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Regular expression to check contains only

EDIT: Thank you all for your inputs. What ever you answered was right.But I thought I didnt explain it clear enough.
I want to check the input value while typing itself.If user is entering any other character that is not in the list the entered character should be rolled back.
(I am not concerning to check once the entire input is entered).
I want to validate a date input field which should contain only characters 0-9[digits], -(hyphen) , .(dot), and /(forward slash).Date may be like 22/02/1999 or 22.02.1999 or 22-02-1999.No validation need to be done on either occurrence or position. A plain validation is enough to check whether it has any other character than the above listed chars.
[I am not good at regular expressions.]
Here is what I thought should work but not.
var reg = new RegExp('[0-9]./-');
Here is jsfiddle.

Your expression only tests whether anywhere in the string, a digit is followed by any character (. is a meta character) and /-. For example, 5x/- or 42%/-foobar would match.
Instead, you want to put all the characters into the character class and test whether every single character in the string is one of them:
var reg = /^[0-9.\/-]+$/
^ matches the start of the string
[...] matches if the character is contained in the group (i.e. any digit, ., / or -).
The / has to be escaped because it also denotes the end of a regex literal.
- between two characters describes a range of characters (between them, e.g. 0-9 or a-z). If - is at the beginning or end it has no special meaning though and is literally interpreted as hyphen.
+ is a quantifier and means "one or more if the preceding pattern". This allows us (together with the anchors) to test whether every character of the string is in the character class.
$ matches the end of the string
Alternatively, you can check whether there is any character that is not one of the allowed ones:
var reg = /[^0-9.\/-]/;
The ^ at the beginning of the character class negates it. Here we don't have to test every character of the string, because the existence of only character is different already invalidates the string.
You can use it like so:
if (reg.test(str)) { // !reg.test(str) for the first expression
// str contains an invalid character
}

Try this:
([0-9]{2}[/\-.]){2}[0-9]{4}

If you are not concerned about the validity of the date, you can easily use the regex:
^[0-9]{1,2}[./-][0-9]{1,2}[./-][0-9]{4}$
The character class [./-] allows any one of the characters within the square brackets and the quantifiers allow for either 1 or 2 digit months and dates, while only 4 digit years.
You can also group the first few groups like so:
^([0-9]{1,2}[./-]){2}[0-9]{4}$
Updated your fiddle with the first regex.

regular expression incorrectly matching % and $

I have a regular expression in JavaScript to allow numeric and (,.+() -) character in phone field
my regex is [0-9-,.+() ]
It works for numeric as well as above six characters but it also allows characters like % and $ which are not in above list.

Even though you don't have to, I always make it a point to escape metacharacters (easier to read and less pain):
[0-9\-,\.+\(\) ]
But this won't work like you expect it to because it will only match one valid character while allowing other invalid ones in the string. I imagine you want to match the entire string with at least one valid character:
^[0-9\-,\.\+\(\) ]+$
Your original regex is not actually matching %. What it is doing is matching valid characters, but the problem is that it only matches one of them. So if you had the string 435%, it matches the 4, and so the regex reports that it has a match.
If you try to match it against just one invalid character, it won't match. So your original regex doesn't match the string %:
> /[0-9\-,\.\+\(\) ]/.test("%")
false
> /[0-9\-,\.\+\(\) ]/.test("44%5")
true
> "444%6".match(/[0-9\-,\.+\(\) ]/)
["4"] //notice that the 4 was matched.
Going back to the point about escaping, I find that it is easier to escape it rather than worrying about the different rules where specific metacharacters are valid in a character class. For example, - is only valid in the following cases:
When used in an actual character class with proper-order such as [a-z] (but not [z-a])
When used as the first or last character, or by itself, so [-a], [a-], or [-].
When used after a range like [0-9-,] or [a-d-j] (but keep in mind that [9-,] is invalid and [a-d-j] does not match the letters e through f).
For these reasons, I escape metacharacters to make it clear that I want to match the actual character itself and to remove ambiguities.

You just need to anchor your regex:
^[0-9-,.+() ]+$
In character class special char doesn't need to be escaped, except ] and -.
But, these char are not escaped when:
] is alone in the char class []]
- is at the begining [-abc] or at the end [abc-] of the char class or after the last end range [a-c-x]

Escape characters with special meaning in your RegExp. If you're not sure and it isn't an alphabet character, it usually doesn't hurt to escape it, too.
If the whole string must match, include the start ^ and end $ of the string in your RegExp, too.
/^[\d\-,\.\+\(\) ]*$/

Regex to validate letters and numbers doesn't work as expected

I'm trying to create a regular expression that contains 17 characters, must have mandatory numbers and letters, and must not contain the letters I, O, Q, Ñ. At the moment I have got:
^(([a-h]|[j-n]|p|[r-z]|[A-H]|[J-N]|P|[R-Z]|[0-9]){17})$
But if I type only numbers or only letters, the regular expression validates it as good.

