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javascript regex : possible to have a range in the quantifier? [duplicate]

I've written a regular expression that matches any number of letters with any number of single spaces between the letters. I would like that regular expression to also enforce a minimum and maximum number of characters, but I'm not sure how to do that (or if it's possible).
My regular expression is:
[A-Za-z](\s?[A-Za-z])+
I realized it was only matching two sets of letters surrounding a single space, so I modified it slightly to fix that. The original question is still the same though.
Is there a way to enforce a minimum of three characters and a maximum of 30?

Yes
Just like + means one or more you can use {3,30} to match between 3 and 30
For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters
From the documentation of the Pattern class
X{n,m} X, at least n but not more than m times
In your case, matching 3-30 letters followed by spaces could be accomplished with:
([a-zA-Z]\s){3,30}
If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)
([a-zA-Z]\s){2,29}[a-zA-Z]
If you'd like whitespaces to count as characters you need to divide that number by 2 to get
([a-zA-Z]\s){1,14}[a-zA-Z]
You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet
If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) at the beginning of the RegExp and removing the other size limitations
All that said, in all honestly I'd probably just test the String's .length property. It's more readable.

This is what you are looking for
^[a-zA-Z](\s?[a-zA-Z]){2,29}$
^ is the start of string
$ is the end of string
(\s?[a-zA-Z]){2,29} would match (\s?[a-zA-Z]) 2 to 29 times..

Actually Benjamin's answer will lead to the complete solution to the OP's question.
Using lookaheads it is possible to restrict the total number of characters AND restrict the match to a set combination of letters and (optional) single spaces.
The regex that solves the entire problem would become
(?=^.{3,30}$)^([A-Za-z][\s]?)+$
This will match AAA, A A and also fail to match AA A since there are two consecutive spaces.
I tested this at http://regexpal.com/ and it does the trick.

You should use
[a-zA-Z ]{20}
[For allowed characters]{for limiting of the number of characters}

Regex - must contain number and must not contain special character

I want to check by regex if:
String contains number
String does not contain special characters (!<>?=+#{}_$%)
Now it looks like:
^[^!<>?=+#{}_$%]+$
How should I edit this regex to check if there is number anywhere in the string (it must contain it)?

you can add [0-9]+ or \d+ into your regex, like this:
^[^!<>?=+#{}_$%]*[0-9]+[^!<>?=+#{}_$%]*$
or
^[^!<>?=+#{}_$%]*\d+[^!<>?=+#{}_$%]*$
different between [0-9] and \d see here

Just look ahead for the digit:
var re = /^(?=.*\d)[^!<>?=+#{}_$%]+$/;
console.log(re.test('bob'));
console.log(re.test('bob1'));
console.log(re.test('bob#'))
The (?=.*\d) part is the lookahead for a single digit somewhere in the input.

You only needed to add the number check, is that right? You can do it like so:
/^(?=.*\d)[^!<>?=+#{}_$%]+$/
We do a lookahead (like peeking at the following characters without moving where we are in the string) to check to see if there is at least one number anywhere in the string. Then we do our normal check to see if none of the characters are those symbols, moving through the string as we go.
Just as a note: If you want to match newlines (a.k.a. line breaks), then you can change the dot . into [\W\w]. This matches any character whatsoever. You can do this in a number of ways, but they're all pretty much as clunky as each other, so it's up to you.

jQuery validation- Input Value combination of number and letter [duplicate]

I have a regex
/^([a-zA-Z0-9]+)$/
this just allows only alphanumerics but also if I insert only number(s) or only character(s) then also it accepts it. I want it to work like the field should accept only alphanumeric values but the value must contain at least both 1 character and 1 number.

Why not first apply the whole test, and then add individual tests for characters and numbers? Anyway, if you want to do it all in one regexp, use positive lookahead:
/^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$/

This RE will do:
/^(?:[0-9]+[a-z]|[a-z]+[0-9])[a-z0-9]*$/i
Explanation of RE:
Match either of the following:
At least one number, then one letter or
At least one letter, then one number plus
Any remaining numbers and letters
(?:...) creates an unreferenced group
/i is the ignore-case flag, so that a-z == a-zA-Z.

I can see that other responders have given you a complete solution. Problem with regexes is that they can be difficult to maintain/understand.
An easier solution would be to retain your existing regex, then create two new regexes to test for your "at least one alphabetic" and "at least one numeric".
So, test for this :-
/^([a-zA-Z0-9]+)$/
Then this :-
/\d/
Then this :-
/[A-Z]/i
If your string passes all three regexes, you have the answer you need.

The accepted answers is not worked as it is not allow to enter special characters.
Its worked perfect for me.
^(?=.*[0-9])(?=.*[a-zA-Z])(?=\S+$).{6,20}$
one digit must
one character must (lower or upper)
every other things optional
Thank you.

