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]+$
Related
I am facing issues with regex pattern for Float number -> that should not end or stars with decimal points..
I have tried following regex patter.. that is
regex = /^\d*\.?\d*$/
// on doing
regex.test(11.)
regex.test(.11)
// it is returning true in checking
// I need to make this as false, comment will be much helpful
thank you.
You should bear in mind that regex only works with strings. When you pass a non-string variable as input to a RegExp, it will first coerce it to a string type.
Have a look:
console.log(11. , 'and', .11); // => 11 and 0.11
So, the actual string values you pass to your ^\d*\.?\d*$ regex are 11 and 0.11 that can be matched with the given pattern. Actually, ^\d*\.?\d*$ is a regex that is usually used for a very loose live number input validation, e.g. see How to make proper Input validation with regex?.
What you want is to implement a final, on-submit validation pattern, so that it could not pass strings like 11. and .11. There have been lots of threads discussing this kind of regex:
Regular expression for floating point numbers
regular expression for finding decimal/float numbers?
Basically, for validation, you will need something like
/^\d+(?:\.\d+)?$/.test(input_string)
/^[0-9]+(?:\.[0-9]+)?$/.test(input_string)
/^[0-9]+(?:\.[0-9]{1,2})?$/.test(input_string) // Some need to only allow 1 or 2 fractional digits
/^[0-9]{1,3}(?:\.[0-9]{2})?$/.test(input_string) // 1-3 digits in the integer part and two required in the fractional part
I am not very familiar with regex and am trying to create a regex code in JavaScript to match a string with only
whole numbers
no decimals/dots
and should be greater than 10,000
So far I have it like the ff. I think I am missing something as it still read through decimal numbers and == 10,000. How do I do that?
[1-9](?!\.)\d[0-9]{3,}
https://regex101.com/r/hG2iU7/61
At the risk of not directly answering the question, JavaScript can already parse numbers. Why bother trying to reimplement this? Especially with RegExp?
Why not just parseFloat(theString) or Number(theString) the entire string?
It will fail/return NaN if what you have isn't a number, and you can test for this with isNaN.
If it doesn't fail, you can then test it to ensure that it's an integral value:
const isIntegral = Math.trunc(theNumber) === theNumber;
and is less than 10000
const isLessThan10000 = theNumber < 10000;
This code is going to be much easier to read and maintain than a regular expression.
You may use
^[1-9][0-9]{4,}$
To exclude 10000 add a (?!10000$) lookahead:
^(?!10000$)[1-9][0-9]{4,}$
^^^^^^^^^^
See the regex demo and the regex graph:
Details
^ - start of string
(?!10000$) - a negative lookahead that cancels the match if the whole string is equal to 10000 (i.e. after start of string (^), there is 10000 and then end of string position follows ($))
[1-9] - a digit from 1 to 9
[0-9]{4,} - any four or more digits
$ - end of string.
I know that dozens of similar questions was asked before, but cannot find solution which fits my needs.
I need a regular expression as a pattern for validation purposes, which is a range between 0.025 to 99.999 but should also match integers and any possible decimal forms up to 3 decimals:
1 - matched
1.0 - matched
1.000 - matched
0.025 - matched
0.03 - matched
0.01 - not matched
0.024 - not matched
So far my regex looks like this:
^(?!0*(\.0+)?$)([1-9]?[0-9]\.[0-9]{0,3}|[1-9]?[0-9])?$
It actually matches all between 0.001 and 99.999, because managed how to exclude 0 with decimal forms, but don't know how in the easiest way exclude 0.001 to 0.024.
in javascript you can just convert the string to number and use logic like this
const str = '1.0'
const test = +str;
if(test >= 0.025 && test <= 99.999) {
console.log('valid');
}
IMO this will be simpler and easier to read
EDIT:
as mentioned in comments this will pass for strings like 3e-2 which is valid syntax in javascript to define numbers, so you might have to handle that case based on your use case.
To do this with a regular exression, you need something a bit complex:
^(?:0\.02[5-9]|0\.0[3-9]\d?|0\.[1-9]\d{0,2}|[1-9][0-9]?(?:\.\d{1,3})?)$
Here it is in Regex101
Here is how this works in JS
const regex = /^(?:0\.02[5-9]|0\.0[3-9]\d?|0\.[1-9]\d{0,2}|[1-9][0-9]?(?:\.\d{1,3})?)$/;
const testValues = [ "0", "1", "99.999", "0.025", "0.024", "0.01" ];
testValues.forEach(value => console.log(value, regex.test(value)));
Explanation for the expression: it's a bunch of ORs that cover the value range. Although you can compress it more, I've tried to keep it logical and readable:
0\.02[5-9] - values from 0.025 to 0.029
0\.0[3-9]\d? - values from 0.03 to 0.099. The last digit is optional.
0\.[1-9]\d{0,2} - values from 0.1 to 0.999. The last two digits are optional. The fractional part in each case up to now is mandatory, so 0 is not valid.
[1-9][0-9]?(?:\.\d{1,3}) - values from 1 to 99.999. Again, the decimal point and fractional part are set as optional. If present, you can have 1-3 of them.
the entire regex is wrapped in a non-capturing group and then nested between ^ and $ anchors to make sure that the ENTIRE string matches.
