This question already has answers here:
Regex phone number to accept + [duplicate]
(4 answers)
Closed 3 years ago.
I want to validate a phone number this format +905555555555.
How can I write a regex expression to test for this?
one "+" and 12 numbers. hint: escape the "+".
^\+(90)\([2-5]{1}\)[0-9]{9}
or not starts with +;
\+(90)\([2-5]{1}\)[0-9]{9}
<script type="text/javascript">
var test = "+905555555555";
var re = new RegExp(/^\+[0-9]{12}$/gi);
if(re.test(test))
{
alert("Number is valid");
} else {
alert("Number is not valid");
}
</script>
Check out this site for testing out your Regular Expressions: http://www.regextester.com/index2.html
And a good starting point to learn is here:
http://www.regular-expressions.info/
if you need it to start with a "+" and a "9" and 11 digits:
^[+]9\d{11}$
I recommend that you understand how regEx work, take a look at this usefull tester:
http://www.sweeting.org/mark/html/revalid.php
At the bottom they explain what each operator means.
Also there are all sort of examples at the internet.
EDIT: after reading the OP's comments, in order for the number to start with "+90" and then have 10 digits, you can use the following expression:
^[+]90\d{10}$
To cover your additional specifications use this:
^\+90\([2-5]\)\d{9}$
You definitely need the anchors ^ and $ to ensure that there is nothing ahead or after your string. e.g. if you don't use the $ the number can be longer as your specified 12 numbers.
You can see it online here on Regexr
If it's exactly that format then you could use something like: /^\+\d{12}\z/
If you'd like to allow some spaces/dashes/periods in it but still keep it at 12 numbers:
/^\+(?:\d[ .-]?){12}\z/
Then you could remove those other chars with:
s/[ .-]+//g (Perl/etc notation)
Adjusted a bit to answer question for a 12 digit regex.
// +905555555555 is +90 (555) 555-5555
let num = '+905555555555';
/^\+90\d{10}$/.test(num);
Related
This question already has an answer here:
javascript regexp replace not working, but string replace works
(1 answer)
Closed 1 year ago.
Hello team I am new to JS so I am trying to use RegEx with replacing to take input from the user and replace it if it doesn't match the RegEx I have to be able to put 7 digits or 6 digits followed with one letter currently I am doing this
someID.replace('^(([0-9]{1,7})|([0-9]{1,6}[a-zA-Z]{1}))$')
I am not able to replace the current string with the RegEx expression if I enter
12345678900 it remain the same in that situation I need to be 1234567 after the replace or if I have 12345678asd to be 123456a. How can I achieve that by only replace function and a RegEx expresion
You need to use a different regex and a dirrent replace function.
You will also need to get rid of $ if you want to be able to successfully match the string, without worrying about how it ends.
const sampleIDs = [
"123456789000",
"123456abc",
];
sampleIDs.forEach(id => {
const clean = id.match(/^\d{6}[\d\D]/);
console.log(clean[0]);
});
This question already has answers here:
How to detect exact length in regex
(8 answers)
Closed 4 years ago.
So I am trying to use a regular expression to check against strings but it doesn't seem to be working properly.
Basically I want it to match a alpha-numeric string that is exactly 3 characters long. The expression I am using below does not seem to be working for this:
const msg = message.content;
const regex = /[A-Za-z0-9]{3}/g;
if (msg.match(regex)) {
// Do something
}
Am I doing something wrong? Any help would be appreciated. Thanks in advance.
You need to add ^ and $ for the start-of-string anchor and end-of-string anchor, respectively - otherwise, for example, for #123, the 123 will match, and it will pass the regex. You also might consider using the i flag rather than repeat A-Za-z, and you can use \d instead of 0-9.
It looks like you just want to check whether the string passes the regex's test or not, in which case .test (evaluates to a boolean) might be a bit more appropriate than .match. Also, either way, there's no need for the global flag if you're just checking whether a string passes a regex:
const regex = /^[a-z\d]{3}$/i;
if (regex.test(msg)) {
// do something
}
This question already has an answer here:
Regular expression to validate US phone numbers? [duplicate]
(1 answer)
Closed 5 years ago.
