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Regex match with 3 condition?

I will creating math program .If i wish to solve i need separate to the equation using with regex.
For example:
`10x-10y2+100x-100k=100`
the format will separate with regex output: "10x" ,"100x" ,"100k" ,"10y2","100"
I have already code for that with separate match function
There are:
for num[a-z]num : ` /([\-+])?\s*(\d+)?([a-z]?(\d+))/g`
for num[a-z] : ` /([\-+])?\s*(\d+)?([a-z]?(\d+))/g`
fon num : `/\b[+-]?[\d]+\b/g`
I need all this three match function within one regex match function.some one help to combine the code with single regex expression
Note :i need match function only not a split beacause i apply that regex expression into parser
Thank You.

Split on symbols:
"10x-10y2+100x-100k=100".split(/[-+=]/);
Output:
["10x", "10y2", "100x", "100k", "100"]
If you need to use match() method I suggest same approach:
"10x-10y2+100x-100k=100".match(/[^-+=]+/g);
Output is the same.

/(?:0|[1-9]\d*)?(?:[a-z]+)?(?:\^(?:0|[1-9]\d*))?/g
// finds:
// vvv vvvvv vvvv vvvv vvvvv vv vvv v vvv
10x-10y^2+100x-100k=100^4+xy+100+100y2+0-1^0
// doesn't find: ^^^^^
(?:0|[1-9]\d*)? 0, OR 1-9 then zero or more numbers. Optional.
(?:[a-z]+)? Optional one or more lowercase letters.
(?:\^[1-9]\d*)? Optional power.
\^ Literal text.
(?:0|[1-9]\d*) Zero, OR 1-9 then zero or more numbers.
If I've missed anything let me know and I'll incorporate it.

Just try with following regex:
/(\d+[a-z]?\d*)/g
example

Try this:
s.match(/\d+([A-Za-z]\d*)?/g)

Allowing a dash in this regex

I'm using this Wordpress plugin called 'Easy contact form' which offers standard validation methods.
It uses the following regex for phone numbers:
/^(\+{0,1}\d{1,2})*\s*(\(?\d{3}\)?\s*)*\d{3}(-{0,1}|\s{0,1})\d{2}(-{0,1}|\s{0,1})\d{2}$/
Just now it allows the following formats (maybe more):
0612345678
+31612345678
But I want it to allow +316-12345678 and 06-12345678 also ... Is this possible? If so, how?
Thanks in advance!

You can use a less complex regex :
^\+?\d{2}(-?\d){8,9}$
This regex allows a + at the beginning of the phone number, then matches two digits, and after that, digits preceded (or not) by a -, for a total of 10 or 11 digits.
Now you can adapt it if the initial + is only for 11-digits phone numbers :
^\+?\d{3}(-?\d){9}|\d{2}(-?\d){8}$
My regex allow you to use - every digit. If that's an issue, it can be changed :
^\+?\d{3}(-?\d{2}){4}|\d{2}(-?\d{2}){4}$
I think this last regex will answer your needs, and it's quite simple !

When you figure out what patterns you actually want to allow and not allow. Then fill them out in this function and if the statement returns true you have your regex.
var regex = /^\+?\d{3}(-?\d{2}){4}|\d{2}(-?\d{2}){4}$/; //this is your regex, based on #Theox third answer
//allowed patterns
['0612345678', '+31612345678', '+316-12345678', '06-12345678'].every(function(test) {
return regex.exec(test) !== null;
}) &&
//disallowed patterns, none right now
[].every(function(test) {
return regex.exec(test) === null;
});

allow input with decimal places upto 3

I want to allow only integers and floats (upto 3 decimal places) in a text box, how can I achieve this using javascript?
Valid values are
1234
12.3
12.314
1.11
0.4
Not valid
1.23456
abcd or any other character

Based on the comment that you need to also match ".1" you need to add a conditional with the first part of the regular expression.
var re = /^(\d+)?(?:\.\d{1,3})?$/;
Rough test suite - jSFiddle

You can use a regular expression to do this:
/^\d+(?:\.\d{1,3})?$/
That's the start of the string (^), one or more digits (\d+), optionally followed by a . and between 1 and 3 digits ((?:\.\d{1,3})), then the end of the string ($).
To compare it to the value of an input, you'd do something like this:
var re = /^\d+(?:\.\d{1,3})?$/;
var testValue = document.getElementById('id-of-input').value;
if(re.test(testValue)) {
// matches - input is valid
}
else {
// doesn't match - input is invalid
}
Take a look at this jsFiddle demo.

use regular expression to validate your input field , regular rexpression is as below
^[0-9]+(?:\.[0-9]{1,3})?$

Try this:
var reg=/^[\d]+(?:\.\d{1,3})?$/;
str=10.2305;
str1=123;
alert(reg.test(str));
alert(reg.test(str1));
Check Fiddle http://jsfiddle.net/8mURL/1

Regex to Validate Three Capital Letters then Six Digits?

