I don't get how hard it is to discern a string containing a number from other strings in JavaScript.
Number('') evaluates to 0, while '' is definitely not a number for humans.
parseFloat enforces numbers, but allow them to be tailed by abitrary text.
isNaN evaluates to false for whitespace strings.
So what is the programatically function for checking if a string is a number according to a simple and sane definition what a number is?
By using below function we can test whether a javascript string contains a number or not. In above function inplace of t, we need to pass our javascript string as a parameter, then the function will return either true or false
function hasNumbers(t)
{
var regex = /\d/g;
return regex.test(t);
}
If you want something a little more complex regarding format, you could use regex, something like this:
var pattern = /^(0|[1-9][0-9]{0,2}(?:(,[0-9]{3})*|[0-9]*))(\.[0-9]+){0,1}$/;
Demo
I created this regex while answering a different question awhile back (see here). This will check that it is a number with atleast one character, cannot start with 0 unless it is 0 (or 0.[othernumbers]). Cannot have decimal unless there are digits after the decimal, may or may not have commas.. but if it does it makes sure they are 3 digits apart, etc. Could also add a -? at the beginning if you want to allow negative numbers... something like:
/^(-)?(0|[1-9][0-9]{0,2}(?:(,[0-9]{3})*|[0-9]*))(\.[0-9]+){0,1}$/;
There's this simple solution :
var ok = parseFloat(s)==s;
If you need to consider "2 " as not a number, then you might use this one :
var ok = !!(+s==s && s.length && s.trim()==s);
You can always do:
function isNumber(n)
{
if (n.trim().length === 0)
return false;
return !isNaN(n);
}
Let's try
""+(+n)===n
which enforces a very rigid canonical way of the number.
However, such number strings can be created by var n=''+some_number by JS reliable.
So this solution would reject '.01', and reject all simple numbers that JS would stringify with exponent, also reject all exponential representations that JS would display with mantissa only. But as long we stay in integer and low float number ranges, it should work with otherwise supplied numbers to.
No need to panic just use this snippet if name String Contains only numbers or text.
try below.
var pattern = /^([^0-9]*)$/;
if(!YourNiceVariable.value.match(pattern)) {//it happen while Name Contains only Charectors.}
if(YourNiceVariable.value.match(pattern)) {//it happen while Name Contains only Numbers.}
This might be insane depending on the length of your string, but you could split it into an array of individual characters and then test each character with isNaN to determine if it's a number or not.
A very short, wrong but correctable answer was just deleted. I just could comment it, besides it was very cool! So here the corrected term again:
n!=='' && +n==n'
seems good. The first term eliminates the empty string case, the second one enforces the string interpretataion of a number created by numeric interpretation of the string to match the string. As the string is not empty, any tolerated character like whitespaces are removed, so we check if they were present.
Related
I want a regular expression for the following values:
1) .1 = False.
2) 0.1 = True.
3) 1.1245 = True.
4) 1.2. = False.
5) 1.24.35 = False.
6) 21.152 = true.
This regex should check that the string represents an integer or is composed of only numbers and a single . representing a decimal point.
I am using: /^[\+\-]?\d*\.?\d+(?:[Ee][\+\-]?\d+)?$/ but it takes "1.2." as true.
You are taking the first digits set as none or more \d*, you should use \d+, then it would work for the cases you want.
Can also look here
Edit: ^[\+\-]?\d+(\.?\d+(?:[Ee][\+\-]?\d+)?)?$
If you use the following regex
(?:\d*.)?\d+
It will match only real numbers.
If the input is like 1.2.3 it will return 1.2.
Hope it helps!
you were puting the dot in the wrong place this should work
[\+\-]?[0-9]+(\.[0-9]+(?:[Ee][\+\-]?[0-9]+)?)?
Your regex seems more complex than it needs to be. In order to live up to your requirements (starts with a number, ends with a number, at most one dot) you can use the following:
(^\d+\.\d+$)|(^\d+$)
Edit:
Or even simpler:
^\d+(\.\d+)?$
So I have a javascript program that solves for 1 variable. I'm coming to a roadblock when selecting numbers that DON'T have a variable associated with them.
Here is my current regex expression:
(\+|-)?([0-9]+)(\.[0-9]+)?(?![a-z])
takes input like 15000.53=1254b+21
and returns [15000.53, 125, +21], when it should return [15000.53, +21] (yes, the + is supposed to be there)
I know why it is happening. The number of digits is optional so the function can handle large numbers and floats, but they are optional, so it is hard to make sure the entire number is selected. The result of this is selecting all the digits of the number EXCEPT the one directly next to the variable.
Anyone know of a way for the number of digits to stay optional, yet still make sure a variable doesn't follow the number? Thanks!
var reg = (\+|-)?([0-9]+)(\.[0-9]+)?(?![a-z]);
var numbers = [];
var equation = '15000.53=1254b+21';
while (aloneInt = reg.exec(side[0])) {
numbers.push(aloneInt[0]);
}
Try the following expression:
(?![+-]?[0-9.]+[a-z])(\+|-)?([0-9]+)(\.[0-9]+)?
The added negative lookahead (?![+-]?[0-9.]+[a-z]) makes sure there isn't one or more optionally signed floating point numbers that are followed by a letter from the alphabet.
In other words, it makes sure there isn't a number followed by a variable name, then it matches the number.
Regex101 Demo
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 have strings which contains thousand separators, however no string-to-number function wants to consume it correctly (using JavaScript). I'm thinking about "preparing" the string by stripping all thousand separators, leaving anything else untoched and letting Number/parseInt/parseFloat functions (I'm satisfied with their behavious otherwise) to decide the rest. But it seems what i have no idea which RegExp can do that!
Better ideas are welcome too!
