So I want to allow A-Z with a length of 8 or 12.
I tried:
^[a-z]{8|12}$
but that doesn't work. What's the correct solution? (without repeating)
You need to use alternation like this:
^([a-z]{8}|[a-z]{12})$
There is no other regex solution that would not involve repeating the [a-z] part. At least you do not have to repeat the ^ and $ anchors if you use a grouping construct.
Alternatively, you may use an optional group, but that is only good when your pattern is static. Actually, the difference is negligent (tested at regexhero):
I'd suggest this solution:
^([a-z]{4}){2,3}$
It means that the group [a-z] of length 4 has to match either 2 (total length: 8) or 3 (total length: 12) times. This way the set of allowed characters has to be defined only once.
As an alternative to the "exactly 8 or exactly 12" types of patterns, here's an "8 and maybe 4 more" type pattern:
^[a-z]{8}([a-z]{4})?$
Try this
^[a-z]{8}$|^[a-z]{12}$
The multiple length field option isn't there, you have to give them out separately including the regex
Related
Regex beginner here. I've been trying to tackle this rule for phone numbers to no avail and would appreciate some advice:
Minimum 6 characters
Maximum 20 characters
Must contain numbers
Can contain these symbols ()+-.
Do not match if all the numbers included are the same (ie. 111111)
I managed to build two of the following pieces but I'm unable to put them together.
Here's what I've got:
(^(\d)(?!\1+$)\d)
([0-9()-+.,]{6,20})
Many thanks in advance!
I'd go about it by first getting a list of all possible phone numbers (thanks #CAustin for the suggested improvements):
lst_phone_numbers = re.findall('[0-9+()-]{6,20}',your_text)
And then filtering out the ones that do not comply with statement 5 using whatever programming language you're most comfortable.
Try this RegEx:
(?:([\d()+-])(?!\1+$)){6,20}
Explained:
(?: creates a non-capturing group
(\d|[()+-]) creates a group to match a digit, parenthesis, +, or -
(?!\1+$) this will not return a match if it matches the value found from #2 one or more times until the end of the string
{6,20} requires 6-20 matches from the non-capturing group in #1
Try this :
((?:([0-9()+\-])(?!\2{5})){6,20})
So , this part ?!\2{5} means how many times is allowed for each one from the pattern to be repeated like this 22222 and i put 5 as example and you could change it as you want .
i need help with a regexp.
It will be allowed to use 0-9 and allowed length is 2 or 3, but not if it begins with 0.
My exp:
^[0-9]{2,3}$
this allows ex. 03 or 033, but it should be disallowed.
just split it in 2 parts: 1st digit and other digits.
^[1-9][0-9]{1,2}$
Take your pick:
^[0-9]|[1-9][0-9]{1,2}$
^[0-9]|[1-9][0-9][0-9]?$
^0|[1-9][0-9]{0,2}$
^0|[1-9][0-9]?[0-9]?$
I'd personally choose the second-to-last or last one.
You can notice only the "0" case needs a particular match.
You can use a very simple regex like this:
^[1-9]\d{0,2}$
Working demo
I need to validate string with 2 groups which are separated with one space with next rules:
Each group needs to be at least 2 character long but less or equal to 15
Both groups together can't be more than 20 chars long (not counting space)
Groups can only contain letters (that's simple, it's [a-zA-Z])
Following these rules, here are some examples
Firstname Lastname (Valid)
Somename T (Invalid, 2nd one is <2)
Somethingsomettt Here (Invalid, first one is > 15)
Somethingsome Somethingsome (Invalid, total > 20)
It'd be simple [a-zA-Z]{2,15} [a-zA-Z]{2,15} if it wasn't for that 2+2<=total<=20 condition.
Is it even possible to limit it this way? If it is - how?
UPDATE
Just for the sake of it, resulting regex was supposed to be ^(?=[a-zA-Z ]{5,21}$)[a-zA-z]{2,15} [a-zA-Z]{2,15}$, #vks was closest one to it. Nevertheless, thanks #popovitsj and #Avinash Raj too.
^(?=.{5,21}$)[a-zA-Z]{2,15} [a-zA-Z]{2,15}$
Try this.See demo.
http://regex101.com/r/nA6hN9/30
This can be done with lookahead. Something like this:
^(?=.{1,20}$)[a-zA-z]{2,14} [a-zA-Z]{2,14}$
You could try the below regex which uses negative lookahead,
(?!^.{22,})^[a-zA-Z]{2,15} [a-zA-Z]{2,15}$
DEMO
I want a regex that checks the following things:
The string starts with an +
After the '+' only numbers can occur
There should be atleast 4 numbers after the +
Does anyone know how to make this?
/^+\d{4,}$/
will meet your requirements.
^ is the anchor for start fo the string
\d is a digit
{4,} says at least 4 of the preceding expression (here the \d). you can add a maximum if needed like {4,20} would allow at least 4 and at most 20 characters.
$ is the anchor for the end of the string
/^((00|\+)[0-9]{2,3}){0,1}[0-9]{4,14}$/
More general than your request, but you can specialize it. Explaining:
((00|\+)[0-9]{2,3})
international code with 00 or + and 2 or 3 digits. Modify the expression according to your needs.
{0,1}
international code is optional - remove it if it is required
[0-9]{4,14}
digits: minimum 4, maximum 14. Change the values according to your needs.
Regards
A.
/\+\d{4,15}/
This should help if 15 is the atmost limit of numbers
OR rather keep the second parameter blank as stema suggested.
I went with this one:
/\A(([+]\d{3,})?\d{6,8})/
I am beginner and have some problems with regexp.
Input text is : something idUser=123654; nick="Tom" something
I need extract value of idUser -> 123456
I try this:
//idUser is already 8 digits number
MatchCollection matchsID = Regex.Matches(pk.html, #"\bidUser=(\w{8})\b");
Text = matchsID[1].Value;
but on output i get idUser=123654, I need only number
The second problem is with nick="Tom", how can I get only text Tom from this expresion.
you don't show your output code, where you get the group from your match collection.
Hint: you will need group 1 and not group 0 if you want to have only what is in the parentheses.
.*?idUser=([0-9]+).*?
That regex should work for you :o)
Here's a pattern that should work:
\bidUser=(\d{3,8})\b|\bnick="(\w+)"
Given the input string:
something idUser=123654; nick="Tom" something
This yields 2 matches (as seen on rubular.com):
First match is User=123654, group 1 captures 123654
Second match is nick="Tom", group 2 captures Tom
Some variations:
In .NET regex, you can also use named groups for better readability.
If nick always appears after idUser, you can match the two at once instead of using alternation as above.
I've used {3,8} repetition to show how to match at least 3 and at most 8 digits.
API links
Match.Groups property
This is how you get what individual groups captured in a match
Use look-around
(?<=idUser=)\d{1,8}(?=(;|$))
To fix length of digits to 6, use (?<=idUser=)\d{6}(?=($|;))