I am using the following regex for detecting negative numbers:
([-]([0-9]*\.[0-9]+|[0-9]+))
But I want to skip the matches which are followed by $.
If i use the folowing regex:
([-]([0-9]*\.[0-9]+|[0-9]+)[^\$])
It will match correctly the positions but will include the following character.
For example in expression:
-0.6+3 - 3.0$
it will match:
-0.6+
I want to match only
-0.6
([-]([0-9]*\.[0-9]+|[0-9]+)(?!\$)
You need a negative lookahead here which will not consume and only make an assertion.
Remove the $ from the group:
([-]([0-9]*\.[0-9]+|[0-9]+))[^\$]
You could use this simplified regex:
(-[0-9]+(?:\.[0-9]+)?)(?!\$)
You can use the regex from Regular-Expressions.info with just minus at the beginning, and a \b added at the end in order to stop before any non-word character:
[-][0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?\b
This regex also captures the numbers with exponent part.
See demo
Related
I know some basics of the RegEx but not a pro in it. And I am learning it. Currently, I am using the following very very simple regex to match any digit in the given sentence.
/d
Now, I want that, all the digits except some patterns like e074663 OR e123444 OR e7736 should be excluded from the match. So for the following input,
Edit 398e997979 the Expression 9798729889 & T900980980098ext to see e081815 matches. Roll over matches or e081815 the expression e081815 for details.e081815 PCRE & JavaScript flavors of RegEx are e081815 supported. Validate your expression with Tests mode e081815.
Only bold digits should be matched and not any e081815. I tried the following without the success.
(^[e\d])(\d)
Also, going forward, some more patterns needs to be added for exclusion. For e.g. cg636553 OR cg(any digits). Any help in this regards will be much appreciated. Thanks!
Try this:
(?<!\be)(?<!\d)\d+
Test it live on regex101.com.
Explanation:
(?<!\be) # make sure we're not right after a word boundary and "e"
(?<!\d) # make sure we're not right after a digit
\d+ # match one or more digits
If you want to match individual digits, you can achieve that using the \G anchor that matches at the position after a successful match:
(?:(?<!\be)(?<=\D)|\G)\d
Test it here
Another option is to use a capturing group with lookarounds
(?:\b(?!e|cg)|(?<=\d)\D)[A-Za-z]?(\d+)
(?: Non capture group
\b(?!e|cg) Word boundary, assert what is directly to the right is not e or cg
| Or
(?<=\d)\D Match any char except a digit, asserting what is directly on the left is a digit
) Close group
[A-Za-z]? Match an optional char a-zA-Z
(\d+) Capture 1 or more digits in group 1
Regex demo
The pattern:
(test):(thestring)
What I want is full match only if there is just one test: before
test:thestring
But in this case there wouldn't be full match:
test:test:thestring
I've tried qualificator, but it didn't work.
Need help
Try this pattern: ^(?!.*((?(?<=^)|(?<=:))test(?=(:|$))).*(?1)).+$.
The main part is ((?(?<=^)|(?<=:))test(?=(:|$))), which matches test if it's preceeded by colon : or is at the beginning of a line and it's followed by colon : or end of the line.
(?(?<=^)|(?<=:)) this is workaround to (?<=(:|^)), but lookbehinds must have fixed length.
Then we have backreference to first capturing group (?1), to see if there are any other test.
This whole pattern is placed in negative lookahead (?!...), to match everything if it doesn't match pattern explained above (test matched more than one time).
Demo
for this very specific case:
(?<!.)(test:thestring)
Regex101
All it does is search for the string test:thestring and ensures that there are no characters before it. (Use Michał Turczyn's regex for an all purpose search!)
^((?!test:).)*(test:thestring)
See in action
If you want a full match and there should be only one time test: before test:string you might assert the start of the string ^, use a negative lookahead (?:(?!test:).) to match any character if what is on the right side is not test:
Then match test:thestring followed by a negative lookahead (?:(?!test:thestring).)* that matches any character if what is on the right side is not test:thestring and assert the end of the string $
^(?:(?!test:).)*test:thestring(?:(?!test:thestring).)*$
Regex demo
I have this regex
(?<=TG00).*?(?=#)
which extracts all strings between TG00 and #. Demo: https://regex101.com/r/04oqua/1
Now, from above results I want to extract only the string which contains TG40 155963. How can I do it?
Try this pattern:
TG00[^#]*TG40 155963[^#]*#
This pattern just says to find the string TG40 155963 in between TG00 and an ending #. For the sample data in your demo there were 3 matches.
Demo
For some reason appending .*? to your lookbehind results in engine error, but works fine with lookahead. Regex below does not match your text exactly, but it does extract it via capture group.
(?<=TG00).*?(TG40 155963)(?=.*?#)
You can use this regex with a lookahead and negated character class:
(?<=TG00)(?=[^#]*TG40 155963)[^#]+(?=#)
RegEx Demo
RegEx Explanation:
(?<=TG00): Assert that we have TG00 at previous position
(?=[^#]*TG40 155963): Lookahead to assert we have string TG40 155963 after 0 or more non-# characters, ahead
[^#]+: Match 1+ non-# characters
I have this regexp:
^[a-z0-9]+([.\-][a-z0-9]+)*$
I need exclude from match only one word "www".
I tried the negative lookahead but without a success.
Use a negative lookahead like this:
^(?!www$)[a-z0-9]+([.-][a-z0-9]+)*$
^^^^^^^^
This will not match a string equal to www.
See the regex demo
If you want to fail a match with strings that contain -www- or .www., use
^(?!.*\bwww\b)[a-z0-9]+([.-][a-z0-9]+)*$
See another regex demo. This pattern contains a (?!.*\bwww\b) lookahead that fails the whole match if there is a www somewhere inside the string and it has no digits or letters round it due to \b word boundaries.
For this example hello.1.2.3.4.world I want to match a result which gives me 1.2.3.4. Number of digits between dots doesn't matter. As long as it follow digit.digit pattern
My part solution was following regular-expression [\d.]+.[^.a-z], which gives me .1.2.3.4 as result. And I strip the first dot by using trim or similar method.
Any regexp master who can tell me how to rid the first dot with one regular expression only?
How about this: \.(\d(?:\.\d)*)\.\D
EDIT:
(\d+(?:\.\d+)*)
Demo
If you want to use your current regex you can put a lookahead at the start, and escape the literal dot when not inside a character group (?=\d)[\d.]+\.[^.a-z]
The lookahead (?=\d) will make sure the first character matched is a digit.
Demo here