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Why this regex fails when there is an extra zero?

I have created this regex to match dollar amounts more than $9,000.00.
\$(?=.{6,11}$)\d{1,3}(?:,\d{3})*
But it fails in cases like this,
$25,000.00. Text Goes here
$1,000,000.00
However it works in cases like this,
$25,000.0. T
$25,000.00
$999,000.00
How to fix this regex?

Some issues in your regex:
The look ahead assertion requires the that match can only start in the final 11 characters of the input string, since it has the $ anchor after at least 6 and at most 11 characters. So it is no surprise that "$25,000.00. Text Goes here" does not match. I suppose you don't want that $ anchor, and then the 11 is not useful anymore either.
The look ahead assertion requires that at least 6 characters follow after the currency symbol, however that could include non-digit characters, and so your regex will match the amount in `$300 oh" (6 characters follow after currency symbol).
There is no provision in your regex for decimals even though you say it works for examples that have decimals. But it will not include those decimals in the match. For instance, for input "$300,000.50" it will only match "$300,000" and not the 50 cents. You would need to accept an optional decimal point followed by one or two digits and then require there are no more decimal digits with a negative look-ahead.
The look-ahead assertion is not the right place to impose a maximum amount, because when you remove the $ (see first point) you must still require that there are no more digits after the 11th position. Instead, just remove the look-ahead assertion and match the patterns you want in more detail. There are just two options: either you have 2 or 3 digits followed by one digit group (for amounts between 10,000 and 999,999.99) or you have 1 to 3 digits followed by two digit groups (for amounts between 1,000,000 and 999,999,999.99). To avoid that more digits follow when no decimal part exists, use a negative look-ahead assertion: (?![,.]?\d).
All this is taken into account in this correction:
\$(?:\d{2,3}(?:,\d{3})|\d{1,3}(?:,\d{3}){2})((?![,.]?\d)|\.\d\d?(?!\d))
On regex101
To allow the same numbers without commas, add \d{5,9} as an option:
\$(?:\d{2,3}(?:,\d{3})|\d{1,3}(?:,\d{3}){2}|\d{5,9})(?:(?![,.]?\d)|\.\d\d?(?!\d))
On regex101

Totally new answer. After closer inspection I see that they have revised
the specifications on this question.
I am submitting this solution based on a $10,000.00 - $999,999,999.00 range
of unacceptable cash amounts. The comma's and decimal are optional.
There cannot be more than 3 consecutive decimal numbers after the period.
Ah, other specifications are dubious.
Note that a text representation of leading zero's is not allowed, which is a
distinction worth investigating as digits \d class is covers characters 0-9.
It is hard, if not impossible to match to infinity.
For example, the OP requested to match cash greater than $9,000 (Ah $10,000).
Regex has no representation of quantifiers representing infinity therefore
#Trincot tried to talk him into a max cash amount number to cap it.
In reality, you can only match the infinite with a negative of the finite.
So it is in the cosmos as it is in regex.
The only real way to match a number greater than another number is to
state that it is not in a finite range. In this case not in the range $0 - $9,999.
In this case they have established a range that the cash cannot be in.
That apparently is this $10,000.00 - $999,999,999.00 range, which
absolutely does not represent all values greater than $10,000.00
My original answer was to match $0 - $9,000 (original minimum) then post that regex in a negative assertion, thereby matching the infinite set of values
greater than $9,000 which was and is the only answer to matching cash values greater than
a fixed amount.
In the end, parsing values is only a preamble to getting it into a float
and there is no way to glean the final value ahead of that conversion.
Therefore, this is really an exercise in futility.
To that end :
$10,000.00 - $999,999,999.00
\$[1-9](?:\d{1,2}(?:,?\d{3}){1,2}|(?:,?\d{3}){2})(?:\.\d{0,2})?(?![,.]?\d)
https://regex101.com/r/1h4XW9/1
\$ [1-9]
(?:
\d{1,2}
(?: ,? \d{3} ){1,2}
| (?: ,? \d{3} ){2}
)
(?: \. \d{0,2} )?
(?! [,.]? \d )

JavaScript Regex not matching mobile number with international code

Am trying to validate a mobile number 254777123456 against a regex /^((254|255)[0-9]+){9,15}$/, the mobile number should be prefixed with the country codes specified but the total length of the mobile number should not be more than 15 characters, doing this via javascript am getting null, can anyone point out what am doing wrong.
PS. Am using way more country codes than the ones I specified, I just put those two as a test before I add the others because they will all be separated by the pipe.

