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jQuery: Unable to use RegExp to search for exactly "+3.00" or "-3.00"

I'm struggling with finding the correct RegExp to match number with plus (+) or minus (-) sign at the end.
I have a select list of number ranging from 0.00- to 40.00- and 0.00+ to 40.00-. I'm using the following RegExp to filter out non matching records:
$("#MySelectBox").change(function() {
var filter = $(this).val()
// If the list item does not contain the numbers phrase fade it out
if ($(this).text().search(new RegExp('\\b'+filter+'\\b', "i")) < 0) {
$(this).hide();
} else {
$(this).show();
}
However, it will show both + and - numbers. For example: if filter = 3.00+ then it will return both 3.00+ and 3.00- values.
Any ideas how to get the exact match?

[\+-]\d{1,2}\.00
Will match + or -, followed by one or two digits (\d{1,2}), followed by .00.
However, RegExes don't have "greater than 40" kind of logic. They only match patterns.
There are useful tools to help you, like Rexegpal

So with your brief:
Check string matches pattern: "+xx.xx" or "-xx.xx"
Only numbers between -40 or +40
Omit results out of bounds
This could be a good way to achieve your desired result.
Notes: (1) Unsure if you wanted to enforce the demical point, (2) there are certainly multiple ways to achieve your result, this is is just one.
const regex = /^[\-\+]\d+\.*\d*$/;
const tests = ["41", "20", "+20", "-20", "+20.00", "-20.20", "20.20"];
const passedTests = tests.filter(number => {
const parsedNumber = Number.parseFloat(number);
const isValid = !!number.match(regex) && parsedNumber > -40 && parsedNumber < 40;
console.log(`Test ${number}, result: ${isValid}`);
return isValid;
});
console.log(passedTests);

To get an exact match for a number with plus (+) or minus (-) sign at the end and from 0.00- to 40.00- and 0.00+ to 40.00+, you can use:
^(?:(?:[0-9]|[123]\d)\.\d\d|40\.00)[+-]$
^ Start of string
(?: Non capture group for the alternation |
(?:[0-9]|[123]\d) Match either a digit 0-9 or a number 10 - 39
\.\d\d Match a . and 2 digits 0-9
| Or
40\.00 Match 40.00
) Close group
[+-] Match either + or -
$ End of string
Regex demo
In Javascript you can use
const regex = /^(?:(?:[0-9]|[123]\d)\.\d\d|40\.00)[+-]$/;
If the value is not the only value in the string, you could start with pattern with a word boundary \b and assert a whitespace boundary at the right (?!\S)
\b(?:(?:[0-9]|[123]\d)\.\d\d|40\.00)[+-](?!\S)
Regex demo

regex: remove leading zeros, but keep single zero

I have an input field to which I have tied a formatting function that is triggered whenever the field loses focus.
What I aim to achieve is that I remove all the leading zeros from an input and I did achieve that with the below line. However, when the user wants to enter a single 0 or something like 0000 I still want that field to end with the value 0 (single). With .replace(/^0+/, '') it would remove every zero and return just an empty string. Someone knows what regex could handle this?
const formatNumber = ($field) => {
var number = $field.val().replace(/\./g, '').replace(/\s/g, '').replace(/^0+/, '');
return number;
};
note: if(number === "") number = "0" is not an option.
edit1:: I noticed there seems to be a bit of confusion. e.g "0009825" need to become 9825 and not 09825. the only instance where i want a 0 up front is when the value is simply zero.

You ay use this regex replacement:
.replace(/^(?:0+(?=[1-9])|0+(?=0$))/mg, '')
RegEx Demo
RegEx Details:
^: Start
(?:: Start capture group
0+(?=[1-9]): Match 1 or more zeroes that must be followed by 1-9
|: OR
0+(?=0$): Match 1 or more zeroes that must be followed by one 0 and end
): End capture group
Replacement is empty string which will leave a single 0 if there are only zeroes in string otherwise will remove leading zeroes.
Alternative solution using a capture group:
str = str.replace(/^0+(0$|[1-9])/mg, '$1');

A simple reg exp with leading zeros and match one digit in a capture group
const cleanZeros = str => str.replace(/^0+(\d)/, '$1')
var tests = ["0009876","0", "0000", "9999", "0090000"]
tests.forEach( s => console.log(s, cleanZeros(s)))

Javascript Regular Expression for numbers

I am trying to make a HTML form that accepts a rating through an input field from the user. The rating is to be a number from 0-10, and I want it to allow up to two decimal places. I am trying to use regular expression, with the following
function isRatingGood()
{
var rating = document.getElementById("rating").value;
var ratingpattern = new RegExp("^[0-9](\.[0-9][0-9]?)?$");
if(ratingpattern.test(rating))
{
alert("Rating Successfully Inputted");
return true;
}
else
{
return rating === "10" || rating === "10.0" || rating === "10.00";
}
}
However, when I enter any 4 or 3 digit number into the field, it still works. It outputs the alert, so I know it is the regular expression that is failing. 5 digit numbers do not work. I used this previous answer as a basis, but it is not working properly for me.
My current understanding is that the beginning of the expression should be a digit, then optionally, a decimal place followed by 1 or 2 digits should be accepted.

