Javascript regex positive less than 10000 than can be decimal - javascript

I need a regex for a number than can be decimal and less than 10000 with max 2 digits after decimal.
I tried
/^([0-9]{1,4})+(\.[0-9]{0,2})$/
but it returns true for 44555.54 for example.

In your regex you are using + which is using for one or more repetition and which leads to match any length of digit and make decimal part non-greedy(using?) to make it optional.
^[0-9]{1,4}(\.[0-9]{0,2})?$
or using \d for digit character class.
^\d{1,4}(\.\d{0,2})?$

Related

angularjs ng-pattern regex for 4 digits and 2 decimal

I am trying to create a Regex for a number with maximum 4 digits and if the input has decimal it has to have 2 digits - .20 and not .1.
tried:
ng-pattern="/^[0-9]+(\.[0-9]{1,2})?$/" //fail for 666666, .10, .1
Examples for valid inputs:
100.10
100
3000.10
Example for invalid:
10000 //has more then 4 digits before decimal
100.1 //has only 1 digit after decimal
.10 //has no digits before decimal
Thanks for any help.
Use {#,#} to limit the number of digits to 1 to 4
Try
^[0-9]{1,4}(\.[0-9][0-9])?$
Use ( )? to make an optional two-digit decimal part
The problem with using the {1,2} is that it allows one or two digits, when you really only want two. And I assume you want to enforce a rule that if they have a ".", they must have two digits?
For example
var patt = /^[0-9]{1,4}(\.[0-9][0-9])?$/i
"1011.11".match(patt)!==null
"1011.1".match(patt)!==null
Returns
true
false
With gratitude to Sebastian Proske and Wiktor Stribiżew
For pointing out the need to escape the .

Regular Expression: Help Matching a number less than 24

so I'm making this regular expression to verify some text boxes on a website that I'm designing for an internship.
The problem is that I'm not so keen on regular expressions, and I'm close to having a working one that matches a number between 0-24 and no more than two decimal places.
This is what I have so far. The pattern is also matching any string; such as, "a" or "az".
var pattern = "^([0-9]{0,2}?.?[0-9]{0,2}|1[0-9].?[0-9]{0,2}|2[0-4].?[0-9]{0,2})$";
To get a number between 0 and 24 (24 excluded) with optional up to two decimal places:
^(\d|1\d|2[0-3])(\.\d{1,2})?$
The decimal part:
\. - match the decimal dot
\d{1,2} - one or two digits
()? - makes it optional
The whole part:
\d - numbers 0-9
1\d - numbers 10-19
2[0-3] - numbers 20-23
(x|y|z) - one of x, y or z
As for the "why is my version matching things like "a" and "az" part" - it's a little complex, but it basically boils down to you using dots (like .?). In regex, a dot means "any one character". To make it match a literal dot, you need to escape it with a slash just like I did.
Minor remark: If you want optional leading zero for single digit numbers, replace 1\d with [01]\d. If you want mandatory leading zero for single digit numbers, replace \d|1\d with [01]\d. If you don't want leading zeroes, leave it as it is.
Assuming you do not want 05 or 5.50
^((?:[0-9]|1[0-9]|2[0-3])(?:\.(?:[1-9]|[0-9][1-9]))?)$
You can try it here
The following is a quick attempt to match a floating point number from 0 to 24.99 with up to two non-zero digits
^(([0-9])|([01][0-9])|(2[0-4]))(\.[0-9]{1,2})?$
I think it might be easier to use math to do this though...
You can see the explanation of the entire regex as well as test it out here. I have also added a few test cases.
^(\d|[01]\d|2[0-3])(\.\d{1,2})?$
Test cases:
Valid:
22
1.29
2.99
9.99
13.24
17.38
20.01
02.15
15.35
23.56
1.1
Invalid:
24.29
235.215
21.256
To get a integer number between 1 and 23: ^([1-9]|1[0-9]|2[0-3])$

Regex to check amount is greator then one

I have a regex
^(?=.*[1-9])\\d{0,5}(?:\\.\\d{0,2})?$
to check amount should be greater then 0 and it can contain maximum five digit before decimal and optional decimal and 2 digits after decimal.
Here (?=.*[1-9]) is to check there should be latest one occurrence of any digit from (1-9) in the string.
But I want modify it to check amount greater then one(1), so I want (?=.*[1-9]) to check only till the occurrence of the decimal point i.e. 0.1 it should return false.
Other condition should also fulfill.
Note : the count of digits before decimal is not fix, it will very from 1 to 5, so we can't modify it to (?=.{0,5}[1-9])
You can fix it by using this negation based regex:
^(?=[^.]*[1-9])\\d{0,5}(?:\\.\\d{0,2})?$
RegEx Demo
[^.]* before [1-9] will match any character except decimal point thus not allowing 0.45 as valid number.

What regex for jQuery satisfies these conditions?

I have a field for which can have:
Up to potentially 2 decimals places
Up to potentially 6 non-decimal places
No more than 8 digits altogether (the max 2 decimals plus max 6 non-decimals)
Be a positive number
So correct inputs would be range from 0.01 to 999999.99, and trailing zeroes isn't an issue, so 4.00 is just as fine as 4.
Try this pattern:
^[0-9]{1,6}(?:\.[0-9]{1,2}0*)?$
If you want to allow leading and trailing whitespace, add \s* to the beginning and the end of the pattern, right after ^ and right before $.
That said, this task is something you might want to accomplish without regex. Why don't you just read the value of the input, parse it and then simply perform a numeric validation? You could even round the input to two decimal places.
You could try something like so:
Up to potentially 2 decimals places: (\.\d{1,2})? - This will match a decimal point followed by a minimum of 1, and a maximum of 2 digits. This is optional.
Up to potentially 6 non-decimal places: \d{1,6} - This will match a minimum of 1 digit and a maximum of 6 digits.
No more than 8 digits altogether (the max 2 decimals plus max 6 non-decimals): You can combing the two above to get this: \d{1,6}(\.\d{1,2})?.
Be a positive number: Change the above to this: ^\d{1,6}(\.\d{1,2})?$. This should make sure that any number you pass to it, does not start with a negative sign. The ^ and $ anchors instruct the regex engine to start matching at the beginning of the string and complete the matching at the end. This should allow you to be sure that the string you are matching is indeed a number.
That being said, you should really be doing numerical range checks using the appropriate mathematical operations which your language (in this case JavaScript) provides. A small change in the numerical range you are after will most likely bring a large change in your regular expression.
This might validate it
# /^(?=.*[1-9].*$)(?=[.]?(?:\d[.]?){1,8}$)(?=\d{0,6}(?:[.]\d{0,2})?$)/
^
(?= # must be a positive number
.* [1-9] .* $
)
(?=
[.]?
(?: # 1 to 8 digits
\d
[.]?
){1,8}
$
)
(?=
\d{0,6} # 0 to 6 non-decimal places
(?:
[.] # 0 to 2 decimal places
\d{0,2}
)?
$
)

Javascript regex for allowing only positive digits

I use this to test for digits
/^\d+$/
But I need to make sure that it's greater than zero, while still allowing 0000123123123 for instance.
You can write:
/^\d*[1-9]\d*$/
(zero or more digits, followed by a nonzero digit, followed by zero or more digits).
It is correct regex for positive digits.
/^[1-9]\d*$/g
The previous answer not correct for 0123.

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