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regular expression to validate a pattern

I am new to regular expression, In my project i am allowing user to put amount in shorthand as well as full digit, i have used material UI TextField for input.
Examples are:
400k - shorthand,
400.2k - shorthand,
4m - shorthand,
500. - should work
500000 - full amount
some pattern user should not be allowed to enter example are:
4.2.k,
.3k,
4...k
300.k
I have written regex which is below but it does allows to enter dot after number.
textValue.match(/^[0-9]*(\.[0-9]{0,2})*([0-9km]{1})$/) && textValue.match(/^[\d]+/)
above code first regex validates the pattern and second regex forces user to put Number because amount cannot start with string, i have wrote two separate regex as i don't understand how to put them in one regex and those regex doesn't accepts dot after number. Please can anyone give a perfect Regex to validate the above pattern in one single regular expression??
Thanks in advance

With alternation (never really the prettiest) it could be done like:
^\d+([km]|\.|\.\d+[km])?$
See the Online Demo
^ - Start string ancor.
d+ - One or more digits.
( - Opening capturing group (you could use non-capturing).
[km] - A single character "k" or "m".
| - Alternation (OR).
\.? - A literal dot.
| - Alternation (OR).
\.\d+[km] - A literal dot followed by at least one digit and a character "k" or "m".
)? - Close capturing group and make it optional
$ - Start string ancor.

About the pattern you tried
Note that you don't need {1}. The character class [0-9km] matches 1 of a char k or m or a digit 0-9. This way the possible digits to match could be 0-3 instead of 0-2.
Using the quantifier * for the group makes it possbile to also match 400.25.22.22.22k
You could use this pattern to validate the examples. The [0-9]+ at the beginning of the pattern makes sure that there has to be at least a single digit present.
If you want to allow 500. you could use:
^[0-9]+(?:(?:\.[0-9]{1,2})?[km]?|\.)$
Explanation
^ Start of string
[0-9]+ Match 1+ digits
(?: Non capture group
(?:\.[0-9]{1,2})? Match an optional decimal part with 2 digits
[km]? Match optional k or m
| Or
\. Match a single dot
)$ End of string
Regex demo
let pattern = /^[0-9]+(?:(?:\.[0-9]{1,2})?[km]?|\.)$/;
[
"400k",
"400.2k",
"4m",
"500000",
"500.",
"300.k",
"4.2.k",
".3k",
"4...k",
].forEach(s => console.log(s + " --> " + pattern.test(s)));
Another option is to only match the dot when not directly followed by k or m
^[0-9]+(?:\.(?![km]))?\d*[km]?$
Regex

You can try:
^\d+\.?(?:\d+)?[KkMm]?(?<!\.[KkMm])$
Explanation of the above regex:
^, $ - Matches start and end of the line respectively.
\d+ - Matches digits 1 or more times.
\.? - Represents 0 or 1 occurrence of ..
[KkMm]? - Matches optional characters from the mentioned character class.
(?<!\.[KkMm]) - Represents a negative look-behind not matching a a character after ..
You can find the demo of the above regex in here.
const regex = /^\d+\.?(?:\d+)?[KkMm]?(?<!\.[KkMm])$/gm;
const str = `400K
4.2.K
4.3K
3.2M
300000
4....K
4K
500.
300.K`;
let m;
while ((m = regex.exec(str)) !== null) {
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`${match}`);
});
}
2nd efficient solution using alternation:
You can probably try this regex for more efficient implementation
^\d+(?:\.$|\.\d+)?[KkMm]?$
Explanation of the above regex:
^, $ - Matches start and end of the line respectively.
\d+ - Matches digits 1 or more times.
(?:\.$|\.\d+)? - Represents a non-capturing group; matching either numbers followed by only . or decimal numbers.
[KkMm]? - Matches one of the mentioned characters zero or 1 time.
You can find the demo of the above regex in here.
const regex = /^\d+(?:\.$|\.\d+)?[KkMm]?$/gm;
const str = `400K
4.2.K
4.3K
3.2M
300000
4....K
4K
500.
300.K`;
let m;
while ((m = regex.exec(str)) !== null) {
// The result can be accessed through the `m`-variable.
m.forEach((match, groupIndex) => {
console.log(`${match}`);
});
}

Formatting a phone number in specific way?