You are using a string literal to created the regex. Inside a string literal, \ is the escape character. The string literal
"^[0-9](\.[0-9][0-9]?)?$"
produces the value (and regex):
^[0-9](.[0-9][0-9]?)?$
(you can verify that by entering the string literal in your browser's console)
\. is not valid escape sequence in a string literal, hence the backslash is ignored. Here is similar example:
> "foo\:bar"
"foo:bar"
So you can see above, the . is not escaped in the regex, hence it keeps its special meaning and matches any character. Either escape the backslash in the string literal to create a literal \:
> "^[0-9](\\.[0-9][0-9]?)?$"
"^[0-9](\.[0-9][0-9]?)?$"
or use a regex literal:
/^[0-9](\.[0-9][0-9]?)?$/

The regular expression you're using will parsed to
/^[0-9](.[0-9][0-9]?)?$/
Here . will match any character except newline.
To make it match the . literal, you need to add an extra \ for escaping the \.
var ratingpattern = new RegExp("^[0-9](\\.[0-9][0-9]?)?$");
Or, you can simply use
var ratingPattern = /^[0-9](\.[0-9][0-9]?)?$/;
You can also use \d instead of the class [0-9].
var ratingPattern = /^\d(\.\d{1,2})?$/;
Demo
var ratingpattern = new RegExp("^[0-9](\\.[0-9][0-9]?)?$");
function isRatingGood() {
var rating = document.getElementById("rating").value;
if (ratingpattern.test(rating)) {
alert("Rating Successfully Inputted");
return true;
} else {
return rating === "10" || rating === "10.0" || rating === "10.00";
}
}
<input type="text" id="rating" />
<button onclick="isRatingGood()">Check</button>

Below find a regex candidate for your task:
^[0-1]?\d(\.\d{0,2})?$
Demo with explanation
var list = ['03.003', '05.05', '9.01', '10', '10.05', '100', '1', '2.', '2.12'];
var regex = /^[0-1]?\d(\.\d{0,2})?$/;
for (var index in list) {
var str = list[index];
var match = regex.test(str);
console.log(str + ' : ' + match);
}

This should also do the job. You don't need to escape dots from inside the square brackets:
^((10|\d{1})|\d{1}[.]\d{1,2})$
Also if you want have max rating 10 use
10| ---- accept 10
\d{1})| ---- accept whole numbers from 0-9 replace \d with [1-9]{1} if don't want 0 in this
\d{1}[.]\d{1,2} ---- accept number with two or one numbers after the coma from 0 to 9
LIVE DEMO: https://regex101.com/r/hY5tG4/7
Any character except ^-]\ All characters except the listed special characters are literal characters that add themselves to the character class. [abc] matches a, b or c literal characters

Just answered this myself.
Need to add square brackets to the decimal point, so the regular expression looks like
var ratingpattern = new RegExp("^[0-9]([\.][0-9][0-9]?)?$");

Extend a regular expression to negative number

I want to extend the following regex to negative numbers.
this.value = this.value.replace(/[^0-9\.]/g, "");
I tried adding minus sign doing something like this, (/[^-0-9.]/g, "") but this allows minus sign to be entered anywhere in the number. I want to allow only one occurance of the minus sign at the beginning of the number. Later occurances of minus should be ignored.
Please help.

Uh, replacing every non-number character makes that a bit harder - it's like "negating" the regex. What I came up with is a negative lookahead, preventing a minus - which would be matched by [^0-9.] - from beeing matched if it is in the beginning of the string (using a start anchor):
….replace(/(?!^-)[^0-9.]/g, "")

Why RegEx? How about:
var temp = parseFloat(this.value)
this.value = isNaN(temp) ? "" : temp;
This should also work:
var temp = +this.value;
this.value = isNaN(temp) ? "" : temp;

Put your dash outside of the character class:
this.value = this.value.replace(/^-?[^0-9\.]/g, "");

The regex in provided by you is faulty for the positive numbers itself as it will convert "100+200=300" to "100200300".
I suggest you try this:
someString.match(/-?[\d]+.?[\d]*/);

It is only to put the negative symbol in front of the 0
this.value = this.value.replace(/[^-0-9\.]/g, "");

What's a good RegExp that will strip out all characters except Integers from a string?

I'm new to using regexp, can someone give me the regexp that will strip out everything but an integer from a string in javascript?
I would like to take the string "http://www.foo.com/something/1234/somethingelse" and get it down to 1234 as an integer.
Thanks

var str = "something 123 foo 432";
// Replace all non-digits:
str = str.replace(/\D/g, '');
alert(str); // alerts "123432"
In response to your edited question, extracting a string of digits from a string can be simple, depending on whether you want to target a specific area of the string or if you simply want to extract the first-occurring string of digits. Try this:
var url = "http://www.foo.com/something/1234/somethingelse";
var digitMatch = url.match(/\d+/); // matches one or more digits
alert(digitMatch[0]); // alerts "1234"
// or:
var url = "http://x/y/1234/z/456/v/890";
var digitMatch = url.match(/\d+/g); // matches one or more digits [global search]
digitMatch; // => ['1234', '456', '890']

This is just for integers:
[0-9]+
The + means match 1 or more, and the [0-9] means match any character from the range 0 to 9.

uri = "http://www.foo.com/something/1234/somethingelse";
alert(uri.replace(/.+?\/(\d+)\/.+/, "$1"))

Just define a character-class that requires the values to be numbers.
/[^0-9]/g // matches anything that is NOT 0-9 (only numbers will remain)