I have requirement to validate telephone number (TN) extension (Just extension only). The extension can be in 3-6 digit length and in 3 digit extension that should not follow 11. And yes other things, the extension should not have special chars and all zeros.
For example: 911, 311 etc.,
We have written the below one.
(?!0+$)[0-9](?!.*11).[0-9]*$
The issue with the above is
For 311, 211 --> Validation is pass.
For 38311, 2311 --> Those are 4 and 5 digit length extension and it can be suffixed with '11'. But the above pattern is not allowing it. How can I achieve that?
You could use:
(?!^((0+)|(\d11))$)(?=^\d{3,6}$).*
(?!^((0+)|(\d11))$) - From start to finish make sure it's not all zeros nor a digit followed by 11
(?=^\d{3,6}$) - From start to finish make sure we are dealing with 3 to 6 digits
.* - If the previous validations passed then it's safe to grab everything
https://regex101.com/r/eIVvvX/1
For checking nonzero you can simply use > operator and for rest of rules you can use this pattern
let data = ['911','311','38311','2311','000000','123111', '112']
data.forEach(v=>{
console.log(v, '\t' , v > 0 && /^(?:(?:(?!11$)\d){3}|\d{4,6})$/.test(v))
})
For checking non zero you can use regex too, but i prefer the above method personally
^(?!^0+$)(?:(?:(?!11$)\d){3}|\d{4,6})$
Related
Requirement is mobile number should start with 61 to 99
like 61xxxxxxxx, 62xxxxxxxxx... , 99xxxxxxxxxx
Need regular expression to match this case.
If mobile no is start with 0 or 11,12 or anything less than 61 then it should be invalid
Mobile no is max 10 digits, no country code needed.
You're probably better off using whatever programming tool you have to evaluate whether the first 2 digits are in range, far simpler and probably performant too. However, if you strictly want to use regex, this will do-
(?:6[1-9]|[7-9][0-9])\d{8}$
Here's the demo
It essentially, checks the first digit, if it's a 6, the next digit should be in range [1-9], if it's a 7, 8 or 9 (i.e range [7-9]), the next digit can be in range [0-9]. Then there should be 8 digits that follow.
Ofcourse, this above is a simple and easy to understand solution. Essentially checking each first digit and then matching the next. However if your regex flavor supports negative lookbehind, you could probably shorten this a bit more (sacrificing readability for brevity) but I do prefer this.
You could generate the prefix for the numbers and add a pattern for the remaining 8 digits.
Something like this
const regexp = new RegExp('('+[...Array(39).keys()].map(key => key + 61).join('|') + ')\\d{8,8}')
I want to setup some validation on an <input> to prevent the user from entering wrong characters. For this I am using ng-pattern. It currently disables the user from entering wrong characters, but I also noticed this is not the expected behavior so I am also planning on creating a directive.
I am using
AngularJS: 1.6.1
What should the regex match
Below are the requirements for the regex string:
Number 0x to xx (example 01 to 93)
Number x to xx (example 9 to 60)
Characters are not allowed
Special characters are not allowed
Notice:
the 'x' is variable and could be any number between 0 and 100.
The number on the place of 'x' is variable so if it is possible to create a string that is easily changeable that would be appreciated!
What I tried
A few regex strings I tried where:
1) ^0*([0-9]\d{1,2})$
--> Does match 01 but not 1
--> Does match 32 where it shouldn't
2) ^[1-9][0-9]?$|^31$
--> Does match 1 but not 01
--> Does match 32 where it shouldn't
For testing I am using https://regex101.com/tests.
What am I missing in my attempts?
If your aim is to match 0 to 100, here's a way, based on the previous solution.
\b(0?[1-9]|[1-9][0-9]|100)\b
Basically, there's 3 parts to that match...
