I'm extracting the phone numbers that begin with 9 followed by other 9 digits from tweets using JavaScript.
Here's the regex pattern I am using:
var numberPattern = /^9[0-9]{9}/;
Here's the pattern matching phase:
var numberstring = JSON.stringify(data[i].text);
if(numberPattern.test(data[i].text.toString()) == true){
var obj={
tweet : {
status : data[i].text
},
phone : numberstring.match(numberPattern)
}
//console.log(numberstring.match(numberPattern));
stringarray.push(obj);
The problem is it is working for few numbers and not all. Also, I want to modify the regex to accept +91 prefix to numbers as well and(or) reject a starting 0 in numbers. I'm a beginner in regex, so help is needed. Thanks.
Example:
#Chennai O-ve blood for #arun_scribbles 's friend's father surgery in few days. Pl call 9445866298. 15May. via #arun_scribbles
Your regex pattern seems to be designed to allow a 9 or 8 at the beginning, but it would be better to enclose that choice in parentheses: /^(9|8)[0-9]{9}/.
To allow an optional "+" at the beginning, follow it with a question mark to make it optional: /^\+?(9|8)[0-9]{9}/.
To allow any character except "0", replace the (9|8) with a construct to accept only 1-9: /^\+?[1-9][0-9]{9}/.
And in your example, the phone number doesn't come at the beginning of the line, so the caret will not find it. If you're looking for content in the middle of the line, you'll need to drop the caret: /\+?[1-9][0-9]{9}/.
var numberPattern = /([+]91)?9[0-9]{9}\b/;
Try this regex pattern: [\+]?[0-9]{1,4}[\s]?[0-9]{10}
It accepts any country code with +, a space, and then 10 digit number.
Related
I'm using this Wordpress plugin called 'Easy contact form' which offers standard validation methods.
It uses the following regex for phone numbers:
/^(\+{0,1}\d{1,2})*\s*(\(?\d{3}\)?\s*)*\d{3}(-{0,1}|\s{0,1})\d{2}(-{0,1}|\s{0,1})\d{2}$/
Just now it allows the following formats (maybe more):
0612345678
+31612345678
But I want it to allow +316-12345678 and 06-12345678 also ... Is this possible? If so, how?
Thanks in advance!
You can use a less complex regex :
^\+?\d{2}(-?\d){8,9}$
This regex allows a + at the beginning of the phone number, then matches two digits, and after that, digits preceded (or not) by a -, for a total of 10 or 11 digits.
Now you can adapt it if the initial + is only for 11-digits phone numbers :
^\+?\d{3}(-?\d){9}|\d{2}(-?\d){8}$
My regex allow you to use - every digit. If that's an issue, it can be changed :
^\+?\d{3}(-?\d{2}){4}|\d{2}(-?\d{2}){4}$
I think this last regex will answer your needs, and it's quite simple !
When you figure out what patterns you actually want to allow and not allow. Then fill them out in this function and if the statement returns true you have your regex.
var regex = /^\+?\d{3}(-?\d{2}){4}|\d{2}(-?\d{2}){4}$/; //this is your regex, based on #Theox third answer
//allowed patterns
['0612345678', '+31612345678', '+316-12345678', '06-12345678'].every(function(test) {
return regex.exec(test) !== null;
}) &&
//disallowed patterns, none right now
[].every(function(test) {
return regex.exec(test) === null;
});
While i type the phone number in a field after 5 digits put a hypen and continue the digits
after hypen. It allows only 10 digits and hypen.
Output : 90000-00000
Can any one help me how to fix the issue.
Thanks
The ReGex for this is really simple, you should have been able to Google this and work it out for yourself. The RegEx you want is as follows:
^[\d]{5}-[\d]{5}$
Using that with javascript:
var input = "90000-00000";
var valid = /^[\d]{5}-[\d]{5}$/.test(input);
//valid is a boolean: true or false
Here is a working example
Greetings overflowers,
I'm trying to write a regular expression to validate phone numbers of the form ########## (10 digits)
i.e. this is these are cases that would be valid: 1231231234 or 1111111111. Invalid cases would be strings of digits that are less than 10 digits or more than 10 digits.
The expression that I have so far is this:
"\d{10}"
Unfortunately, it does not properly validate if the string is 11+ digits long.
Does anyone know of an expression to achieve this task?
You need to use ancors, i.e.
/^\d{10}$/
You need to anchor the start and the end too
/^\d{10}$/
This matches 10 digits and nothing else.
