...
should suffice, since .* means any character, any amount of times.
Tested with notepad++ 6.3.2:
find: (<g_n><!\[CDATA\[)([^]]+)(?=]]></g_n>)
replace: $1WhatYouWant
You can replace + by * in the pattern to match void CDATA:
<g_n><![CDATA[]]></g_n>
I want to remove words inside bracket,
I'm currently using this
【.*】
【remove this】preserve this 【remove this】
but it removes everything for this because there is another bracket
How can I solve this? it also happens on comma
◆.*、
◆ remove this、preserve this、
that regex removes everything because I have 2 commas
Use non-greedy matching with ?, and also escape the brackets, which are special characters:
\[.*?\]
You can try two solutions:
Lazy Operators (this might not work on your RegEx parser)
\[.*?\]
.*?,
Or replace . by a negation list to match any element but the end delimiter:
\[[^]]*\]
[^,]*,
use a specified character group
\[[^\]]+\]