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I am trying to looking for regex which allows any comma separated values. Like this.
23,23,23
233-2-3,23,23
23/23/2333,22-22-2222,23
I have tried couple of things but nothing fits 100%.
[^,;]+
/(?!,)(?:(?:,|^)([-+]?(?:\d*\.)?\d+))*$/
(\d+, ?)+(\d+)?
My use case is , if my string pass through above regex, I will parse all the values and store it into an array. If string doesn't have any single , than leave that string.
thanks
As stated in the comments, we have come to the following solution:
(?m)^(?:\d+[\/,-]){2,}\d+$
Explanation:
(?m) : make ^ and $ match start and end of line respectively.
^ : Start of line
(?: : Ignore group
\d+ : Match a digit 1 or more times
[\/,-] : Followed by a / or , or -
) : Closing parenthesis for the ignore group
{2,} : Match two times or more
\d+ : Match a digit 1 or more times
$ : End of line
Online demo
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I'm working on a regex to take #users
In italian, sometime there is something like l'#Orazio.
I can't find the right way to take this.
I'm using this line /(?<=^|\s)(#(\S+))/
this is my online tester
https://regex101.com/r/s5BTm0/12
as you can see I have an issue with the case 4
Any tips?
You may use
(?<!\S)(?:\w+['’])?#(\S+)
(?<!\w)(?:\w+['’])?#(\S+)
See the regex demo
Details
(?<!\S) - whitespace or start of string must appear immediately to the left of the current location
(?<!\w) - a location the is not immediately preceded with a word char
(?:\w+['’])? - an optinal sequence of 1+ word chars and then ' or ’
# - a # char
(\S+) - Capturing group 1: one or more non-whitespace chars.
Use an optional group:
(?<=^|\s)(?:\w?'?)(#(\S+))
https://regex101.com/r/s5BTm0/5
For more than 1 letter in front, use:
(?<=^|\s)(?:(\w*')?)(#(\S+))
https://regex101.com/r/s5BTm0/8
And to match all your updated cases:
https://regex101.com/r/s5BTm0/10
For different types of apostrophes:
(?<=^|\s)(?:(\w*['´])?)(#(\S+))
https://regex101.com/r/s5BTm0/13
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I have a string with alphanumerical characters like this:
hajshajs12 +1212121 12 1AAsdsd 5665789 +987 ++789 aBcD
Can I extract only the numeric characters with length >= 4 starting with + or without + using regex ?
So in the end I should have +1212121 5665789 +987 ++789
You can use this regex
([+](?:[+]\d{2,}|\d{3,})|\d{4,})\b
Regex Breakdown
( #Capturing group
[+] #Match + sign
(?: #Non-capturing group
[+]\d{2,} #Match another + followed by at least 2 digits
| #Alterantion (OR)
\d{3,} #Match at least 3 digits
)
| #Alterantion (OR)
\d{4,} #Match at least 4 digits
)
\b #Word boundary. Number shouldn't be in between alphabets
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I want to match a string where there are 5 characters where first four characters are A-Z and 5th is a digit. Also the first and the fourth character should be same.
I have a regex: [A-Z]{4}\d
However this wont check if 1st and 4th character are same. Please help
Regex: ^([A-Z])[A-Z]{2}\1\d$
1. ^ start of string.
2. ([A-Z]) capture first character.
3. [A-Z]{2} match next two character which can be in A-Z
4. \1 using captured group which contains first character of string.
5. \d a digit which can be 0-9
6. $ for end of string.
Regex demo
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Im looking for a regular expression for a textbox thats accepts either 10 digits, or 8 digits, followed by a dash, followed by 2 digits. Examples:
1212345678
or
12345678-91
If you just want to check if the input is valid without matching any content, this one should be enough:
^\d{8}-?\d\d$
Beginning with 8 digits, followed (or not) by an optional dash, and another 2 digits up to the end.
This Should Work.
((?:^[0-9]{10})|(?:^[0-9]{8}(?:\-\d{2})*))$
Input:
1212345678
12345678-91
12345678901112-10
1234567890-12
Output:
1212345678
12345678-91
See: https://regex101.com/r/AzbVXt/2
^\d{8}(\d{2})?(-\d{2})?$
^ and $ : match from the begging to the end
\d{8} : 8 digits
(\d{2})? : optional two other digits
(-\d{2})? : optional dash, with two other digits
This will match
1212345678
121234567800
Not 1212345678001
1212345678-
1212345678-00
Not 1212345678-001
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I have in Notepad++ plenty of datetime texts like:
'20160310 17:59:00'
And I want to add '-' character to separate numbers in the date like
'2016-03-10 17:59:00'
How I would be a RegEx expression to do that?
The following regex expression:
('\d{4})(\d{2})(\d{2} \d{2}:\d{2}:\d{2}')
Combined with the following replace expression:
$1-$2-$3
will achieve what you want.
Explanation:
('\d{4}) capture group 1: an apostrophe followed by 4 digits
(\d{2}) capture group 2: 2 digits
(\d{2} \d{2}:\d{2}:\d{2}') capture group 3: 2 digits, a space, 2 digits, a colon, 2 digits, a colon and 2 digits, followed by an apostrophe
$1-$2-$3 - capture group 1, followed by a -, followed by capture group 2, followed by a -, followed by capture group 3