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Regular Expressions - Match all alphanumeric characters except individual numbers

I would like to create a RegEx to match only english alphanumeric characters but ignore (or discard) isolated numbers in Ruby (and if possible in JS too).
Examples:
1) I would like the following to be matched:
4chan
9gag
test91323432
asf5asdfaf35edfdfad
afafaffe
But not:
92342424
343424
34432
and so on..
The above is exactly what I would want.
Edit: I deleted the second sub-question. Just focus on the first one, thank you very much for your answers!!
Sorry, my regex skills aren't that great (hence this question!)
Thank you.

You can try the following expression (works both in Ruby and Javascript):
^(?!^\d+$)[[:alnum:]]+$
This first ensures the string is not just digits by using a negative look ahead (?!^[0-9]+$), then it matches one or more alphanumeric character, Unicode characters are supported which means this works with French letters too.
EDIT: If you only want English alphabet:
^(?!^\d+$)\w+$
Rubular Demo

For any Latin letters:
/(?=.*\p{Alpha})\p{Alnum}+/

I'm pretty sure that you can't do what you want to do with one regex. A single alpha character, anywhere in a group of numbers, will make it a valid match, and there is no way to represent that in regex, because what you are really saying is something along the lines of "a letter is required at the front of this word, but only if there isn't a letter in the middle or at the end", and regex won't do that.
Your best bet is to do two passes:
one that matches your alphanumeric, plus special "French" characters (pattern: TBD, based on what special characters you want to accept), and
one that matches numbers only (pattern: would include [0-9]+ . . . need more information about the specific situation to give you a final, complete regex)
The values that you want in the end would need to pass the first regex and fail the second one.
Also . . .
To give you a better answer, we'll need to know a couple of things:
Are you testing that an entire string matches the pattern?
Are you trying to capture a single instance of the pattern in a bigger string?
Are you trying to capture all of the instances of the pattern in a bigger string?
The answers to those questions have a big impact on the final regex pattern that you will need.
And, finally . . .
A note on the "French" characters . . . you need to be very specific about which special characters are acceptable and which aren't. There are three main approaches to special character matching in regex: groups, additive, and subtractive
groups - these are characters that represent a preset group of characters in the version of regex that you are using. For example, \s matches all whitespaces
additive - this is the process of listing out each acceptable character (or range of characters) in your regex. This is better when you have a small group of acceptable characters
subtractive - this is the process of listing out each UNacceptable character (or range of characters) in your regex. This is better when you have a large group of acceptable characters
If you can clear up some of these questions, we should be able to give you a better answer.

Maybe this ^(?![0-9]+$)[a-zA-Z0-9\x80-\xa5]+$
Edit - fixed cut&paste error and added Extended character range \x80-\xa5
which includes the accent chars (depending on locale set, the figures may be different)

regex for alphaspecialnumeric

I would like to check few of my text boxes that must satisfy the following conditions:
Alphabets i meant are from a-z(uppercase and lower case) numbers 0-9 and special characters are ~`!##$%^&*()-_+={}[];:'",.<>/?
It can contain only alphabets
It cannot contain only numbers
It cannot contain only special characters
It cannot contain only numbers and special characters
It can contain alphabets,numbers and special characters
It can contain alphabets and numbers
It can contain alphabets and special charcters
I found a solution but seems not working for me:
/^[a-z0-9/. -!##$%^&*(){}:;"',/?]+$/i
I am checking it as:
var alpha=/^[a-z0-9/. -!##$%^&*(){}:;"',/?]+$/i;
if (!alpha.test(username.value))
{
alert('Invalid username');
document.theForm.username.focus();
return false;
}

The problem can be restated as that of matching a string containing ONLY the characters
A-Za-z0-9~`!##$%^&*()-_+={}[];:'",.<>/?
such that at least one of them is a letter.
Fortunately, you've covered all the printable characters in the range U+0021 to U+007F, so that the desired regex is simply
[!-~]*[A-Za-z][!-~]*
EDIT: On closer reading, I noticed you did not allow the backslash! If you want to allow the backslash, the regex above is okay; if not, you should modify it like so:
[!-\[\]-~]*[A-Za-z][!-\[\]-~]*
It's a bit uglier, because to exclude the backslash we have to say
All characters in the range ! to [ union characters in the range ] to ~, and the explicit mention of [ and ] requires escaping with, you guessed it, the \.
Hopefully you meant to allow the \ so you can use the simpler regex above.
EDIT 2
To make the regex more efficient, you should use a reluctant quantifier (as kcsoft did):
[!-~]*?[A-Za-z][!-~]*
Also for JavaScript, but not for Java if you are using matches, you should anchor the regex to match the whole string, giving this in JavaScript:
/^[!-~]*?[A-Za-z][!-~]*$/
And, as you did in your question, you can shorten it a bit more by using the i modifier:
/^[!-~]*?[A-Z][!-~]*$/i