While the accepted answer is correct, I find this regex a lot easier to read:
REGEX = "([A-Za-z]+[0-9]|[0-9]+[A-Za-z])[A-Za-z0-9]*"

This solution accepts at least 1 number and at least 1 character:
[^\w\d]*(([0-9]+.*[A-Za-z]+.*)|[A-Za-z]+.*([0-9]+.*))

And an idea with a negative check.
/^(?!\d*$|[a-z]*$)[a-z\d]+$/i
^(?! at start look ahead if string does not
\d*$ contain only digits | or
[a-z]*$ contain only letters
[a-z\d]+$ matches one or more letters or digits until $ end.
Have a look at this regex101 demo
(the i flag turns on caseless matching: a-z matches a-zA-Z)

Maybe a bit late, but this is my RE:
/^(\w*(\d+[a-zA-Z]|[a-zA-Z]+\d)\w*)+$/
Explanation:
\w* -> 0 or more alphanumeric digits, at the beginning
\d+[a-zA-Z]|[a-zA-Z]+\d -> a digit + a letter OR a letter + a digit
\w* -> 0 or more alphanumeric digits, again
I hope it was understandable

What about simply:
/[0-9][a-zA-Z]|[a-zA-Z][0-9]/
Worked like a charm for me...
Edit following comments:
Well, some shortsighting of my own late at night: apologies for the inconvenience...
The - incomplete - underlying idea was that only one "transition" from a digit to an alpha or from an alpha to a digit was needed somewhere to answer the question.
But next regex should do the job for a string only comprised of alphanumeric characters:
/^[0-9a-zA-Z]*([0-9][a-zA-Z]|[a-zA-Z][0-9])[0-9a-zA-Z]*$/
which in Javascript can be furthermore simplified as:
/^[0-9a-z]*([0-9][a-z]|[a-z][0-9])[0-9a-z]*$/i
In IMHO it's more straigthforward to read and understand than some other answers (no backtraking and the like).
Hope this helps.

If you need the digit to be at the end of any word, this worked for me:
/\b([a-zA-Z]+[0-9]+)\b/g
\b word boundary
[a-zA-Z] any letter
[0-9] any number
"+" unlimited search (show all results)

Regular Expression: Allow letters, numbers, and spaces (with at least one letter not number)

I'm currently using this regex (/^[A-Za-z0-9 _]*[A-Za-z0-9][A-Za-z0-9 _]*$/) to accept letters, numbers, spaces and underscores. I want to change it like this that it takes the combination of number and the character but not only the number.

If i understand correctly you want to allow a string that begins with at least one letter and optionally is followed by number or underscore or space.
Try this: /^(?:[A-Za-z]+)(?:[A-Za-z0-9 _]*)$/ at this online regex tester.
This should work.
Cheers!

Try this:
/^[A-Za-z0-9 _]*[A-Za-z]+[A-Za-z0-9 _]*$/
This allows for 0 or more of each of the outside groups, and 1 or more of the inner group (letters only). A string of only digits will fail.

/^[\w\s]+$/
\w allows letters, numbers and underscores
\s allow spaces

Try this:
^(?![0-9]*$)[a-zA-Z0-9\s_]+$
This expression has a negative lookahead to verify that the string is not only numbers. See it in action with regexr

javascript regex float number which ends with string

Learning regex but this one gives me a headache. I need to match a float number (with either . or , as decimal point) and it MUST end with the following characters: €/g.
Valid matches should be for example:
40€/g
43.33€/g
40,2€/g
40.2€/g
38.943€/g
Appreciate help..

The regex will look like:
\d+(?:[.,]\d+)?€/g
In Javascript, as a regex object (note that the forward slash needs to be escaped):
/\d+(?:[.,]\d+)?€\/g/
Here's a breakdown of what each part does:
\d+ # one or more digits
(?: # ... don't capture this group separately
[.,] # decimal point
\d+ # one or more digits
)? # make the group optional
€/g # fixed string to match
If you want to allow something like .123€/g to be valid as well, you can use:
(?=[.,]|\d)(?:\d+)?(?:[.,]\d+)?€/g
That is, both the groups of digits are optional, but at least one must be present (this uses lookahead, which is a bit more tricky).
Note that this will also match constructions like 'word2€/g'. If you want to prevent this, start the regex with (?<=^|\s) (matches if preceded by a space or the start of the string) and end it with (?=$|\s) (matches if followed by a space or the end of the string).
Full-blown version:
(?<=^|\s)(?=[.,]|\d)(?:\d+)?(?:[.,]\d+)?€/g(?=$|\s)

\d+([.,]\d+)?€/g
should work, I guess.

Are you really sure you need a regex for this? It might be easier to instead leverage the builtin floating point parsing that is available: take whatever comes before the euro sign, normalize commas to decimals (or vice versa, whatever ends up working) and then try to parse it with the Number function. Note that you would need to check if the conversion worked with the Number.isNaN function.
Another possibility is to just use the parseFloat function. Since it ignores any characters after the numbers then it would parse "40€ as 40.0. However, it might not be what you want since it would also allow things like "40a" and "40b" as well.