This is not a duplicate, the linked thread does not explain how to achieve this.
I'm looking to get a phone number in a specific format.
+xx (x) xxx xxx xxxx
Country code.
Space.
Zero in brackets.
Space.
3 digits.
Space.
3 digits.
Space.
4 digits.
The user could type anything in (but should always be a +61 number). So far I have tried the below.
Removing spaces and non numeric characters.
If starting with a zero, remove.
If starting with 610, remove.
If starting with 61, remove.
Re add country code in specific format and format rest of phone number is a 3,3,4 format.
My question, is - is there a way to simply the below to perhaps one expression?
value = value.replace(/\D/g,'');
value = value.startsWith(0) ? value.substring(1) : value;
value = value.startsWith('610') ? value.substring(3) : value;
value = value.startsWith('61') ? value.substring(2) : value;
value = '+61 (0) ' + value.replace(/\d{3,4}?(?=...)/g, '$& ');

To expand and explain on #splash58's comment they propose using two regular expressions to do the full replacement you desire. The first(/\D|(0+|610|61)/gi) will remove all unwanted characters within the string. The second (/(\d{3})(\d{3})(\d{4})/gi) will take the remaining digits and capture the desired groupings so you can format them as desired. I highly suggest looking at the regex101 links they provided as that site will fully explain how and why a given expressions matches what it does on the right.
Short version:
/\D|(0+|610|61)/gi will match any NON-digit character OR a string of 0s, "610" or "61". Replace this with nothing to remove
/(\d{3})(\d{3})(\d{4})/gi will match a string of 10 digits and capture groups, that's what the parentheses are, of 3 digits, 3 digits and 4 digits. These can be referenced in the replacement as identifiers $1, $2 and $3 according to their position.
Putting it all together:
// look in a string and return formatted phone number only
function phone(str) {
str = str.replace(/\D|(0+|610|61)/gi, '');
str = str.replace(/(\d{3})(\d{3})(\d{4})/gi, '+61 (0) $1 $2 $3');
return str;
}
console.log(phone('xgsh6101231231234vvajx'));
console.log(phone('+6101231231234'));
I would also recommend first doing a search on the entire input string for a series of numbers or whitespace so that you end up with less false positives. This can be done with a regular expression like /[\d\s]+/

You might match the number using:
^.*?\+?0*610?(\d{3})(\d{3})(\d{4})(?!\d).*$
Regex demo
And replace with:
+61 (0) $1 $2 $3
Explanation
^ Assert the start of the string
.*? Match 0+ characters non greedy
\+? Match an optional plus sign
0*610? Match 0+ times a zero, 61 with optional zero
(\d{3})(\d{3})(\d{4}) match 3 groups with 3, 3, and 4 digits
(?!\d) Negative lookahead to assert what follows is not a digit
.* Match 0+ characters
$ Assert the end of the string
const strings = [
"xgsh6101231231234vvajx",
"xgsh06101231231234vvajx",
"xgsh000006101231231234vvajx",
"+6101231231234",
"xgsh61012312312345vvajx",
"xgsh5101231231234vvajx",
"xgsh00000101231231234vvajx",
"xgsh6143545626455345601231231234vvajx"
];
let pattern = /^.*?\+?0*610?(\d{3})(\d{3})(\d{4})(?!\d).*$/;
strings.forEach((s) => {
console.log(s.replace(pattern, "+61 (0) $1 $2 $3"));
});

How does the following code mean two consecutive numbers?

This is from an exercise on FCC beta and i can not understand how the following code means two consecutive numbers seeing how \D* means NOT 0 or more numbers and \d means number, so how does this accumulate to two numbers in a regexp?
let checkPass = /(?=\w{5,})(?=\D*\d)/;

