The general format is:
YYYY/MM/DD/INFO
Only the separators / are mandatory.
Each part is optional.
YYYY - exactly 4 numbers.
MM - exactly 2 numbers.
DD - exactly 2 numbers.
INFO - any sequence of letters, spaces or hyphens.
So these are valid strings:
2020/06/25/XYZConf
2020///XYZConf
2020//25/XYZConf
2020/06//XYZConf
//25/XYZConf
///
I'm really struggling to come up with a regex that validates optional parts while maintaining the integrity of the string as a whole.
How would you write this regular expression?
PS: This needs to be a regular expression as it will be part of a third-party lexer that doesn't accept anything else.
Relevant posts:
Regex to validate comma-separated numbers with optional fractional parts
Regex for dates that matches for every stage of valid date entry
You could try something like:
^(?:\d{4})?\/(?:(?:\d\d)?\/){2}(?:[A-Za-z\s-]+)?$
See the Online Demo
I believe that you are looking for optional (non)capturing groups. The pattern above matches:
^ - Start string ancor.
(?: - Open 1st non-capturing group.
\d{4} - Match 4 digits.
)? - Close 1st non-capturing group and make it optional.
\/ - Match a forward slash.
(?: - Open 2nd non-capturing group.
(?: - Open 3rd non-capturing group.
\d\d - Match two digits.
)? - Close 3rd non-capturing group and make it optional.
\/ - Match a forward slash.
){2} - Close 2nd non-capturing group and make it match twice.
(?: - Open 4th non-capturing group.
[A-Za-z\s-]+ - Match upper- and lowercase letters, a space and hyphen at least one time (in any sequence as per your OP).
)? - Close 4th non-capturing group and make it optional.
$ - End string ancor.
Related
I am trying to create regex where user have to enter exactly the same thing no extra no less
Here is my regex;
/[a-zA-Z0-9][a-zA-Z0-9\-]*\.myshopify\.com/
when I test this with, for example, myshop.myshopify.coma it returns true or myshop.myshopify.com myshop123.myshopify.com still returns true
What I am trying to get is if user enters myshop.myshopify.coma or myshop321.myshopify.com myshop123.myshopify.coma it shouldn't be match.
It should only match when the entire input is exactly like this [anything except ()=>%$ etc].myshopify.com
what should I include in my regex to strictly test exactly one thing.
you can use boundary-type assertions to match the beginning of an input (^) and an end ($) - to make sure your input matches fully.
const pattern = /^[a-zA-Z0-9][a-zA-Z0-9\-]*\.myshopify\.com$/
console.log(pattern.test('myshop.myshopify.com')) // true
console.log(pattern.test('myshop.myshopify.coma')) // false
console.log(pattern.test('myshop.myshopify.com myshop123.myshopify.com')) // false
You'd currently allow for input like "A---", so besides the good point about start and end line anchors, you'd maybe want to reconsider your pattern. Maybe something like:
^[a-z\d]+(?:-[a-z\d]+)*\.myshopify\.com$
See the online demo
^ - Start line anchor.
[a-z\d]+ - 1+ any alnum character.
(?: - Open non-capture group:
-[a-z\d]+ - A literal hyphen followed by 1+ alnum chars.
)* - Close non-capture group and match it zero or more times.
\.myshopify\.com - Match a ".myshopify.com" literallyy.
$ - End line anchor.
A 2nd option would be to use a negative lookahead to achieve the same concept:
^(?!-|.*-[-.])[a-z\d-]+\.myshopify\.com$
See the online demo
^ - Start line anchor.
(?! - Negative lookahead for:
- - A leading hypen
| - Or:
.*-[-.] - Any character other than newline zero or more times up to an hypen with either another hypen or a literal dot.
) - Close negative lookahead.
[a-zA-Z\d]+ - 1+ any alnum character.
\.myshopify\.com - Match a ".myshopify.com" literallyy.
$ - End line anchor.
In both cases I used both the global and case-insensitive flags: /<pattern>/gi. See a sample below:
const patt1 = /^[a-z\d]+(?:-[a-z\d]+)*\.myshopify\.com$/gi
console.log(patt1.test('myshop.myshopify.com'))
console.log(patt1.test('myshop-.myshopify.com'))
const patt2 = /^(?!-|.*-[-.])[a-z\d-]+\.myshopify\.com$/gi
console.log(patt2.test('myshop.myshopify.com'))
console.log(patt2.test('myshop-.myshopify.com'))
I have an input that I want to apply validation to. User can type any integer (positive or negative) numbers separated with a comma. I want to
Some examples of allowed inputs:
1,2,3
-1,2,-3
3
4
22,-33
Some examples of forbidden inputs:
1,,2
--1,2,3
-1,2,--3
asdas
[]\%$1
I know a little about regex, I tried lots of ways, they're not working very well see this inline regex checker:
^[-|\d][\d,][\d]
You can use
^(?:-?[0-9]+(?:,(?!$)|$))+$
https://regex101.com/r/PAyar7/2
-? - Lead with optional -
[0-9]+ - Repeat digits
(?:,(?!$)|$)) - After the digits, match either a comma, or the end of the string. When matching a comma, make sure you're not at the end of the string with (?!$)
As per your requirements I'd use something simple like
^-?\d+(?:,-?\d+)*$
at start ^ an optional minus -? followed by \d+ one or more digits.
followed by (?:,-?\d+)* a quantified non capturing group containing a comma, followed by an optional hyphen, followed by one or more digits until $ end.