Generally in such cases you want to use positive lookahead in order to assert that the input satisfies additional conditions. Since you have two conditions here (must contain at least one number, must contain at least one letter), this translates to two different lookaheads.
Together with extra whitespace and "comments" for readability the regex should look like this:
^
(?=.*[a-hj-npr-zA-HJ-NPR-Z].*) // assert the input contains at least one letter
(?=.*[0-9].*) // assert the input contains at least one digit
[a-hj-npr-zA-HJ-NPR-Z0-9]{17} // existing condition (17 allowed chars exactly)
$

Use positive and negative look-aheads to require and disallow characters:
/^(?=.*[0-9])(?=.*[A-Za-z])(?!.*[ioqñIOQÑ])[0-9a-zA-Z]{17}$/.test(s);
Demo: http://jsfiddle.net/2mNyg/
Description:
(?=.*[0-9]) - requires a digit
(?=.*[A-Za-z]) - requires a letter
(?!.*[ioqñIOQÑ]) - disallows all characters in the set
[0-9a-zA-Z]{17} - allow basic set and require 17 characters
Note: The look-ahead that requires a letter will also be satisfied by the disallowed characters, but the look-ahead that disallows the characters will still stop it. That way you can make the expression simpler.

Remove entire word from string if it contains numeric value

What I'm trying to accomplish is to auto-generate tags/keywords for a file upload, basing these keywords from the filename.
I have accomplished auto-generating titles for each upload, as shown here:
But I have now moved on to trying to auto-generate keywords. Similar to titles, but with more formatting. First, I run the string through this to remove commonly used words from the filename (such as this,that,there... etc)
I am happy with it, but I need to not include words that have numbers in it. I have not found a solution on how to remove a word entirely if it contains a number. The solutions I have found like here only works for a certain match, while this one removes numbers alone. I would like to remove the entire word if it contains ANY numeric digit.

To remove all words which contain a number, use:
string = string.replace(/[a-z]*\d+[a-z]*/gi, '');

Try this expression:
var regex = /\b[^\s]*\d[^\s]*\b/g;
Example:
var str = "normal 5digit dig555it digit5 555";
console.log( str.replace(regex,'') );​ //Result-> normal

Apply a simple regular expression to you current filename strings, replacing all occurrences with the empty string. The regular expression matches "words" containing any digits.
Javascript example:
'asdf 8bit jawesome234 mayhem 234'.replace(/\s*\b\w*\d\w*\b/g, '')
Evaluates to:
"asdf mayhem"
Here the regular expression is /\s*\b\w*\d\w*\b/g, which matches maximal sequences consisting of zero or more whitespace characters (\s*) followed by a word-boundary transition (\b), followed by zero or more alphanum characters (\w*), followed by a digit (\d), followed by zero or more alphanum characters, followed by a word-boundary transition (\b). \b matches the empty string at the transition to an alphanumeric character from either the beginning or end of the word or a non-alphanumeric character. The g after the final / of the regular expression means replace all occurrences, not just the first.
Once the digit-words are removed, you can split the string into keywords however you want (by whitespace, for example).
"asdf mayhem".split(/\s+/);
Evaluates to:
["asdf", "mayhem"]

('Apple Cover Photo 23s423 of your 543634 moms').match(/\b([^\d]+)\b/g, '')
returns
Apple Cover Photo , of your , moms
http://jsfiddle.net/awBPX/2/

use this to Remove words containing numeric :
string.replace("[0-9]","");
hope this helps.
Edited :
check this :
var str = 'one 2two three3 fo4ur 5 six';
var result = str.match(/(^[\D]+\s|\s[\D]+\s|\s[\D]+$|^[\D]+$)+/g).join('');

regular expression using javascript

I have a regular expression ^(?=.*?[A-Za-z])\S*$ which indicates that the input should contain alphabets and can contain special characters or digits along with the alphabets. But it is not allowing white spaces since i have used \S.
Can some one suggest me a reg exp which should contain alphabets and it can contain digits or special characters and white space but alphabets are must and the last character should not end with a white space

Quite simply:
^(?=.*?[A-Za-z]).*$
Note that in JavaScript . doesn't match new lines, and there is no dot-all flag (/s). You can use something like [\s\S] instead if that is an issue:
^(?=[\s\S]*?[A-Za-z])[\s\S]*$
Since you only have a single lookahead, you can simplify the pattern to:
^.*[A-Za-z].*$
Or, even simpler:
[A-Za-z]
[A-Za-z] will match if it finds a letter anywhere in the string, you don't really need to search the rest from start to end.
To also validate the last character isn't a whitespace, it is probably easiest to use the lookahead again (as it basically means AND in regular expressions:
^(?=.*?[A-Za-z]).*\S$