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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 .

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})?$

Regular Expression For Decimal with precision 5 and scale 2

I am trying to validate a text box on key press and here is my criteria.
I need a regular expression which allows total of 5 digits and a decimal point.
If there's no decimal point then it should allow to enter 5 digits.
And if there's a decimal point then it should allow only 2 digits after the decimal point.
However the total number of digits should not exceed 5 digits excluding decimal point and it can be less than 5 digits.
And below is my regex
/^(?:\d{1,2}(?:\.\d{0,6})?)?$/
I hope I am clear and let me know if any clarifications needed
Please help Thanks in advance.

Here's a regex that just lists out alternatives separated by |:
/^\d{1,5}|\d{1,4}\.\d|\d{1,3}\.\d{2}$/
i.e.
1-5 digits with no decimal point OR 1-4 digits followed by a decimal point and one digit after the decimal OR 1-3 digits followed by a decimal point and two digits after the decimal.

Regular expression to enforce 2 digits after decimal point

I need to validate a numeric string with JavaScript, to ensure the number has exactly two decimal places.
The validation will pass only if
the number has precisely two decimal places
there is at least one digit before the decimal point. (could be zero)
the number before the decimal point can not begin with more than one zero.
Valid numbers:
0.01
0.12
111.23
1234.56
012345.67
123.00
0.00
Invalid numbers:
.12
1.1
0.0
00.00
1234.
1234.567
1234
00123.45
abcd.12
12a4.56
1234.5A
I have tried the regular expression [0-9][\.][0-9][0-9]$, but it allows letters before decimal point like 12a4.56.

. matches any character, it does not do what you think it does. You have to escape it. Also, you have two more errors; try
^[0-9]+\.[0-9][0-9]$
instead, or even better, use \d for decimal digits:
^\d+\.\d\d$

This covers all requirements:
^(0|0?[1-9]\d*)\.\d\d$
the number has precisely two decimal places
Trivially satisfied due to the non-optional \.\d\d$
The other two conditions can be restated as follows:
The number before the decimal points is either a zero
or a number with exactly one zero, then a number that does not start with zero
This is covered in these two cases:
0
0?[1-9]\d*

You don't need regular expressions for this.
JavaScript has a function toFixed() that will do what you need.
var fixedtotwodecimals = floatvalue.toFixed(2);

i used this
^[1-9][1-9]*[.]?[1-9]{0,2}$
0 not accept
123.12 accept but 123.123 not accept
1 accept
12213123 accept
sdfsf not accept
15.12 accept
15#12 not accept
15&12 not accept

var values='0.12';
document.write(values.match(/\d+[.]+\d+\d/));
change value as you want and check it

Here it is:
^(0[.]+\d{2})|^[1-9]\d+[.]+\d{2}$

Try This Code
pattern="[0-9]*(\.?[0-9]{1,2}$)?"
1 Valid
1.1 Valid
1.12 Valid
1.123 not Valid
only number Valid
pattern="[0-9]*(.?[0-9]{2}$)?"
1 Valid
1.1 not Valid
1.12 Valid
1.123 not Valid
only number Valid

Regex which allows 3 digit before precision and 2 digits after precision

I have written a Regex which will take only integers but I need to rewrite this regex to only allow 3 digit before precision and 2 digits after precision
How to do that?

If it must always have 3 digits, a decimal point, and 2 digits, e.g., 412.88, then:
/^\d{3}\.\d{2}$/
If it can be up to 3 digits before and up to 2 after (possibly no decimal point at all) then maybe something like:
/^\d{1,3}(\.\d{1,2})?$/

In c#
#"^\d{3}\.\d{2}$"
//in c# we need to use verbatim string `#""` to treat escape sequences as normal literals instead of giving it a special meaning..
In javascript
/^\d{3}\.\d{2}$/