I am trying to write a regex for US phone numbers where user can enter the number and dash comes automatically.
But here the drawback is if user enters "123" and then "-" the regex breaks and instead of 123-456-7890 it becomes 123-4567890
Here is the regex code:
$('#AccountFrm_telephone').attr('maxlength', '12');
$('#AccountFrm_telephone').keyup(function(){
$(this).val($(this).val().replace(/^(\d{3})(\d{3})(\d)+$/, "$1-$2-$3"));
});
Maybe There is something that we add in regex so that user can not type dash?
for completes:
you have 2 issues
first this part (\d)+ should be (\d+) or you wont capture the last group on numbers correctly.
the second part is that you aren't handling possible dashes in the input, so try something like:
.replace(/^(\d{3})-?(\d{3})-?(\d+)$/, "$1-$2-$3")
the question marks (?) denote 0 or 1 times, meaning the user can input the dashes if he wants
Maybe try the following:
$('#AccountFrm_telephone').attr('maxlength', '12');
$('#AccountFrm_telephone').keyup(function(){
$(this).val($(this).val().replace(/(\d{3})(\d{3})(\d{4})/, "$1-$2-$3"));
});
Try this:
$("#input1").on('keyup', e => {
e.target.value = e.target.value.replace(/(\d{3})-?(\d{3})-?(\d+)/, '$1-$2-$3')
})
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<input id="input1">
This question already has answers here:
Numeric validation with RegExp to prevent invalid user input
(3 answers)
Closed 7 years ago.
I need a regex for any number OR float.
I have used this but doesn't work:
/(^[0-9]+*[.][0-9]+)$|^[\d+]$/
Why ?
Try this regex:
/^[+-]*[0-9]+[.][0-9]+|[+-]*[0-9]+$/g
You can use http://www.regexr.com/ to test and create regexes
Try this one:
^[-+]?[0-9]*\.?[0-9]+$
In case this might help you. This online Regex Tools is actually very helpful:
http://www.regexr.com/
You have a few issues with your regex:
You cannot have a quantifier after + so change +* to just + (1 or more matches)
Move your start identifier ^ outside of the group, for consistency
hmmm, you seem to have edited your regex: but [^\d+] was looking for NOT a number or + symbol.
One possible solution would be as follows:
^([0-9]+[.][0-9]+)$|^\d+$
Here is a working example
For more example, see here
You expression doesn't work because it contains errors.
This is the website I use for RegEx testing - It is good to test them on a website like this that gives you feedback
You have the beginning inside a capturing group, but the end outside - (^...)$
You also have double operators - +* - Use only one
And the or not plus sign and digit is not needed - ?[^\d+]
I believe this expression will do what you want: ^-?\d+(\.\d+)?$
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is a regular expression for a MAC Address?
I would like to validate a string to ensure that it is a valid MAC Address.
Any ideas in Jquery or Javascript?
I have the following:
var mystring= '004F78935612' - This type of MAC Address
var rege = /([0-9a-fA-F][0-9a-fA-F]){5}([0-9a-fA-F][0-9a-fA-F])/;
alert(rege.test(mystring));
But its not all that accurate.
Ie. My tissue box is a valid MAC Address?!?
Thanks!
Taking the regular expression from this question, you would implement it like so:
var mystring= 'Hello';
var regex = /^([0-9A-F]{2}[:-]){5}([0-9A-F]{2})$/;
alert(regex.test(mystring));
http://jsfiddle.net/s2WDq/
This regular expression searches for the beginning of the string ^, then TWO hexidecimal digits [0-9A-F]{2}, then a colon or dash [:-], five times over (...){5}, then a final group of TWO hexidecimal digits [0-9A-F]{2}, and finally the end of the string $.
Edit: in response to your comment, Pinch, that format is not considered a valid MAC address. However, if you wanted to support it, simply add a ? in the right place:
/^([0-9A-F]{2}[:-]?){5}([0-9A-F]{2})$/
// question mark ^ allows the colon or dash to be optional