See your updated demo at regex101
Another perhaps harder to understand one which might be a bit less efficient:
^(?:(?:\B-)?\d+,?)+\b$
The quantified non capturing group contains another optional non capturing group with a hyphen preceded by a non word boundary, followed by 1 or more digits, followed by optional comma.
\b the word boundary at the $ end ensures, that the string must end with a word character (which can only be a digit here).
You can test this one here at regex101
Need some help in designing regular expression to validate hyphen separated floating point numbers in Javascript. So far I have managed to achieve this RegEx:
(^((\\d)+(\.[0-9]+)?)(\-)?((\\d)+(\.[0-9]+)?)$)|^(\\d+)$
It matches the following:
1) 2
2) 2.10
3) 3.10-3.14
The problem with this one is that its also matching "3.103.310" which is wrong number. Much appreciate any help in fixing this issue.
The problem comes from the first alternative that matches 1 or more digits with an optional fractional part ((\d)+(\.[0-9]+)?) and then matches a hyphen and again 1+ digits and again an optional fractional part. Thus, 2 dots are allowed.
You may fix the pattern like this:
^\d+(?:\.\d+)?(?:-\d+(?:\.\d+)?)*$
See the regex demo
Details
^ - start of string
\d+ - 1+ digits
(?:\.\d+)? - an optional non-capturing group:
\. - a dot
\d+ - 1+ digits
(?:-\d+(?:\.\d+)?)* - an non-capturing group matching 0+ occurrences of
- - a hyphen
\d+(?:\.\d+)? - 1+ digits and 1 or 0 occurrences of . and 1+ digits
$ - end of string
jQuery Regex
/((\b([a-zA-Z]{0,15})\b)([^a-z0-9\$_]))/g
My Attempt So Far: https://regex101.com/r/d3VUpG/1
Example test string:
(options.method==="
|options.method==="
=options.method==="HEAD"
options.method.options.method==="HEAD"
What I'm Trying TO Achieve
Returned as $1 the value of any connected words such as:
options.method - Would = $1
options.method.options.method - Would also = $1
Question
How can I find all words connected with a dot (.) to then wrap in a span like the below example;
.replace(//gi,'<span class="join">$1</span>')
You can use the following expression:
/((?:\w+\.)+\w+)/g
Explanation:
( - Start of capturing group 1
(?: - Start of a non-capturing group
\w+\. - Match [a-zA-Z0-9_] characters one or more times followed by a literal . character
)+ - End of the non-capturing group; match the group one or more times
\w+ - Match [a-zA-Z0-9_] characters one or more times
) - End of capturing group 1
So in other words, the non-capturing group, (?:\w+\.)+, will match a substring like option. one or more times followed by a final \w+ which will match the final word without a literal . character following it. Since there is only one capturing group wrapping everything, you can wrap your span tag around the first group, $1.
Live Example
string.replace(/((?:\w+\.)+\w+)/g, '<span class="join">$1</span>');
As mentioned above, \w includes underscore, numbers and letters ([a-zA-Z0-9_]), so if you only want to match letter characters, then you could swap out \w with [a-z] and use the case-insensitive flag:
/((?:[a-z]+\.)+[a-z]+)/gi
I am using regex to add a survey to pages and I want to include it on all pages except payment and signin pages. I can't use look arounds for the regex so I am attempting to use the following but it isn't working.
^/.*[^(credit|signin)].*
Which should capture all urls except those containing credit or signin
[ indicates the start of a character class and the [^ negation is per-character. Thus your regular expression is "anything followed by any character not in this class followed by anything," which is very likely to match anything.
Since you are using specific strings, I don't think a regular expression is appropriate here. It would be a lot simpler to check that credit and signin don't exist in the string, such as with JavaScript:
-1 === string.indexOf("credit") && -1 === string.indexOf("signin")
Or you could check that a regular expression does not match
false === /credit|signin/.test(string)
Whitelisting words in regex is generally pretty easy, and usually follows a form of:
^.*(?:option1|option2).*$
The pattern breaks down to:
^ - start of string
.* - 0 or more non-newline characters*
(?: - open non-capturing group
option1|option2 - | separated list of options to whitelist
) - close non-capturing group
.* - 0 or more non-newline characters
$ - end of string
Blacklisting words in a regex is a bit more complicated to understand, but can be done with a pattern along the lines of:
^(?:(?!option1|option2).)*$
The pattern breaks down to:
^ - start of string
(?: - open non-capturing group
(?! - open negative lookahead (the next characters in the string must not match the value contained in the negative lookahead)
option1|option2 - | separated list of options to blacklist
) - close negative lookahead
. - a single non-newline character*
) - close non-capturing group
* - repeat the group 0 or more times
$ - end of string
Basically this pattern checks that the values in the blacklist do not occur at any point in the string.
* exact characters vary depending on the language, so use caution
The final version:
/^(?:(?!credit|signin).)*$/