This does not match two numbers. It doesn't really match anything except an empty string, as there is nothing preceding the lookup.
If you want to match two digits, you can do something like this:
(\d)(\d)
Or if you really want to do a positive lookup with the (?=\D*\d) section, you will have to do something like this:
\d(?=\D*\d)
This will match against the last digit which is followed by a bunch of non-digits and a single digit. A few examples (matched numbers highlighted):
2 hhebuehi3
^
245673
^^^^^
2v jugn45
^ ^
To also capture the second digit, you will have to put brackets around both numbers. Ie:
(\d)(?=\D*(\d))
Here it is in action.
In order to do what your original example wants, ie:
number
5+ \w characters
a non-number character
a number
... you will need to precede your original example with a \d character. This means that your lookups will actually match something which isn't just an empty string:
\d(?=\w{5,})(?=\D*\d)
IMPORTANT EDIT
After playing around a bit more with a JavaScript online console, I have worked out the problem with your original Regex.
This matches a string with 5 or more characters, including at least 1 number. This can match two numbers, but it can also match 1 number, 3 numbers, 12 numbers, etc. In order to match exactly two numbers in a string of 5-or-more characters, you should specify the number of digits you want in the second half of your lookup:
let regex = /(?=\w{5,})(?=\D*\d{2})/;
let string1 = "abcd2";
let regex1 = /(?=\w{5,})(?=\D*\d)/;
console.log("string 1 & regex 1: " + regex1.test(string1));
let regex2 = /(?=\w{5,})(?=\D*\d{2})/;
console.log("string 1 & regex 2: " + regex2.test(string1));
let string2 = "abcd23";
console.log("string 2 & regex 2: " + regex2.test(string2));
My original answer was about Regex in a vacuum and I glossed over the fact that you were using Regex in conjunction with JavaScript, which works a little differently when comparing Regex to a string. I still don't know why your original answer was supposed to match two numbers, but I hope this is a bit more helpful.

?= Positive lookahead
w{5,} matches any word character (equal to [a-zA-Z0-9_])
{5,}. matches between 5 and unlimited
\D* matches any character that\'s not a digit (equal to [^0-9])
* matches between zero and unlimited
\d matches a digit (equal to [0-9])
This expression is global - so tries to match all
You can always check your expression using regex101

matching any characters but a number (negative, decimal)

I need your help as I'm stuck on a regular expression.
The regular expression needs to match any characters but the first number.
This first number can be an integer, negative, decimal.
so I have the RegExp for that:
var b = /[-]?[0-9]+([\.][0-9]+)?/;
but when I do that in JavaScript:
var a = 'ab123ab45',
b = /[-]?[0-9]+([\.][0-9]+)?/;
a.replace(b, '');
it obviously return: abab45
But what I need, as you may understood, is the other way around.
Here are some examples.
123 -> 123
123a -> 123
a123a -> 123
123ab45 -> 123
ab123ab45 -> 123
a1b2c3 -> 1
a1.2b -> 1.2
a1,2b -> 1
And I need to get that using only 1 regular expression with the replace function.

If you need it with replace (not with match):
var a = 'ab123ab45',
b = /.*?([-]?[0-9]+([\.][0-9]+)?).*/;
a.replace(b, '$1');

Try:
m = a.match(b);
console.log(m[0]);

Try this;
var a = "a1b2c3";
a = a.replace(/^.*?([.,\d]+).*?$/, "$1");
alert(a);
LIVE DEMO
Regex Explanation
^.*?([.,\d]+).*?$
Assert position at the beginning of the string «^»
Match any single character that is not a line break character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Match the regular expression below and capture its match into backreference number 1 «([.,\d]+)»
Match a single character present in the list below «[.,\d]+»
Between one and unlimited times, as many times as possible, giving back as needed (greedy) «+»
One of the characters “.,” «.,»
A single digit 0..9 «\d»
Match any single character that is not a line break character «.*?»
Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
Assert position at the end of the string (or before the line break at the end of the string, if any) «$»

Regex needed to split a string by "."

I am in need for a regex in Javascript. I have a string:
'*window.some1.some\.2.(a.b + ")" ? cc\.c : d.n [a.b, cc\.c]).some\.3.(this.o.p ? ".mike." [ff\.]).some5'
I want to split this string by periods such that I get an array:
[
'*window',
'some1',
'some\.2', //ignore the . because it's escaped
'(a.b ? cc\.c : d.n [a.b, cc\.c])', //ignore everything inside ()
'some\.3',
'(this.o.p ? ".mike." [ff\.])',
'some5'
]
What regex will do this?