0?[1-9] Addresses numbers 1 to 9, by mentionning that 0 migh be present
[1-9][0-9] covers number 10 to 99, the [1-9] representing the tens
100 covers for 100
Here's an example of it
Where you to require to set the higher boundary to 42, the middle part of the expression would become [1-3][0-9] (covering 10 to 39) and the last part would become 4[0-2] (covering 40 to 42) like so:
\b(0?[1-9]|[1-3][0-9]|4[0-2])\b
This should work:
^(0?[1-9]|[12][0-9]|3[01])$
https://regex101.com/r/BYSDwz/1
i need a regular expression for decimal/float numbers like 12 12.2 1236.32 123.333 and +12.00 or -12.00 or ...123.123... for using in javascript and jQuery.
Thank you.
Optionally match a + or - at the beginning, followed by one or more decimal digits, optional followed by a decimal point and one or more decimal digits util the end of the string:
/^[+-]?\d+(\.\d+)?$/
RegexPal
The right expression should be as followed:
[+-]?([0-9]*[.])?[0-9]+
this apply for:
+1
+1.
+.1
+0.1
1
1.
.1
0.1
Here is Python example:
import re
#print if found
print(bool(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0')))
#print result
print(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0').group(0))
Output:
True
1.0
If you are using mac, you can test on command line:
python -c "import re; print(bool(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0')))"
python -c "import re; print(re.search(r'[+-]?([0-9]*[.])?[0-9]+', '1.0').group(0))"
You can check for text validation and also only one decimal point validation using isNaN
var val = $('#textbox').val();
var floatValues = /[+-]?([0-9]*[.])?[0-9]+/;
if (val.match(floatValues) && !isNaN(val)) {
// your function
}
This is an old post but it was the top search result for "regular expression for floating point" or something like that and doesn't quite answer _my_ question. Since I worked it out I will share my result so the next person who comes across this thread doesn't have to work it out for themselves.
All of the answers thus far accept a leading 0 on numbers with two (or more) digits on the left of the decimal point (e.g. 0123 instead of just 123) This isn't really valid and in some contexts is used to indicate the number is in octal (base-8) rather than the regular decimal (base-10) format.
Also these expressions accept a decimal with no leading zero (.14 instead of 0.14) or without a trailing fractional part (3. instead of 3.0). That is valid in some programing contexts (including JavaScript) but I want to disallow them (because for my purposes those are more likely to be an error than intentional).
Ignoring "scientific notation" like 1.234E7, here is an expression that meets my criteria:
/^((-)?(0|([1-9][0-9]*))(\.[0-9]+)?)$/
or if you really want to accept a leading +, then:
/^((\+|-)?(0|([1-9][0-9]*))(\.[0-9]+)?)$/
I believe that regular expression will perform a strict test for the typical integer or decimal-style floating point number.
When matched:
$1 contains the full number that matched
$2 contains the (possibly empty) leading sign (+/-)
$3 contains the value to the left of the decimal point
$5 contains the value to the right of the decimal point, including the leading .
By "strict" I mean that the number must be the only thing in the string you are testing.
If you want to extract just the float value out of a string that contains other content use this expression:
/((\b|\+|-)(0|([1-9][0-9]*))(\.[0-9]+)?)\b/
Which will find -3.14 in "negative pi is approximately -3.14." or in "(-3.14)" etc.
The numbered groups have the same meaning as above (except that $2 is now an empty string ("") when there is no leading sign, rather than null).
But be aware that it will also try to extract whatever numbers it can find. E.g., it will extract 127.0 from 127.0.0.1.
If you want something more sophisticated than that then I think you might want to look at lexical analysis instead of regular expressions. I'm guessing one could create a look-ahead-based expression that would recognize that "Pi is 3.14." contains a floating point number but Home is 127.0.0.1. does not, but it would be complex at best. If your pattern depends on the characters that come after it in non-trivial ways you're starting to venture outside of regular expressions' sweet-spot.