Your regex ^((254|255)[0-9]+){9,15}$ means, that pick at least 4 digits (of which first 3 should be either 254 or 255) and whole of them must occur at least 9 times to max 15 times, which will mean the minimum length of string that will match should be of 36 characters. Which obviously you don't want. Your regex needs little correction where you need to take [0-9] part out and have {9,12} quantifier separately. Correct regex to be used should be this,
^(?:(?:254|255)[0-9]{9,12})$
This regex will match 254 or 255 separately and will restrict remaining number to match from 9 to 12 (as you want max number to be matched of length 15 where 3 numbers we have already separated out)
Demo
var nums = ['254777123456','255777123456','255777123456123','2557771234561231']
for (n of nums) {
console.log(n + " --> " + /^(?:(?:254|255)[0-9]{9,12})$/g.test(n));
}

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])$

Regex for number with decimals and thousand separator

I need regex to validate a number that could contain thousand separators or decimals using javascript.
Max value being 9,999,999.99
Min value 0.01
Other valid values:
11,111
11.1
1,111.11
INVALID values:
1111
1111,11
,111
111,
I've searched all over with no joy.

/^\d{1,3}(,\d{3})*(\.\d+)?$/
About the minimum and maximum values... Well, I wouldn't do it with a regex, but you can add lookaheads at the beginning:
/^(?!0+\.00)(?=.{1,9}(\.|$))\d{1,3}(,\d{3})*(\.\d+)?$/
Note: this allows 0,999.00, so you may want to change it to:
/^(?!0+\.00)(?=.{1,9}(\.|$))(?!0(?!\.))\d{1,3}(,\d{3})*(\.\d+)?$/
which would not allow a leading 0.
Edit:
Tests: http://jsfiddle.net/pKsYq/2/

((\d){1,3})+([,][\d]{3})*([.](\d)*)?
It worked on a few, but I'm still learning regex as well.
The logic should be 1-3 digits 0-1 times, 1 comma followed by 3 digits any number of times, and a single . followed by any number of digits 0-1 times

First, I want to point out that if you own the form the data is coming from, the best way to restrict the input is to use the proper form elements (aka, number field)
<input type="number" name="size" min="0.01" max="9,999,999.99" step="0.01">
Whether "," can be entered will be based on the browser, but the browser will always give you the value as an actual number. (Remember that all form data must be validated/sanitized server side as well. Never trust the client)
Second, I'd like to expand on the other answers to a more robust (platform independent)/modifiable regex.
You should surround the regex with ^ and $ to make sure you are matching against the whole number, not just a subset of it. ex ^<my_regex>$
The right side of the decimal is optional, so we can put it in an optional group (<regex>)?
Matching a literal period and than any chain of numbers is simply \.\d+
If you want to insist the last number after the decimal isn't a 0, you can use [1-9] for "a non-zero number" so \.\d+[1-9]
For the left side of the decimal, the leading number will be non-zero, or the number is zero. So ([1-9]<rest-of-number-regex>|0)
The first group of numbers will be 1-3 digits so [1-9]\d{0,2}
After that, we have to add digits in 3s so (,\d{3})*
Remember ? means optional, so to make the , optional is just (,?\d{3})*
Putting it all together
^([1-9]\d{0,2}(,?\d{3})*|0)(\.\d+[1-9])?$

Tezra's formula fails for '1.' or '1.0'. For my purposes, I allow leading and trailing zeros, as well as a leading + or - sign, like so:
^[-+]?((\d{1,3}(,\d{3})*)|(\d*))(\.|\.\d*)?$

In a recent project we needed to alter this version in order to meet international requirements.
This is what we used: ^-?(\d{1,3}(?<tt>\.|\,| ))((\d{3}\k<tt>)*(\d{3}(?!\k<tt>)[\.|\,]))?\d*$
Creating a named group (?<tt>\.|\,| ) allowed us to use the negative look ahead (?!\k<tt>)[\.|\,]) later to ensure the thousands separator and the decimal point are in fact different.

I have used below regrex for following retrictions -
^(?!0|\.00)[0-9]+(,\d{3})*(.[0-9]{0,2})$
Not allow 0 and .00.
','(thousand seperator) after 3 digits.
'.' (decimal upto 2 decimal places).