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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 .

Regex with inputmask for commas

$("#xyz").inputmask('Regex', {regex: "^[0-9]{1,8}(\\.\\d{1,2})?$"});
Above expression allows me to enter 8 digits and 2 decimals but it doesn't automatically handle commas. Can any one help me with commas.

Your regex is a bit wrong, this should work instead:
^[0-9]{1,8}([,.][0-9]{1,2})$
It will accept up to 8 digits and then a comma or dot, followed by up to another 2 digits.
Edit: This regex requires commas, decimals are optional. Maximum is 999,999,999, minimum is 0.00
^([0-9]{0,3}|0)(,[0-9]{3})?(,[0-9]{3})?(\.[0-9]{1,2})?$

Javascript regex positive less than 10000 than can be decimal

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})?$

Regex for non negative and non zero for the format ###.##

I have a requirement to validate some inputs which should be in format ###.##
Invalid inputs are:
-11.10 ==> no negative
000.00 or 0 ==> 0 not allowed should be positive
Valid inputs are:
1
11
111
1.1
11.11
111.11
I have tried with the following regex ^([^-]\d{0,2}(.\d{1,2})?)$ which fulfills my requirements except it's accepting 0 which I don't want. How I can modify my regex so only 0's do not get matched?
Thanks For Help

Try
^(?=.*[1-9])\d{1,3}(?:\.\d\d?)?$
It should do it for you.
It starts with a positive look-ahead to make sure there's a digit other than 0 present.
Then it matches 1-3 digits, optionally followed by a . and 1-2 digits.
Your regex101 updated.

([0-9]){1,3}(\.[0-9]{1,2})? is the expression you are searching for.
([0-9]){1,3} = any number sequence with a length from 1 up to 3
(\.[0-9]{1,2})? = "?" maybe there is a float tail
(\.[0-9]{1,2}) = float tail must start with a dot and decimal numbers have to be up to 2
There is a way to except all zero sequences but it will waste your time for no reason, as you can simply check it with if myNum > 0.
It will work for negative values also, but this regex excludes them too.

^[1-9][0-9]*(\.[0-9]+)?$|^0\.[0-9]+$
This will work for you. It accepts all valid positive decimal numbers excluding 0.

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])$