Paulpro and lbsweek answers led me to this:
re=/^[+-]?(?:\d*\.)?\d+$/;
>> /^[+-]?(?:\d*\.)?\d+$/
re.exec("1")
>> Array [ "1" ]
re.exec("1.5")
>> Array [ "1.5" ]
re.exec("-1")
>> Array [ "-1" ]
re.exec("-1.5")
>> Array [ "-1.5" ]
re.exec(".5")
>> Array [ ".5" ]
re.exec("")
>> null
re.exec("qsdq")
>> null
For anyone new:
I made a RegExp for the E scientific notation (without spaces).
const floatR = /^([+-]?(?:[0-9]+(?:\.[0-9]+)?|\.[0-9]+)(?:[eE][+-]?[0-9]+)?)$/;
let str = "-2.3E23";
let m = floatR.exec(str);
parseFloat(m[1]); //=> -2.3e+23
If you prefer to use Unicode numbers, you could replace all [0-9] by \d in the RegExp.
And possibly add the Unicode flag u at the end of the RegExp.
For a better understanding of the pattern see https://regexper.com/.
And for making RegExp, I can suggest https://regex101.com/.
EDIT: found another site for viewing RegExp in color: https://jex.im/regulex/.
EDIT 2: although op asks for RegExp specifically you can check a string in JS directly:
const isNum = (num)=>!Number.isNaN(Number(num));
isNum("123.12345678E+3");//=> true
isNum("80F");//=> false
converting the string to a number (or NaN) with Number()
then checking if it is NOT NaN with !Number.isNaN()
If you want it to work with e, use this expression:
[+-]?[0-9]+([.][0-9]+)?([eE][+-]?[0-9]+)?
Here is a JavaScript example:
var re = /^[+-]?[0-9]+([.][0-9]+)?([eE][+-]?[0-9]+)?$/;
console.log(re.test('1'));
console.log(re.test('1.5'));
console.log(re.test('-1'));
console.log(re.test('-1.5'));
console.log(re.test('1E-100'));
console.log(re.test('1E+100'));
console.log(re.test('.5'));
console.log(re.test('foo'));
Here is my js method , handling 0s at the head of string
1- ^0[0-9]+\.?[0-9]*$ : will find numbers starting with 0 and followed by numbers bigger than zero before the decimal seperator , mainly ".". I put this to distinguish strings containing numbers , for example, "0.111" from "01.111".
2- ([1-9]{1}[0-9]\.?[0-9]) : if there is string starting with 0 then the part which is bigger than 0 will be taken into account. parentheses are used here because I wanted to capture only parts conforming to regex.
3- ([0-9]\.?[0-9]): to capture only the decimal part of the string.
In Javascript , st.match(regex), will return array in which first element contains conformed part. I used this method in the input element's onChange event , by this if the user enters something that violates the regex than violating part is not shown in element's value at all but if there is a part that conforms to regex , then it stays in the element's value.
const floatRegexCheck = (st) => {
const regx1 = new RegExp("^0[0-9]+\\.?[0-9]*$"); // for finding numbers starting with 0
let regx2 = new RegExp("([1-9]{1}[0-9]*\\.?[0-9]*)"); //if regx1 matches then this will remove 0s at the head.
if (!st.match(regx1)) {
regx2 = new RegExp("([0-9]*\\.?[0-9]*)"); //if number does not contain 0 at the head of string then standard decimal formatting takes place
}
st = st.match(regx2);
if (st?.length > 0) {
st = st[0];
}
return st;
}
Here is a more rigorous answer
^[+-]?0(?![0-9]).[0-9]*(?![.])$|^[+-]?[1-9]{1}[0-9]*.[0-9]*$|^[+-]?.[0-9]+$
The following values will match (+- sign are also work)
.11234
0.1143424
11.21
1.
The following values will not match
00.1
1.0.00
12.2350.0.0.0.0.
.
....