var string = '*window.some1.some\\.2.(a.b + ")" ? cc\\.c : d.n [a.b, cc\\.c]).some\\.3.(this.o.p ? ".mike." [ff\\.]).some5';
var pattern = /(?:\((?:(['"])\)\1|[^)]+?)+\)+|\\\.|[^.]+?)+/g;
var result = string.match(pattern);
result = Array.apply(null, result); //Convert RegExp match to an Array
Fiddle: http://jsfiddle.net/66Zfh/3/
Explanation of the RegExp. Match a consecutive set of characters, satisfying:
/ Start of RegExp literal
(?: Create a group without reference (example: say, group A)
\( `(` character
(?: Create a group without reference (example: say, group B)
(['"]) ONE `'` OR `"`, group 1, referable through `\1` (inside RE)
\) `)` character
\1 The character as matched at group 1, either `'` or `"`
| OR
[^)]+? Any non-`)` character, at least once (see below)
)+ End of group (B). Let this group occur at least once
| OR
\\\. `\.` (escaped backslash and dot, because they're special chars)
| OR
[^.]+? Any non-`.` character, at least once (see below)
)+ End of group (A). Let this group occur at least once
/g "End of RegExp, global flag"
/*Summary: Match everything which is not satisfying the split-by-dot
condition as specified by the OP*/
There's a difference between + and +?. A single plus attempts to match as much characters as possible, while a +? matches only these characters which are necessary to get the RegExp match. Example: 123 using \d+? > 1 and \d+ > 123.
The String.match method performs a global match, because of the /g, global flag. The match function with the g flag returns an array consisting of all matches subsequences.
When the g flag is omitted, only the first match will be selected. The array will then consist of the following elements:
Index 0: <Whole match>
Index 1: <Group 1>

The regex below :
result = subject.match(/(?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))/g);
Can be used to acquire the desired results. Group 1 has the results since you want to omit the .
Use this :
var myregexp = /(?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))/g;
var match = myregexp.exec(subject);
while (match != null) {
for (var i = 0; i < match.length; i++) {
// matched text: match[i]
}
match = myregexp.exec(subject);
}
Explanation :
// (?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))
//
// Match the regular expression below «(?:(\(.*?[^'"]\)|.*?[^\\])(?:\.|$))»
// Match the regular expression below and capture its match into backreference number 1 «(\(.*?[^'"]\)|.*?[^\\])»
// Match either the regular expression below (attempting the next alternative only if this one fails) «\(.*?[^'"]\)»
// Match the character “(” literally «\(»
// Match any single character that is not a line break character «.*?»
// Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
// Match a single character NOT present in the list “'"” «[^'"]»
// Match the character “)” literally «\)»
// Or match regular expression number 2 below (the entire group fails if this one fails to match) «.*?[^\\]»
// Match any single character that is not a line break character «.*?»
// Between zero and unlimited times, as few times as possible, expanding as needed (lazy) «*?»
// Match any character that is NOT a “A \ character” «[^\\]»
// Match the regular expression below «(?:\.|$)»
// Match either the regular expression below (attempting the next alternative only if this one fails) «\.»
// Match the character “.” literally «\.»
// Or match regular expression number 2 below (the entire group fails if this one fails to match) «$»
// Assert position at the end of the string (or before the line break at the end of the string, if any) «$»

It is notoriously difficult to use a Regex to do balanced parenthesis matching, especially in Javascript.
You would be way better off creating your own parser. Here's a clever way to do this that will utilize the strength of Regex's:
Create a Regex that matches and captures any "pattern of interest" - /(?:(\\.)|([\(\[\{])|([\)\]\}])|(\.))/g
Use string.replace(pattern, function (...)), and in the function, keep a count of opening braces and closing braces.
Add the matching text to a buffer.
If the split character is found and the opening and closing braces are balanced, add the buffer to your results array.
This solution will take a bit of work, and requires knowledge of closures, and you should probably see the documentation of string.replace, but I think it is a great way to solve your problem!
Update:
After noticing the number of questions related to this one, I decided to take on the above challenge.
Here is the live code to use a Regex to split a string.
This code has the following features:
Uses a Regex pattern to find the splits
Only splits if there are balanced parenthesis
Only splits if there are balanced quotes
Allows escaping of parenthesis, quotes, and splits using \
This code will work perfectly for your example.

not need regex for this work.
var s = '*window.some1.some\.2.(a.b + ")" ? cc\.c : d.n [a.b, cc\.c]).some\.3.(this.o.p ? ".mike." [ff\.]).some5';
console.log(s.match(/(?:\([^\)]+\)|.*?\.)/g));
output:
["*window.", "some1.", "some.", "2.", "(a.b + ")", "" ? cc.", "c : d.", "n [a.", "b, cc.", "c]).", "some.", "3.", "(this.o.p ? ".mike." [ff.])", "."]

So, was working with this, and now I see that #FailedDev is rather not a failure, since that was pretty nice. :)
Anyhow, here's my solution. I'll just post the regex only.
((\(.*?((?<!")\)(?!")))|((\\\.)|([^.]))+)
Sadly this won't work in your case however, since I'm using negative lookbehind, which I don't think is supported by javascript regex engine. It should work as intended in other engines however, as can be confirmed here: http://gskinner.com/RegExr/. Replace with $1\n.