This question already has answers here:
Regular expression to match strings that do NOT contain all specified elements
(2 answers)
Closed 2 years ago.
I am trying to find the inverse of the following string but with no success:
(?i)(?s)^(?=.*?word1)(?=.*?word2)
I built this but it is for one word only and it does not even function properly, considering that if I test it with regex101 I got one match while I should get no match:
(?i)(?s)(?!.*?word2)^.*$
Please see the following link: https://regex101.com/r/qS7yN9/72
Hope you guys can help to build the right string.
You can effectively negate the result by wrapping a negative look-ahead assertion around the whole thing:
(?i)(?s)^(?!(?=.*?word1)(?=.*?word2))
Related
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 3 years ago.
I need to capture strings containing only one dot. String will mostly contains domain names like
test.com, fun.test.com, lesh.test.com.
I need to check only the first one and to ignore the string that has more than one dots.
How can I do this using regex?
Like this :
^[^.]+\.[^.]+$
Check explanations https://regex101.com/r/mn7Ccr/1
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
How to extract a substring using regex
(14 answers)
Closed 3 years ago.
I'm trying to find the word: <*hasburnt*> in the string below using the this regex: <\*.*(bur).*\*>
But it gives me both <*hasburnt*> <*electrical*>. How do I just get <*hasburnt*> ?
bench testedstarter, starter just makes noise, and <*hasburnt*>
<*electrical*> smell.
Try this: /<.*?(bur).*?>/
Regex101 demo
The reason for ? here is because .* tries to match as much characters as possible, so it also matches <electrical. .*? makes it lazy - trying to match as little as possible, and as such ending the match at <hasburnt>.
EDIT: using ? for the first .* would make <hasburnt> independent of positions of similar strings.
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 4 years ago.
I have a question about scala regex
The thing I need to do is given a string, I need to find a sub-string up to a specific word given. For example, my regular expression looks like following
val x= "(?s)^(.*)(?=(foo|bar)".r
Then given a string, I need to find the longest sub-string until before foo or bar. This works perfectly but I would like to get the whole string if the string does not contain foo or bar at all.
Right now if I do
x.findAllIn("hello nice to meet you").toArray
it gives me an empty string but I would like to get
"hello nice to meet you" when I do that.
Does anyone have an idea how to implement that?
You can add an end-of-string assertion to the alternative:
(?s)^(.*?)(?=(foo|bar|$))
Demo
This question already has answers here:
Regular expressions: Ensuring b doesn't come between a and c
(4 answers)
Closed 4 years ago.
I'd like to find these three strings in any order and the result may have all these three strings including any character between them with the shortest length.
strings are: "ACT", "AGT" and "CGT".
Sample input: "ACTACGTTTAGTAACTCGTCT"
I tried but the regex returns the first occurrence matched which is "ACTACGTTTAGTAACTCGT"
/(ACT.*AGT.*CGT)|(ACT.*CGT.*AGT)|(AGT.*ACT.*CGT)|(AGT.*CGT.*ACT)|(CGT.*ACT.*AGT)|(CGT.*AGT.*ACT)/g
Output has to be "AGTACTCGT"
You can't return separate bits of a string already concatenated in one go.
See here: Regular expression to skip character in capture group
You can first match each bit, using parentheses to group them, and then put them together in a separate step
This question already has answers here:
Matching numbers with regular expressions — only digits and commas
(10 answers)
regular expression for finding decimal/float numbers?
(9 answers)
Closed 7 years ago.
I need to do a regular expression to match the floats only, what i got is the following :
[\-\+]?[0-9]*(\.[0-9]+)?
But this match also the below
123123132 ,
05/03/1994
I only need want to match the number with the decimal point
Your regex is almost correct for your purpose.
It finds 123123132, because the last part is optional. Removing the ? solves that.
[-+]?[0-9]*(\.[0-9]+)
With that adjustment, it might still find matches in strings like .12/39/3239, if you don't want that to happen, insert enforce matching over the complete string by inserting ^ and $:
^[-+]?[0-9]*(\.[0-9]+)$
How about:
([+-]?[0-9]*\.[0-9]*)
You can see it working here
Here is a regexp handling also existing exponents:
[-+]?[0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?
Debuggex Demo
Additionally you should force the hole string to be matched to avoid matchings within your date values.
^[-+]?[0-9]*\.?[0-9]+([eE][-+]?[0-9]+)?$
By the way here is a nice tutorial about matching floating point numbers using regular expressions: http://www.regular-expressions.info/floatingpoint.html.