Javascript regular expression if no digit specified before the period (.) - javascript

I have tried following expression which works fine to allow only 2 decimal places:
/^[0-9]+(\.[0-9]{1,2})?$/
But it's not considering the value like ".34" or ".5" as there isn't any digit before the period.
How can I update my expression so that, it should take digit "0" before period if nothing is specified before period.
so .34 => 0.34 with valid 2 decimal places expression.

You can use this regex:
/^\d*\.?\d{1,2}$/
RegEx Demo
If you want to format decimal numbers up to 2 decimal points then use toFixed:
var n = ".34";
var str = parseFloat(n).toFixed(2) // will do rounding as well
//=> 0.34

Use * instead of +
^[0-9]*(\.[0-9]{1,2})?$
+: Matches 1 or more of the preceding token.
*: Matches 0 or more of the preceding token.

Related

How to create a Regular Expression with the following conditions

Numerical input should not exceed 8 digits ( counting numbers on both the side of the decimal). Example: 123456.78 or 0.12 or 1.23
Numerical values should be able to accept negative numbers. Example: -0.8 , or -1.23
Max of 2 decimal number allowed
Numerical value should not accept any alpahbets in the input box
I tried Creating the following regular expression but i am not able to figure it out quite efficiently
^-?[0-9]\d*(\.\d+)?[,8]$
Regular exp
To match the required number of digits and no more, lookahead at the beginning of the pattern for (?:\d\.?) 1 to 8 times, and then match \d*(?:\.\d{1,2})?$ to match a number which, if containing decimals, contains at maximum 2 decimal characters:
^-?(?=(?:\d\.?){1,8}$)\d*(?:\.\d{1,2})?$
https://regex101.com/r/rQMRVX/5
(unless you need to capture the decimal part, it can be a non-capturing group like above)
You can try this
^-?\d{1,8}.?\d{1,2}$
Link : https://regex101.com/r/x7yw5M/2

Regular expression for length limit 10 digit before decimal till 2 decimal place with Number only?

i want a regular expression for length limit 10 digit with 2 decimal place numbers with only numbers allowed.10 digit before decimal
allowed
1
1111111111
111
1111111111.22
.2
1.2
1.22
Not allowed
4.
.
-1
abc
Null
Emptystring
""
1.222
-1.22
111111111111
tried but not working
^[0-9]*(\.[0-9]{0,2})?$
You're almost there - all you have to do is include a check that the string is not empty (which can be accomplished with a positive lookahead for .{1,10} right after the ^), and check that its first digit string has at most 10 characters (just use a {0,10} quantifier for the digits). Also note that [0-9] simplifies to \d.
In order to also exclude trailing dots, repeat the digit after the dot with {1,2} instead of {0,2}:
^(?!$)\d{0,10}(?:\.\d{1,2})?$
https://regex101.com/r/Ah8dNu/5
I have also prepared below RegEx.
Seems this will also work.
^[0-9]{1,10}((\.)[0-9]{0,2}){0,1}$

Regex to validate decimals

I need to validate the integers with the below pattern. Integer part can have up to 5 characters and the fraction value can have up to 2 chars. (fraction value is optional). Leading + or - is also optional.
(+/-) mmmmm (5).nn (2)
Test data
Valid
-1
+1
1
+.1
-.1
.1
+11111.11
-11111.11
11111
Invalid
+111111.11
-111111.11
+11111.111
-11111.111
11111.111
111111.11
+1.
-1.
1.
This is what I currently use
[+-]?\d{0,5}\.?\d{1,2}?
Is this correct? Am I missing something obvious?
Here is the test case.
EDIT
If there is an additional constraint to have the number of digits from the scale is included in the precision.
For example DECIMAL(5, 2) defines numbers of the form 1234.5 whereas DECIMAL(5, 5) defines numbers of the form 1.2345. How should I change this?
Do:
^[+-]?(?:[0-9]{1,5})?(?:\.[0-9]{1,2})?$
^[+-]? matches + or - at start, optional
(?:[0-9]{1,5})? matches one to five digits, optional
(?:\.[0-9]{1,2})?$ matches a literal dot, followed by one or two digits at the end, optional. As the literal . is inside the non-captured group with the digits pattern following, it will only be matched when there are required digits afterwards
Demo
In Javascript, you can validate a number using
Number(n) === Number(n)
For any non-number values of n, like "abc" or [1, 2, 3], Number(n) will return NaN. Given that NaN !== NaN in any case, you can easily tell if n is a number.
Number constructor works fine with almost any form of number representation:
Number(123) // -> 123
Number('123') // -> 123
Number('+12e+3') // -> 12000
Number('-1.2e-3') // -> -0.0012
Number('--123') // -> NaN
However, if you are constrained to using regular expressions, it's very close to what you described:
Optional "+" or "-".
0 to 5 digits.
Optional "." and then 1 to 2 digits in decimal part. The whole part is optional, meaning that a single dot with no digits is invalid
And no exponential form.
That would be
/^[+-]?\d{0,5}(\.\d{1,2})?$/
so that
/^[+-]?\d{0,5}(\.\d{1,2})?$/.test(123) // -> true
/^[+-]?\d{0,5}(\.\d{1,2})?$/.test('123') // -> true
/^[+-]?\d{0,5}(\.\d{1,2})?$/.test('+123.45') // -> true
/^[+-]?\d{0,5}(\.\d{1,2})?$/.test('--123') // -> false
/^[+-]?\d{0,5}(\.\d{1,2})?$/.test('-123.456') // -> false
Please note that the part (\.\d{1,2})? is whole optional. In your case, either dot or two decimal digits are optional, so "123." would be a valid number. In Javascript, it is valid, though, so there shouldn't be any problem with that.
A slight modification in your regex works :
^[+-]?[0-9]{0,5}?(\.[0-9]{1,2})?$
or
^[+-]?\d{0,5}?(\.\d{1,2})?$
Demo1
Demo2

currency regex with decimal allowed with 0

With reference to my previous question currency regex
I want to add an condition, i want to allow decimal only if it starts with 0
example
0.25 should be allowed
1.25 not allowed
current regex is as following
/^(?:\d{1,3}(?:,\d{3})+|\d+)(?:\.\d+)?$/
which matches comma trailing etc.
This will match all the currency matches from before without decimals or decimals with 0.x*
/^((?:\d{1,3}(?:,\d{3})+|\d+)|(?:0\.\d+))$/
If you only want to match 0.xx instead of an arbitrary number of decimal places use
/^((?:\d{1,3}(?:,\d{3})+|\d+)|(?:0\.\d{2}))$/
This one changes \d+, one or more digits, to \d{2}, exactly 2 digits.

RegEx to filter out all but one decimal point [duplicate]

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]+$

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