I'm just working on some JavaScript to accept some user input via jQuery. What I'm trying to validate is 3 capital characters followed by 6 integers.
Anyone know how this can be done using a regular expression in JavaScript?

A simplified version could be /^[A-Z]{3}\d{6}$/.
A more 'compatible' version would be to use /^\p{Lu}{3}\pN{6}$/.

Simple regex could be:
/^[A-Z]{3}\d{6}$/
or, if you want to be unicode compatible:
/^\p{Lu}{3}\pN{6}$/

Easy like this:
var match = yourString.match(/^\p{Lu}{3}\pN{6}$/);
if(match) {
// tada!
} else { alert("not matched"); }

Javascript Regular Expressions - Replace non-numeric characters

This works:
var.replace(/[^0-9]+/g, '');
That simple snippet will replace anything that is not a number with nothing.
But decimals are real too. So, I'm trying to figure out how to include a period.
I'm sure it's really simple, but my tests aren't working.

Simply: var.replace(/[^\d.-]+/g, '');

Replacing something that is not a number is a little trickier than replacing something that is a number.
Those suggesting to simply add the dot, are ignoring the fact that . is also used as a period, so:
This is a test. 0.9, 1, 2, 3 will become .0.9123.
The specific regex in your problem will depend a lot on the purpose. If you only have a single number in your string, you could do this:
var.replace(/.*?(([0-9]*\.)?[0-9]+).*/g, "$1")
This finds the first number, and replaces the entire string with the matched number.

Try this:
var.replace(/[^0-9\\.]+/g, '');

there's a lot of correct answers already, just pointing out that you might need to account for negative signs too.. "\-" add that to any existing answer to allow for negative numbers.

Try this:
var.replace(/[0-9]*\.?[0-9]+/g, '');
That only matches valid decimals (eg "1", "1.0", ".5", but not "1.0.22")

If you don't want to catch IP address along with decimals:
var.replace(/[^0-9]+\\.?[0-9]*/g, '');
Which will only catch numerals with one or zero periods

How about doing this:
var numbers = str.gsub(/[0-9]*\.?[0-9]+/, "#{0} ");

Sweet and short inline replacing of non-numerical characters in the ASP.Net Textbox:
<asp:TextBox ID="txtJobNo" runat="server" class="TextBoxStyle" onkeyup="this.value=this.value.replace(/[^0-9]/g,'')" />
Alter the regex part as you'ld like. Lots and lots of people complain about the cursor going straight to the end when using the arrow keys, but people tend to deal with this without noticing it for instance, arrow... arrow... arrow... okay then... backspace back space, enter the new chars.

Here are a couple of jQuery input class types I use:
$("input.intgr").keyup(function (e) { // Filter non-digits from input value.
if (/\D/g.test($(this).val())) $(this).val($(this).val().replace(/\D/g, ''));
});
$("input.nmbr").keyup(function (e) { // Filter non-numeric from input value.
var tVal=$(this).val();
if (tVal!="" && isNaN(tVal)){
tVal=(tVal.substr(0,1).replace(/[^0-9\.\-]/, '')+tVal.substr(1).replace(/[^0-9\.]/, ''));
var raVal=tVal.split(".")
if(raVal.length>2)
tVal=raVal[0]+"."+raVal.slice(1).join("");
$(this).val(tVal);
}
});
intgr allows only numeric - like other solutions here.
nmbr allows only positive/negative decimal. Negative must be the first character (you can add "+" to the filter if you need it), strips -3.6.23.333 to -3.623333
I'm putting nmbr up because I got tired of trying to find the way to keep only 1 decimal and negative in 1st position

This one just worked for -ve to +ve numbers
<input type="text" oninput="this.value = this.value.replace(/[^0-9\-]+/g, '').replace(/(\..*)\./g, '$1');">

I use this expression to exclude all non-numeric characters + keep negative numbers with minus sign.
variable.replace(/[^0-9.,\-]/g,'')