Can you give some input examples. Can you try this?
/.*?[a-zA-Z]+.*/
Or if you need to specify the list of special characters:
/[list of chars]*?[a-zA-Z]+[list of chars]*/

(js)regular expression for matching a words only

I'm making a dictionary application and need an regexp that check if the users input is only letters and spaces eventually. This is probably the most easiest regexp but i can figure it out. So far i have
/^[\w\D]$/
which is not working :/
sorry guys, forgot to mention that will need to exclude all spec characters also.

You seem to want this one :
/^[\u00C0-\u1FFF\u2C00-\uD7FFa-zA-Z\s]+$/
It should accept only characters (including "not English" characters like the ones you have in Spanish and Cyrillic) as well as spaces, but exclude digits.
Example :
/^[\u00C0-\u1FFF\u2C00-\uD7FFa-zA-Z\s]+$/.test("переполнения стека")
returns true

Your regular expression matches exactly one such character.
You can add the + modifier to match one or more characters.

To match a string consisting only of letters and whitespace characters, you can use:
/^[a-zA-Z\s]+$/

RegEx a name with special characters in javascript

I'm relative new to RegEx and I've encountered a problem. I want to regex a name. I want it to be max 100 characters, contain at least 2 alphabetic characters and it will allow the character '-'.
I have no problem to only check for alphabetic characters or both alphabetic characters and hyphen but I dont't want a name that potantially can be '---------'.
My code without check for hyphens is
var nameRegExp = /^([a-z]){2,100}$/;
An explanation for the code is appreciated as well.
Thanks!

I guess
/^(?=.*[a-z].*[a-z])[a-z-]{1,100}$/
the lookahead part (^(?=.*[a-z].*[a-z])) checks if there are at least two letters. This pattern ("start of string, followed by...") is a common way to express additional conditions in regexes.
You can limit the number of - by adding a negative assertion, as #Mike pointed out:
/^(?=.*[a-z].*[a-z])(?!(?:.*-){11,})[a-z-]{1,100}$/ // max 10 dashes
however it might be easier to write an expression that would match "good" strings instead of trying to forbid "bad" ones. For example, this
/^[a-z]+(-[a-z]+)*$/
looks like a good approximation for a "name". It allows foo and foo-bar-baz, but not the stuff like ---- or foo----bar----.

To limit the number of - you could add a negative look-ahead, where the number 3 is one more than the maximum number you want to allow
/^(?!(?:[a-z]*-){3,})(?=-*[a-z]-*[a-z])[a-z-]{2,100}$/

Regex to validate brazilian money using Javascript

I need to validate some form fileds that contain brazilian money (its name is "Real") using Javascript. It has the following format:
0,01
0,12
1,23
12,34
123,45
1.234,56
12.235,67
123.456,78
1.234.567,89
12.345.678,90
123.456.789,01
1.234.567.890,12
My regex knowledge is weak, can somebody help me please?

Does this do what you want?
^\d{1,3}(?:\.\d{3})*,\d{2}$
That says "1 to 3 digits, optionally followed by any number of groups of three digits preceded by a period, followed by a comma and two more digits." If you want to allow the leading whitespace present in your example, add \s* to the front:
^\s*\d{1,3}(?:\.\d{3})*,\d{2}$
EDIT: As #ElRonnoco pointed out, the above regular expression accepts leading zeroes (e.g. 010.100,00). To disallow those, you may use this longer version:
^\s*(?:[1-9]\d{0,2}(?:\.\d{3})*|0),\d{2}$
EDIT 2 The above regular expressions all match a string containing a single monetary amount and nothing else. It's not clear from the question if that's the intent.
EDIT 3 To allow numbers that have no decimal part, or only one decimal digit, change it like this:
^\s*(?:[1-9]\d{0,2}(?:\.\d{3})*|0)(?:,\d{1,2})?$

I would give this regex a try:
\d+(?:\.\d{3})*?,\d{2}
What it says is:
- match digits until
a. a dot followed by 3 digits is found (and this step can be repeated several times)
b. or a comma followed by 2 digits is found
EDIT:
- thanks for the comments, I forgot about the constraint for the first value
updated regex
\d{1,3}(?:\.\d{3})*?,\d{2}

Complementing Mark's reply:
Who needs "." in the string and not "," to count cents. And need find the values in middle a text :
(?:[1-9]\d{0,2}(?:\,\d{3})*|0)(?:.\d{1,2})?
https://regexr.com/61166