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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}`);
});
}

Regex in JS: How to match a string that starts with "api" with exceptions

I need to match all string that starts with /api/v except those ends with /user/logout.
Ex:
/api/v2/segments (match)
/api/v2/user (match)
/api/v2/user/logout (NO match)
I'm trying with this regex but it doesn't work
/.*\/api\/v.*(^\/user\/logout)$/

You may use
/^\/api\/v(?!.*\/user\/logout$).*/
See the regex demo
Details
^ - start of string
\/api\/v - a /api/v string
(?!.*\/user\/logout$) - (a negative lookahead) no 0+ chars other than line break chars, as many as possible, followed with /user/logout at the end of string are allowed immediately to the right of the current location
.* - any 0+ chars other than line break chars as many as possible.
You need no .* at the end if you just test the string with RegExp#test for a match. If you want to use the match value, you need .* at the end.

How to match string with hypens in JavaScript?

I want to match certain parts of a URL that has the following form:
http://example.com/somepath/this-is-composed-of-hypens-3144/someotherpath
More precisely, I want to match only the part that is composed of all hypens and ends in numbers. So, I want to extract the part this-is-composed-of-hypens-3144 in the above URL. I have something like this:
const re = /[a-z]*-[a-z]*-[0-9]*/gis;
const match = re.exec(url);
return (match && match.length) ? match[0] : null;
However, this works only if there are 2 hypens, however, the number of hypens in my case can be arbitrary. How can I make my regex work for arbitrary number of hypens?

You may use
/\/([a-z]+(?:-[a-z]+)*-[0-9]+)(?:\/|$)/i
See the regex demo
Details
\/ - a / char
([a-z]+(?:-[a-z]*)*-[0-9]+) - Capturing group 1:
[a-z]+ - 1+ ASCII letters
(?:-[a-z]+)* - 0+ occurrences of - followed with 1+ ASCII letters
(?:\/|$) - either / or end of string.
If there can be any word chars, not just ASCII letters, you may replace each [a-z] with \w.
var s = "http://example.com/somepath/this-is-composed-of-hypens-3144/someotherpath";
var m = s.match(/\/([a-z]+(?:-[a-z]+)*-[0-9]+)(?:\/|$)/i);
if (m) {
console.log(m[1]);
}

Javascript in regexp not matching something

I want to match everything except the one with the string '1AB' in it. How do I do that? When I tried it, it said nothing is matched.
var text = "match1ABmatch match2ABmatch match3ABmatch";
var matches = text.match(/match(?!1AB)match/g);
console.log(matches[0]+"..."+matches[1]);

Lookarounds do not consume the text, i.e. the regex index does not move when their patterns are matched. See Lookarounds Stand their Ground for more details. You still must match the text with a consuming pattern, here, the digits.
Add \w+ word matching pattern after the lookahead. NOTE: You may also use \S+ if there can be any one or more non-whitespace chars. If there can be any chars, use .+ (to match 1 or more chars other than line break chars) or [^]+ (matches even line breaks).
var text = "match100match match200match match300match";
var matches = text.match(/match(?!100(?!\d))\w+match/g);
console.log(matches);
Pattern details
match - a literal substring
(?!100(?!\d)) - a negative lookahead that fails the match if, immediately to the right of the current location, there is 100 substring not followed with a digit (if you want to fail the matches where the number starts with 100, remove the (?!\d) lookahead)
\w+ - 1 or more word chars (letters, digits or _)
match - a literal substring
See the regex demo online.

Regex to validate hypen(-) at start and end of the string

I'm validating a string("-test-") whether it contains hypens(-) at start and end of the string using regex. So i found an regex to restrict hypen at start and end of regex.
/^(?!-)[a-zA-Z0-9-' ]+[^-]$/i
This regex was validating as expected when the string contains more than one char("aa") with or without hypen. But its not working as expected when i'm simply passing one character string("a") without hypen.
And also these need to allow special characters and alphanumeric characters like "$abcd&". Need to restirct oly hypen at start and end of the string.
Could you guys help out of this..

The pattern you have matches a string that consists of at least 2 chars because [a-zA-Z0-9-' ]+ needs 1 char to match and [^-] requires another char to be present.
You may revamp the lookahead to also fail a string that ends with -:
/^(?!-)(?!.*-$).+$/
^^^^^^^^
See the regex demo
Details
^ - start of a string
(?!-)(?!.*-$) - negative lookaheads that fail the match if the string starts with - or ends with -
.+ - any 1 or more chars other than line break chars (use [\s\S] to match any char)
$ - end of string.
An unrolled version for this pattern would be
^[^-]+(?:-+[^-]+)*$
See this regex demo
Details
^ - start of string
[^-]+ - 1 or more chars other than -
(?:-+[^-]+)* - 0+ sequences of
-+ - 1+ hyphens
[^-]+ - 1 or more chars other than -
$ - end of string.

To allow any character but only disallow hyphen at start and end:
^(?!-).*[^-]$
^ start of string
(?!-) look ahead if there is no hyphen
.* match any amount of any character
[^-] match one character, that is not a hyphen
$ at the end
See demo at regex101