UPDATE:
Sorry, answers enlightened me how badly formulated question it is. What i'm triyng to achieve is: 1) to strip thousand separators only if any, but 2) to not disturb original string much so i will get NaNs in the cases of invalid numerals.
MORE UPDATE:
JavaScript is limited to English locale for parsing, so lets assume thousand separator is ',' for simplicity (naturally, it never matches decimal separator in any locale, so changing to any other locale should not pose a problem)
Now, on parsing functions:
parseFloat('1023.95BARGAIN BYTES!') // parseXXX functions just "gives up" on invalid chars and returns 1023.95
Number('1023.95BARGAIN BYTES!') // while Number constructor behaves "strictly" and will return NaN
Sometimes I use rhw loose one, sometimes strict. I want to figure out the best approach for preparing string for both functions.
On validity of numerals:
'1,023.99' is perfectly well-formed English number, and stripping all commas will lead to correct result.
'1,0,2,3.99' is broken, however generic comma stripping will give '1023.99' which is unlikely to be a correct result.
welp, I'll venture to throw my suggestion into the pot:
Note: Revised
stringWithNumbers = stringwithNumbers.replace(/(\d+),(?=\d{3}(\D|$))/g, "$1");
should turn
1,234,567.12
1,023.99
1,0,2,3.99
the dang thing costs $1,205!!
95,5,0,432
12345,0000
1,2345
into:
1234567.12
1023.99
1,0,2,3.99
the dang thing costs $1205!!
95,5,0432
12345,0000
1,2345
I hope that's useful!
EDIT:
There is an additional alteration that may be necessary, but is not without side effects:
(\b\d{1,3}),(?=\d{3}(\D|$))
This changes the "one or more" quantifier (+) for the first set of digits into a "one to three" quantifier ({1,3}) and adds a "word-boundary" assertion before it. It will prevent replacements like 1234,123 ==> 1234123. However, it will also prevent a replacement that might be desired (if it is preceded by a letter or underscore), such as A123,789 or _1,555 (which will remain unchanged).
A simple num.replace(/,/g, '') should be sufficient I think.
Depends on what your thousand separator is
myString = myString.replace(/[ ,]/g, "");
would remove spaces and commas.
This should work for you
var decimalCharacter = ".",
regex = new RegExp("[\\d" + decimalCharacter + "]+", "g"),
num = "10,0000,000,000.999";
+num.match(regex).join("");
To confirm that a numeral-string is well-formed, use:
/^(\d*|\d{1,3}(,\d{3})+)($|[^\d])/.test(numeral_string)
which will return true if the numeral-string is either (1) just a sequence of zero or more digits, or (2) a sequence of digits with a comma before each set of three digits, or (3) either of the above followed by a non-digit character and who knows what else. (Case #3 is for floats, as well as your "BARGAIN BYTES!" examples.)
Once you've confirmed that, use:
numeral_string.replace(/,/g, '')
which will return a copy of the numeral-string with all commas excised.
You can use s.replaceAll("(\\W)(?=\\d{3})","");
This regex gets all alpha-numeric character with 3 characters after it.
Strings like 4.444.444.444,00 € will be 4444444444,00 €
I have used the following in a commercial setting, and it has worked often:
numberStr = numberStr.replace(/[. ,](\d\d\d\D|\d\d\d$)/g,'$1');
In the above example, thousands can be marked with a decimal, a comma, or a space.
In some cases ( like a price of 1000,5 Euros) the above doesn't work. If you need something more robust, this should work 100% of the time:
//convert a comma or space used as the cent placeholder to a decimal
$priceStr = $priceStr.replace(/[, ](\d\d$)/,'.$1');
$priceStr = $priceStr.replace(/[, ](\d$)/,'.$1');
//capture cents
var $hasCentsRegex = /[.]\d\d?$/;
if($hasCentsRegex.test($priceStr)) {
var $matchArray = $priceStr.match(/(.*)([.]\d\d?$)/);
var $priceBeforeCents = $matchArray[1];
var $cents = $matchArray[2];
} else{
var $priceBeforeCents = $priceStr;
var $cents = "";
}
//remove decimals, commas and whitespace from the pre-cent portion
$priceBeforeCents = $priceBeforeCents.replace(/[.\s,]/g,'');
//re-create the price by adding back the cents
$priceStr = $priceBeforeCents + $cents;
Firstly apologies for being a tad dim. I need to create a test to check the if the value of an input field.
I currently use /[^A-Za-z0-9 ]/.test(document.form.Serial.value) to test to see if the value of Serial is alphanumeric only.
Now, if an additional field is set, Serial must either being with 'i' or 'I', then the remaining characters must all be numbers. I had considered doing this with substrings, but it seems a bit long and unnecessary.
Any advice people can give would be very much appreciated!
If you want to test if a string begins with i or I, and then only contain numbers, you could use a regular expression such as this one :
/^[iI][0-9]+$/
Or, for a case-insensitive match :
/^i[0-9]+$/i
Basically, this will match :
Beginning of string : ^
an i
any character between 0 and 9 : [0-9]
one or more time : [0-9]+
end of string : $
You may try the code below
var test_value = false
if (document.form.Additional_Field.value) {
test_value = /^(i|I)[0-9]+/.test(document.form.Serial.value) }
else {
test_value = /[A-Za-z0-9]+/.test(document.form.Serial.value) }
it will result in test_value set to true if Serial is either alphanumeric or if Additional_Field has value true and Serial begins with i or I fallowed by any number of numbers, and test_value set to false otherwise.
Why not break the problem down. You have two valid inputs, so a pattern for syntactically checking the input would be:
/^([iI][0-9]+)|([A-Za-z0-9]+)$/
Then you have a separate, and simpler, problem of determining whether the validated input is appropriate based on the state of the other controls on the form.