The conditions of the regex are as follows:
Starts with either digits or a '+' sign and ends with digits.
This is going to be used to validate a certain type of number. What I got so far is:
/^\d*|\+\d*$/
This regex seems to match any string though. How would a regex that matches my conditions look like?
The regex will be used in a JavaScript function.
I think you want something like this,
^(?:[+\d].*\d|\d)$
^ Asserts that we are at the start.
[+\d] Matches a plus symbol or a digit.
.* Matches any character zero or more times.
\d Matches a digit.
| OR
\d A single digit.
$ Asserts that we are at the end.
Use this if you want to match also a line which has a single plus or digit.
^[+\d](?:.*\d)?$
DEMO
You need to use anchors ^ and $ on both sides of your regex and make first part + or digit) optional.
You can use this regex:
^([+\d].*)?\d$
RegEx Demo
Related
I have a special requirement, where i need the achieve the following
No Special Character is allowed except _ in between string.
string should not start or end with _, . and numeric value.
underscore should not be allowed before or after any numeric value.
I am able to achieve most of it, but my RegEx pattern is also allowing other special characters.
How can i modify the below RegEx pattern to not allow any special character apart from underscore that to in between strings.
^[^0-9._]*[a-zA-Z0-9_]*[^0-9._]$
What you might do is use negative lookaheads to assert your requirements:
^(?![0-9._])(?!.*[0-9._]$)(?!.*\d_)(?!.*_\d)[a-zA-Z0-9_]+$
Explanation
^ Assert the start of the string
(?![0-9._]) Negative lookahead to assert that the string does not start with [0-9._]
(?!.*[0-9._]$) Negative lookahead to assert that the string does not end with [0-9._]
(?!.*\d_) Negative lookahead to assert that the string does not contain a digit followed by an underscore
(?!.*_\d) Negative lookahead to assert that the string does not contain an underscore followed by a digit
[a-zA-Z0-9_]+ Match what is specified in the character class one or more times. You can add to the character class what you would allow to match, for example also add a .
$ Assert the end of the string
Regex demo
Keep it simple. Only allow underscore and alphanumeric regex:
/^[a-zA-Z0-9_]+$/
Javascript es6 implementation (works for React):
const re = /^[a-zA-Z0-9_]+$/;
re.test(variable_to_test);
Your opening and closing sections; [^0-9._], say match ANY character other than those.
So you need to change it to be what you can match.
/^[A-Z][A-Z0-9_]*[A-Z]$/i
And since you now said one character is valid:
/^[A-Z]([A-Z0-9_]*[A-Z])?$/i
I have this ^[a-zA-Z0-9 #&$]*$, but not working for me in few cases.
If someone types
A string that only consists of digits (e.g. 1234567)
A string starting with a special character (e.g. &123abc)
need to be rejected. Note that a special char can be in the middle and at the end.
You seem to need to avoid matching strings that only consist of digits and make sure the strings start with an alphanumeric. I assume you also need to be able to match empty strings (the original regex matches empty strings).
That is why I suggest
^(?!\d+$)(?:[a-zA-Z0-9][a-zA-Z0-9 #&$]*)?$
See the regex demo
Details
^ - start of string
(?!\d+$) - the negative lookahead that fails the match if a string is numeric only
(?:[a-zA-Z0-9][a-zA-Z0-9 #&$]*)? - an optional sequence of:
[a-zA-Z0-9] - a digit or a letter
[a-zA-Z0-9 #&$]* - 0+ digits, letters, spaces, #, & or $ chars
$ - end of string.
you can do it with the following regex
^(?!\d+$)\w+\S+
check the demo here
I am trying to get a regular expression to work but am stumped. What I want is to do the inverse of this:
/(\w)\1{5,}/
This regex does the exact opposite of what I'm trying to do. I would like to get everything but a string that has 6 repeating numbers i.e. 111111 or 999999.
Is there a way to use a negative look-around or something with this regex?
You can use this rgex:
/^(?!.*?(\w)\1{5}).*$/gm
RegEx Demo
(?!.*?(\w)\1{5}) is a negative lookaahead that will fail the match if there are 6 consecutive same word characters in it.
I'd rather go with the \d shorthand class for digits since \w also allows letters and an underscore.
^(?!.*(\d)\1{5}).*$
Regex explanation:
^ - Start of string/line anchor
(?!.*(\d)\1{5}) - The negative lookahead checking if after an optional number of characters (.*) we have a digit ((\d)) that is immediately followed with 5 identical digits (\1{5}).
.* - Match 0 or more characters up to the
$ - End of string/line.
See demo. This regex will allow
I would like to use regex in javascript to put a zero before every number that has exactly one digit.
When i debug the code in the chrome debugger it gives me a strange result where only every second match the zero is put.
My regex
"3-3-7-3-9-8-10-5".replace(/(\-|^)(\d)(\-|$)/g, "$10$2$3");
And the result i get from this
"03-3-07-3-09-8-10-05"
Thanks for the help
Use word boundaries,
(\b\d\b)
Replacement string:
0$1
DEMO
> "3-3-7-3-9-8-10-5".replace(/(\b\d\b)/g, "0$1")
'03-03-07-03-09-08-10-05'
Explanation:
( starting point of first Capturing group.
\b Matches between a word character and a non word character.
\d Matches a single digit.
\b Matches between a word character and a non word character.
) End of first Capturing group.
You can use this better lookahead based regex to prefix 0 before every single digit number:
"3-3-7-3-9-8-10-5".replace(/\b(\d)\b(?=-|$)/g, "0$1");
//=> "03-03-07-03-09-08-10-05"
Reason why you're getting alternate prefixes in your regex:
"3-3-7-3-9-8-10-5".replace(/(\-|^)(\d)(\-|$)/g, "$10$2$3");
is that rather than looking ahead you're actually matching hyphen after the digit. Once a hyphen has been matched it is not matched again since internal regex pointer has already moved ahead.
use a positive lookahead to see the one digit numbers :
"3-3-7-3-9-8-10-5".replace(/(?=\b\d\b)/g, "0");
I want to restrict the user from entering same consecutive digits in a text field e.g User can't enter string like John22 or 22John or jo22hn....He can enter string like Joh2n2 , 2Joh2n and so on...All this has to be done in Javascript (Using regular expressions would be a better option)...Please help
Test a string for consecutive digits:
/(\d)\1/.test(string)
You can do this by using a negative lookahead.
^(?!.*(\d)\1).*$
See it here at Regexr
The ^ and the $ anchor the match at the start and the end of the string.
.* Will match everything (except newline characters)
The important part here is the Negative lookahead (?!.*(\d)\1) it will check the whole string for a digit \d put it in a capture group because of the brackets (\d) and reuse the value using the backreference \1 and the whole thing fails it there is a digit followed by the same digit.
The following regex should help:
/[0-9]{2,}/
Or
/[\d]{2,}/
Although, you can match for all instances using the /g flag:
/[0-9]{2,}/g
See it at this JSFiddle