Need a Reg Expression to find a decimal range - javascript

I am currently needing a reg expression which will evaluate a decimal range.
The requirements are as below
1) Can allow only 1 or 2 decimal places after dot or can as well allow whole numbers (e.g) 1234 , 123.4, 1245.78 are valid
2) The range should be within 9999 (e.g) 9999.0 , 9998.99 , 9999.00 - Valid | 9999.01,10000.00 - not Valid
3)Do not require leading or trailing zeros
So far i have tried to achieve till writing this reg expression
/^[0-9]\d{1,4}(\.\d{1,2})?$/.test(value);
... but unable to proceed with setting range till digit 9999 (since 9999.01 also not valid )can you help.

Why don't just apply regular expression to determine is your string a valid digit with dots float, then typecast it to Number and find wether it is bigger than 9999 or not.
Regexp for your needs caould be very complex and take too much CPU from client.

Here's something quick and dirty that should work for you: http://regex101.com/r/vK1jM3
/^(?(?=9999)9999(?:\.0+)?|\d{1,4}(?:\.\d{1,2})?)$/gm
I merely handle the special case of 9999

This works as far as I can see:
^(9999(?!\.[1-9])(?!\.0[1-9])\.[0-9]{1,2}|9999|(?!9999)[0-9]{1,4}|(?!9999)[0-9]{1,4}\.[0-9]{1,2})$
Testing it out:
var monstrosity = /^(9999(?!\.[1-9])(?!\.0[1-9])\.[0-9]{1,2}|9999|(?!9999)[0-9]{1,4}|(?!9999)[0-9]{1,4}\.[0-9]{1,2})$/;
console.log(monstrosity.test("9999.00")); // true
console.log(monstrosity.test("9999.01")); // false
console.log(monstrosity.test("9999")); // true
console.log(monstrosity.test("9998.4")); // true
console.log(monstrosity.test("0")); // true
console.log(monstrosity.test("0.5")); // true
If you add something like this to the codebase, future maintenance programmers will hunt you down with pitchforks. Try to solve the range check without regex, as webbandit suggested.

Why a regexp? Just do
x > 0 && x <= 9999 && (x*100 - Math.floor(x*100) == 0)

Related

regex check for Float number -> not ending and stars with decimal point

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

Match number with no decimal and greater than 10,000

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.

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]+$

JavaScript Number preserve leading 0

I have a problem, I build a very simple javascript search for postal codes.
I am using JS Numbers because I want to check if the passed number (search term) is less||equal or more||equal to the max and min.
value >= splitZips[0] && value <= splitZips[1]
But the Javascript Number var type deletes leading 0, which is a problem because I have postal codes like 01075 and also postal codes like 8430. So it can not find the small 4 digit codes.
Any idea how to fix this?
Represent them as a String. Outside of strict mode, a leading zero denotes an octal number otherwise.
Also, why would a leading zero have any significance when calculating numbers? Just use parseInt(num, 10) if you need to.
Store and display the postcodes as strings, thus retaining the leading zeros. If you need to make a numerical comparison convert to number at the time. The easiest way to convert is with the unary plus operator:
var strPC = "01745",
numPC = +strPC;
alert(numPC === +"01745"); // true
+value >= +splitZips[0] && +value <= +splitZips[1];
// etc.
Before you start comparing you might want to ensure the entered value actually is numeric - an easy way to be sure it is a four or five digit code with or without leading zeros is with a regex:
/^\d{4,5}$/.test(searchTerm) // returns true or false
Instead a parseInt you could use type casting :)
"0123">"122" // false
+"0123">"122" // true | that means: 123>"122"
Btw, what more you can use a each of bitwise operators :
~~"0123"
"0123"|0
"0123"&"0123"
"0123">>0
"0123"<<0
With the same effect :)

Regex to match from 0.0 to 150.0 (including floating point)

I have a problem: i should match values from 0.0 to a specific double value (for example, i should match from 0.0 to 150.00 including value as 12, 21.23213, 149.111)
anyone can help me?
i tried everything.
i used this regexp for match by 0.0 to 60.0 but it doesn't work
(^0(\.[0-9]+)?$|^[1-9]{1}(\.[0-9]+)?$|^[1-5]{1}[0-9]{1}(\.[0-9]+)?$|^60$)
with 123 it doesn't work
thank you in advance
Marco
Don't use a regex - use Number, check it's a number with isNaN, then compare values using <= and >=.
e.g.
var your_val = "3.05";
var your_val_num = Number(your_val);
if (!isNaN(your_val_num) && your_val_num >= 0 && your_val_num <= 150) {
// do something
}
N.B. I've changed my answer to use Number rather than parseFloat, per AndyE's comment, and to check for NaN before doing numerical comparisons, per lincolnk's comment.
I agree with the other answers: regex is a poor way to do numeric comparisons.
If you really have to, either:
because a dumb framework you're stuck with only allows regex checks, or
you need extra decimal precision that a JavaScript Number can't provide (as JavaScript has no built-in Decimal type)... this won't be the case for comparing against the whole numbers 0 and 150 though
then:
^0*( // leading zeroes
150(\.0+)?| // either exactly 150
1[0-4]\d(\.\d+)?| // or 100-149.9*
\d{0,2}(\.\d+)? // or 0-99.9*
)$
(newlines/comments added for readability, remove to use.)
This doesn't support E-notation (150=1.5E2) but should otherwise allow what normal JS Number parsing would.
forget regex - just check if(parseFloat(x)=<150 && parseFloat(x)>=0)

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