How it works
The (?! regex) means NOT operation
let's break down the regex by | operator which is same as logical OR operator
^[+-]?0(?![0-9]).[0-9]*(?![.])$
This regex is to check the value starts from 0
First Check + and - sign with 0 or 1 time ^[+-]
Then check if it has leading zero 0
If it has,then the value next to it must not be zero because we don't want to see 00.123 (?![0-9])
Then check the dot exactly one time and check the fraction part with unlimited times of digits .[0-9]*
Last, if it has a dot follow by fraction part, we discard it.(?![.])$
Now see the second part
^[+-]?[1-9]{1}[0-9]*.[0-9]*$
^[+-]? same as above
If it starts from non zero, match the first digit exactly one time and unlimited time follow by it [1-9]{1}[0-9]* e.g. 12.3 , 1.2, 105.6
Match the dot one time and unlimited digit follow it .[0-9]*$
Now see the third part
^[+-]?.{1}[0-9]+$
This will check the value starts from . e.g. .12, .34565
^[+-]? same as above
Match dot one time and one or more digits follow by it .[0-9]+$
I'm trying to write a RegEx to test if a number is valid and for valid I mean any number that matches country calling codes but also where the format of telephone numbers is standardized by ITU-T in the recommendation E.164. This specifies that the entire number should be 15 digits or shorter, and begin with a country prefix as said here so I did this:
^\+\d{2}|\d{3}([0-9])\d{7}$
But it's not working. In my case (VE numbers can't match the RegEx since this one are validated in another way) this input is valid:
+1420XXXXXXXXXXX // Slovakia - X is a digit and could be more, tough, 5 minimum
001420XXXXXXXXXX // Slovakia - I've changed from + to 00
420XXXXXXXXXXXXX // Slovakia - I've removed the 00 o + but number still being valid
+40XXXXXXXXXXXXX // Romania
Invalid numbers are the one that doesn't match the RegEx and the one started with +58 since they are from VE. So, resuming, a valid number should have:
+XX|+XXX plus 12|11 digits (5 minimum) where XX|XXX is the country code and then since maximum is 15 digits then should be 12 or 11 digits depending on the country format
Can any help me with this? It's a one I called complex
Few strange things going on with your regexp:
\d is shorthand for [0-9] - fine to use both, but I'm wondering why they're mixed
what you are searching with you OR (|) is "something that starts with +XX" i.e. plus and two numbers (^\+\d{2}) OR "something that ends with XXXXXXXXXXX" i.e. 11 numbers (\d{3}([0-9])\d{7}$)
You need to group (with brackets) the OR choices, otherwise it is everything to the left or everything to the right (simplistically)
^\+(\d{2}|\d{3})([0-9])\d{7}$
There is, however, another way of giving the number of occurrences : {m,n} means occurs between m and n times. So you could say ^\+\d{7,15}$ (where 7 is your minimum 5 + the minimum country code of 2).
To really do this, however, you might want to take a look here (https://code.google.com/p/libphonenumber/ 1) where there is a complete validation and formatting for all phone numbers available as javascript.
I'm using plain vanilla JavaScript and need some help with my regex. Money in the following formats has to be allowed, and in these formats only (with no limit on the number of 0s (tens, hundreds, thousands, etc.) for the dollar amounts allowed):
$25,000
$25000
25,000
25000
25000.01
25,000.99
2000.99
50.00
50
1.95
1 .99
0.25
$0.25
0.2
2.3
2000.5
.75
var regex = /^\$?.?[1-9][0-9,]*(.[0-9]{0,2})?$/;
Currently, it's not allowing amounts like 0.99 to be entered.
Try this
^\$?(?:\d+|\d{1,3}(?:,\d{1,3})*)(?:\.\d{2})?$
See it here on Regexr
The only thing that is is not matching is your third last example, it has a space before the dot. Is that valid?
Edit:
My frist solution has the restriction, that it would accept numbers starting with 0, like 001. This solution uses a negative lookahead to avoid this:
^\$?(?!0\d)(?:\d+|\d{1,3}(?:,\d{1,3})*)(?:\.\d{2})?$
See it here on Regexr
Solution without lookahead
^\$?(?:0|[1-9]\d*|[1-9]\d{0,2}(?:,\d{1,3})*)(?:\.\d{2})?$
See it on Regexr
^\$?(?:[1-9]\d?\d?(?:(?:,\d{3})*|(?:\d{3})*)|0)(?:\.\d\d?)?$
Altho I wouldn't use such strict input restrictions.