This expression work for google form 10 digit phone number like below:
(123) 123 1234 or 123-123-1234 or 123123124
(\W|^)[(]{0,1}\d{3}[)]{0,1}[\s-]{0,1}\d{3}[\s-]{0,1}\d{4}(\W|$)
I included the option to use dashes (xxx-xxx-xxxx) for a better user experience (assuming this is your site):
var regex = /^\d{3}-?\d{3}-?\d{4}$/g
window.alert(regex.test('1234567890'));
http://jsfiddle.net/bh4ux/279/
I usually use
phone_number.match(/^[\(\)\s\-\+\d]{10,17}$/)
To be able to accept phone numbers in formats 12345678, 1234-5678, +12 345-678-93 or (61) 8383-3939 there's no real convention for people entering phone numbers around the world. Hence if you don't have to validate phone numbers per country, this should mostly work. The limit of 17 is there to stop people from entering two many useless hyphens and characters.
In addition to that, you could remove all white-space, hyphens and plus and count the characters to make sure it's 10 or more.
var pureNumber = phone_number.replace(/\D/g, "");
A complete solution is a combination of the two
var pureNumber = phone_number.replace(/\D/g, "");
var isValid = pureNumber.length >= 10 && phone_number.match(/^[\(\)\s\-\+\d]{10,17}$/) ;
Or this (which will remove non-digit characters from the string)
var phoneNumber = "(07) 1234-5678";
phoneNumber = phoneNumber.replace(/\D/g,'');
if (phoneNumber.length == 10) {
alert(phoneNumber + ' contains 10 digits');
}
else {
alert(phoneNumber + ' does not contain 10 digits');
}
I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}
By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.
If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.
Friends,
I'm new to both Javascript and Regular Expressions and hope you can help!
Within a Javascript function I need to check to see if a comma(,) appears 1 or more times. If it does then there should be one or more numbers either side of it.
e.g.
1,000.00 is ok
1,000,00 is ok
,000.00 is not ok
1,,000.00 is not ok
If these conditions are met I want the comma to be removed so 1,000.00 becomes 1000.00
What I have tried so is:
var x = '1,000.00';
var regex = new RegExp("[0-9]+,[0-9]+", "g");
var y = x.replace(regex,"");
alert(y);
When run the alert shows ".00" Which is not what I was expecting or want!
Thanks in advance for any help provided.
strong text
Edit
strong text
Thanks all for the input so far and the 3 answers given. Unfortunately I don't think I explained my question well enough.
What I am trying to achieve is:
If there is a comma in the text and there are one or more numbers either side of it then remove the comma but leave the rest of the string as is.
If there is a comma in the text and there is not at least one number either side of it then do nothing.
So using my examples from above:
1,000.00 becomes 1000.00
1,000,00 becomes 100000
,000.00 is left as ,000.00
1,,000.00 is left as 1,,000.00
Apologies for the confusion!
Your regex isn't going to be very flexible with higher orders than 1000 and it has a problem with inputs which don't have the comma. More problematically you're also matching and replacing the part of the data you're interested in!
Better to have a regex which matches the forms which are a problem and remove them.
The following matches (in order) commas at the beginning of the input, at the end of the input, preceded by a number of non digits, or followed by a number of non digits.
var y = x.replace(/^,|,$|[^0-9]+,|,[^0-9]+/g,'');
As an aside, all of this is much easier if you happen to be able to do lookbehind but almost every JS implementation doesn't.
Edit based on question update:
Ok, I won't attempt to understand why your rules are as they are, but the regex gets simpler to solve it:
var y = x.replace(/(\d),(\d)/g, '$1$2');
I would use something like the following:
^[0-9]{1,3}(,[0-9]{3})*(\.[0-9]+)$
[0-9]{1,3}: 1 to 3 digits
(,[0-9]{3})*: [Optional] More digit triplets seperated by a comma
(\.[0-9]+): [Optional] Dot + more digits
If this regex matches, you know that your number is valid. Just replace all commas with the empty string afterwards.
It seems to me you have three error conditions
",1000"
"1000,"
"1,,000"
If any one of these is true then you should reject the field, If they are all false then you can strip the commas in the normal way and move on. This can be a simple alternation:
^,|,,|,$
I would just remove anything except digits and the decimal separator ([^0-9.]) and send the output through parseFloat():
var y = parseFloat(x.replace(/[^0-9.]+/g, ""));
// invalid cases:
// - standalone comma at the beginning of the string
// - comma next to another comma
// - standalone comma at the end of the string
var i,
inputs = ['1,000.00', '1,000,00', ',000.00', '1,,000.00'],
invalid_cases = /(^,)|(,,)|(,$)/;
for (i = 0; i < inputs.length; i++) {
if (inputs[i].match(invalid_cases) === null) {
// wipe out everything but decimal and dot
inputs[i] = inputs[i].replace(/[^\d.]+/g, '');
}
}
console.log(inputs); // ["1000.00", "100000", ",000.00", "1,,000.00"]