I have a string and I want to achieve to remove all zeros between the characters -s and the first number.
1v-s001v => 1v-s1v
2v-s030r => 2v-s30r
3v-s021v => 3v-s21v
I'm trying with:
\w+-s0*(\d)
but it does not match the subject string.
You may use
(-s)0+(\d)
and replace with $1$2. You may replace \d with [0-9] in case the \d is not supported by your regex flavor.
See the regex demo
Details
(-s) - Capturing group 1 (later referred to with $1 placeholder/replacement backreference from the replacement pattern): a -s substring
0+ - one or more 0 chars
(\d) - Capturing group 2 (later referred to with $2 placeholder/replacement backreference from the replacement pattern): any one digit
Related
I have the following regular expressions that extract everything after first two alphabets
^[A-Za-z]{2})(\w+)($) $2
now I want to the extract nothing if the data doesn't start with alphabets.
Example:
AA123 -> 123
123 -> ""
Can this be accomplished by regex?
Introduce an alternative to match any one or more chars from start to end of string if your regex does not match:
^(?:([A-Za-z]{2})(\w+)|.+)$
See the regex demo. Details:
^ - start of string
(?: - start of a container non-capturing group:
([A-Za-z]{2})(\w+) - Group 1: two ASCII letters, Group 2: one or more word chars
| - or
.+ - one or more chars other than line break chars, as many as possible (use [\w\W]+ to match any chars including line break chars)
) - end of a container non-capturing group
$ - end of string.
Your pattern already captures 1 or more word characters after matching 2 uppercase chars. The $ does not have to be in a group, and this $2 should not be in the pattern.
^[A-Za-z]{2})(\w+)$
See a regex demo.
Another option could be a pattern with a conditional, capturing data in group 2 only if group 1 exist.
^([A-Z]{2})?(?(1)(\w+)|.+)$
^ Start of string
([A-Z]{2})? Capture 2 uppercase chars in optional group 1
(? Conditional
(1)(\w+) If we have group 1, capture 1+ word chars in group 2
| Or
.+ Match the whole line with at least 1 char to not match an empty string
) Close conditional
$ End of string
Regex demo
For a match only, you could use other variations Using \K like ^[A-Za-z]{2}\K\w+$ or with a lookbehind assertion (?<=^[A-Za-z]{2})\w+$
I am trying to put together a regex expression that matches a word (only one per line) that starts and ends with the same three characters.
I was able to write a solution for words that are at least 6 characters long (meaning there is no overlap), but I am unsure how to do it for overlapping starts and ends such as "heheh".
This is what I have, nice and simple:
^(...).*\1$
I am inclined to believe that this might have something with lookahead and lookbehind but I am not sure.
Any help would be appreciated, thank you!
You will need lookarounds since they are non-consuming patterns, i.e. the regex index is not advanced when the lookaround pattern is matched.
For example, you may do this with GNU grep:
grep -P '^(?=(...)).+\1$' file
grep -P '^(?=(\S{3})).+\1$' file # To avoid counting in spaces
grep -P '^(?=(\w{3})).+\1$' file # Or only allowing letters/digits/underscores
grep -P '^(?=(\p{L}{3})).+\1$' file # Or only allowing letters
See the regex demo
Details
^ - start of string
(?=(...)) - a positive lookahead with a capturing group inside that matches any 3 chars
.+ - any 1+ chars other than line break chars as many as possible
\1 - Group 1 value
$ - the end of string.
To extract words, you may use \w shorthand (that matches letters, digits and underscores) and word boundaries \b:
grep -oP '\b(?=(\w{3}))\w+\1\b' file
See another demo.
Details
\b - a word boundary (start of word here, because it is followed with word chars)
(?=(\w{3})) - a positive lookahead making sure there are 3 word chars while capturing them into Group 1
\w+ - 1+ word chars (not 0 or more because otherwise a 3-char word would be matched)
\1 - Group 1 value
\b - end of word here (as it is preceded with word chars).
I have regular expression like this in ruby
%r{
(ST)
([A-Z]) ?
(#{A_VAL})
-?
(T)?
}x
Now ,I don't want my regex to accept any string that ends with "-" .So, for example it should accept
1)"STCA1-T"
2)"STCA1T"
But it shouldn't accept "STCA1-"
You may use
/\AST[A-Z]?(#{A_VAL})(?:-?T)?\z/
Details
\A - start of string
ST - an ST substring
[A-Z]? - an optional ASCII letter
(#{A_VAL}) - Group 1 (if there is a single alternative, just one string, and you do not need this value later, you may omit the capturing parentheses): a pattern inside A_VAL variable
(?:-?T)? - an optional non-capturing group that matches an optional - and an obligatory T (i.e. it matches -T or T 1 or 0 times)
\z - end of string.
Hello i want to match with regex this word
(Parc Installé)
from this text:
31/1/2017 17:19:23,4245986,ct0001#Intotel.int,Parc Installé,100.100.30.100
I did this regex ',[A-Za-zA-zÀ-ú+ \/\w+0-9._%+-]+,'
But the result is : 4245986 ans Parc Installé.
How can i match only Parc Installé
You may try a regex based on a lookahead that will require a comma and digits/commas after it up to the end of string:
[^,]+(?=\s*,[\d.]+$)
See this regex demo
Details:
[^,]+ - 1 or more chars other than ,
(?=\s*,[\d.]+$) - a lookahead requiring
\s* - zero or more whitespaces
, - a comma
[\d.]+ - 1+ digits or dots up to...
$ - ... the end of string
To make it a bit more restrictive, you may replace the lookahead with (?=\s*,\d+(?:\.\d+){3}$) to require 4 sequences of dot-separated 1+ digits. See this regex demo.
If a lookahead is not supported (case with a RE2 engine), you might want to use a capturing group based solution:
([^,]+)\s*,[\d.]+$
Here, the part within (...) will be captured into Group 1 and will be accessible via a backreference or a function like =REGEXEXTRACT in Google Spreasheets that only retrieves the contents of a capturing group if the latter is present in the pattern.
I'm trying to turn cast(("Sparkles"), GetBitmapData); to GetBitmapData("Sparkles");
I've got this for my find code:
cast\(\(\"\.*\"\),\ .*\);
but this replace doesn't work:
$2\(\"$1\"\);
What do I need to do to make this work?
You regex does not contain capturing groups and you try to access them with numbered backreferences. Besides, you escaped the dot, and \.* just matches 0+ dot symbols.
You may use the following regex replacement:
Find what: cast\(\("(.*?)"\),\s*(\w+)\);
Replace with: $2("$1");
Here is a .NET regex demo (FlashDevelop S&R feature uses .NET regex flavor).
Pattern details:
cast\(\(" - a cast((" substring
(.*?) - Group 1 (referred to with $1) capturing any 0+ chars as few as possible up to the first...
"\), - a "), substring
\s* - 0+ whitespaces
(\w+) - Group 2 (referred to with $2) capturing 1+ word chars (letters/digits/_)
\); - a ); substring.