Let's say I have the following text:
new_item['uid']
And I want to capture everything within the [ ... ]. So in this case grab the 'uid'. Normally I could use something like:
\[([^\]]+)]
To match this (start with the opening bracket and get everything until the closing bracket). But without the character classes, or negated character class in vim, how would I do something similar?
If you want to have a capture group with (..), you need the verymagic mode, otherwise you have to escape the ( and ), similar to the BRE.
So both give the matched part in \1:
\[\([^]]*\)
and (\v tells vim to match in verymagic mode)
\v\[([^]]*)
You could use .\{-} in place of .*? to make a lazy dot match:
\[(.\{-})\]
in addition to the other answers, you could use \v\[\zs.{-}\ze\] to only highlight the text within \zs and \ze, see :h \ze
Related
I am using TextMate to replace expression [my_expression] consisting in characters between open and closed brackets by {my_expression}; so I tried to replace
\[[^]]*\]
by
{$1}
The regex matches the correct expression, but the replacement gives {$1}, so that the variable is not recognised. Can someone has an idea ?
You forgot to escape a character, [^]] should be [^\]].
You also need a capture group. $1 is back-referencing the 1st Capture Group, and you had no capture groups, so use the following Regex:
\[([^\]]*)\]
This adds () around [^\]]*, so the data inside the [] is captured. For more info, see this page on Capture Groups
However, this RegEx is shorter:
\[(.*?)\]
Also substituting with {$1}
Live Demo on Regex101
Use a capturing group (...):
\[([^\]]*)\]
The $1 is a backreference to the text enclosed with [...].
Here is the regex demo and also Numbered Backreferences.
Also, the TextMate docs:
1. Syntax elements
(...) group
20.4.1 Captures
To reference a capture, use $n where n is the capture register number. Using $0 means the entire match.
And also:
If you want to use [, -, ] as a normal character in a character class, you should escape these characters by \.
Consider the below line for example
'{"place":"buddy's home"}'
I want to replace the single quote in buddy's only. Single quotes at the start and end of line had to be intact. So the resulting line would look like.
'{"place":"buddy\'s home"}'
There could be multiple lines with multiple occurrences of such single quotes in each line. I have to escape all of them except at the start and end of line.
I'm able to find out such pattern using vim regex :/.'. This pattern ensures that single quote is surrounded by two characters and is not at start or at the end of line. But I'm having trouble how to replace the y's into y\'s at all places.
If the regex .'. is accurate enough then you can substitute all occurrences with:
:%s/.\zs'\ze./\\'/g
Instead of using \ze and \zs you could use groups (...) as well. However I find this version slightly more readable.
See :h /\zs and :h /\ze for further information.
:%s/\(.\)'\(.\)/\1\\'\2/gc
:%s/ substitute over the whole buffer (see :help range to explain the %)
\(.\) match a character and save it in capture group 1 (see :help \()
' a literal '
\(.\) match a character and save it in capture group 2
/ replace by
\1 capture group 1 (see :help \1)
\\' this is a \' (you need to escape the backslash)
\2 capture group 2
/gc replace globally (the whole line) and ask for confirmation (see :help :s_flags)
You can omit the c option if you are sure all replaces are legit.
As kongo2002 says in his answer you could replace the capture groups by \zs and \ze:
\zs will start a match and discard everything before
\ze will end a match and discard everything after
See :help \ze and :help \zs.
I am using TextMate to replace expression [my_expression] consisting in characters between open and closed brackets by {my_expression}; so I tried to replace
\[[^]]*\]
by
{$1}
The regex matches the correct expression, but the replacement gives {$1}, so that the variable is not recognised. Can someone has an idea ?
You forgot to escape a character, [^]] should be [^\]].
You also need a capture group. $1 is back-referencing the 1st Capture Group, and you had no capture groups, so use the following Regex:
\[([^\]]*)\]
This adds () around [^\]]*, so the data inside the [] is captured. For more info, see this page on Capture Groups
However, this RegEx is shorter:
\[(.*?)\]
Also substituting with {$1}
Live Demo on Regex101
Use a capturing group (...):
\[([^\]]*)\]
The $1 is a backreference to the text enclosed with [...].
Here is the regex demo and also Numbered Backreferences.
Also, the TextMate docs:
1. Syntax elements
(...) group
20.4.1 Captures
To reference a capture, use $n where n is the capture register number. Using $0 means the entire match.
And also:
If you want to use [, -, ] as a normal character in a character class, you should escape these characters by \.
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'd like to replace this
y[100] with this Ith(y,100) in Sublime3.
I've got the regular expression \by[\d+] in find what and Ith(y,$1) in replace with, but it doesn't work. It finds what to replace correctly but just replaces it with Ith(y, )
You need to pick the \d+ inside () to capture in group($1)
\by[(\d+)]
Also, you may need to to escape the [ and ] characters here.
\by\[(\d+)\]
You have to capture the data you want to backreference: to do that you have to use unescaped parenthesis (...), ie capturing groups. Also, [...] are character classes, a special character in regex that you need to escape.
Try replacing
\by\[(\d+)\]
with
Ith(y, \1)
You need to escape the special characters and capture the number in a capturing group group. Try this regex:
\by\[(\d+)]\b
And replace with
Ith(y, $1)
Online explanation and demonstration: http://regex101.com/r/nX3yJ9