Notable exclusions with this regular expression:
exponential form/scientific notation. For example: 3e1 or 3e-1
numbers that lead with zero. For example: 01, 02.345
numbers that start with a sign. For example: +10
values that start from the decimal part. For example: 0.2, 0.123
values that end in a dot. For example 1., 2.
strings that contain whitespace. For example: " 1" will fail to match. The easiest way to solve this is by ensuring the values are trimmed before being tested.
I've got following formats, that are acceptable
1200000,00
1200000.00
1,200,000.00
1 200 000.00
1 200 000,00
1 200 000,0000
-1 200 000.00
At the moment I was able to verify only ^-?\\d+$, ^-?\\d+[\\,\\.]\\d{2}$, ^-?\\d+[\\,\\.]\\d{2,}$. Two last format are separate, so that I would know is rounding needed or not. All three format use gm flags to check string from start ^ to end $.
Those regular expressions cover only first two elements in list. Other elements, that use commas and spaces for thousand separation are not verified yet and I'm not sure how to achieve that.
Also there is a "beautifier" expression (\\d)(?=(\\d{3})+(?!\\d)), that will take this 1200000,00 and turn it into 1 200 000,00 with such usage '1200000,00'.replace(('(\\d)(?=(\\d{3})+(?!\\d))', 'g'), '$1 ').
So question states, what would be a correct regular expression to validate such format 1 200 000.00 or 1,200,000.00? Since I assume difference with \s\, could be easily done in same expression.
Thank you.
For validating the last two numbers, you can use the following:
^-?\d{1,3}(?:[\s,]\d{3})*(?:\.\d+)?$
1 2 3 4 5
Optional minus sign
1..3 digits
Zero or more fragments that consist of
comma or space
3 digits
optional fraction part consisting of a dot followed by 1 or more digits.
This doesn't directly solve the problem due to me misreading. But it might still be useful to someone so I'll let it stay.
Stop trying to solve every problem with regex. Regex is great when you have one or two very well defined strings. Not a million formats.
This can be solved with minimal regex. Magic is in the bold part.
var numbers = [
"1200000,00",
"1200000.00",
"1,200,000.00",
"1 200 000.00",
"1 200 000,00",
"1 200 000,0000",
"-1 200 000.00"
];
var parseWeirdNumber = function(numberString) {
//Split numbers to parts. , . and space are all valid delimiters.
var numberParts = numberString.split(/[.,\s]/);
//Remove the last part. **This means that all input must have fraction!!**
var fraction = numberParts.pop();
//Rejoin back without delimiters, and reapply the fraction.
//parseFloat to convert to a number
var number = parseFloat(numberParts.join('') + "." + fraction);
return number;
}
numbers = numbers.map(parseWeirdNumber);
console.log(numbers);
A thought struck me as I was writing a piece of JavaScript code that processed some floating point values. What is the decimal point symbol in JavaScript? Is it always .? Or is it culture-specific? And what about .toFixed() and .parseFloat()? If I'm processing a user input, it's likely to include the local culture-specific decimal separator symbol.
Ultimately I'd like to write code that supports both decimal points in user input - culture-specific and ., but I can't write such a code if I don't know what JavaScript expects.
Added: OK, Rubens Farias suggests to look at similar question which has a neat accepted answer:
function whatDecimalSeparator() {
var n = 1.1;
n = n.toLocaleString().substring(1, 2);
return n;
}
That's nice, it lets me get the locale decimal point. A step towards the solution, no doubt.
Now, the remaining part would be to determine what the behavior of .parseFloat() is. Several answers point out that for floating point literals only . is valid. Does .parseFloat() act the same way? Or might it require the local decimal separator in some browser? Are there any different methods for parsing floating point numbers as well? Should I roll out my own just-to-be-sure?
According to the specification, a DecimalLiteral is defined as:
DecimalLiteral ::
DecimalIntegerLiteral . DecimalDigitsopt ExponentPartopt
. DecimalDigits ExponentPartopt
DecimalIntegerLiteral ExponentPartopt
and for satisfying the parseFloat argument:
Let inputString be ToString(string).
Let trimmedString be a substring of inputString consisting of the leftmost character that is not a StrWhiteSpaceChar and all characters to the right of that character.(In other words, remove leading white space.)
If neither trimmedString nor any prefix of trimmedString satisfies the syntax of a StrDecimalLiteral (see 9.3.1), return NaN.
Let numberString be the longest prefix of trimmedString, which might be trimmedString itself, that satisfies the syntax of a StrDecimalLiteral.
Return the Number value for the MV
So numberString becomes the longest prefix of trimmedString that satisfies the syntax of a StrDecimalLiteral, meaning the first parseable literal string number it finds in the input. Only the . can be used to specify a floating-point number. If you're accepting inputs from different locales, use a string replace:
function parseLocalNum(num) {
return +(num.replace(",", "."));
}
The function uses the unary operator instead of parseFloat because it seems to me that you want to be strict about the input. parseFloat("1ABC") would be 1, whereas using the unary operator +"1ABC" returns NaN. This makes it MUCH easier to validate the input. Using parseFloat is just guessing that the input is in the correct format.
use:
theNumber.toLocaleString();
to get a properly formatted string with the right decimal and thousands separators
As far as I'm aware, javascript itself only knows about the . separator for decimals. At least one person whose judgement I trust on JS things concurs:
http://www.merlyn.demon.co.uk/js-maths.htm#DTS