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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)

Regex for first and last name

I'm trying to make a regexp where it has to match a name like: John Smith. The only rules are first and last name should start with capital letter and has to be at least 2 characters long. Also the last name has a limit of 20 characters maximum and there's a coma or white space between the names. So far I have this:
/[A-Z][a-z]+(\s|,)[A-Z][a-z]{19}/
It doesn't work when I tested it in this site: http://www.regular-expressions.info/javascriptexample.html. I'm not sure what I missed. Any ideas?

Change the {19} to {1,19}. By itself, {19} means "match exactly 19 of the previous character". {1,19} means "match between 1 and 19 of the previous character".
/[A-Z][a-z]+(\s|,)[A-Z][a-z]{1,19}/
UPDATE: People are commenting that this does not meet your requirements. As you described them, it's possibly a naive implementation of your requirements, but it is just your original implementation with the bug fixed. If you are actually looking for names, a less naive implementation might be:
/^[A-Z][-'a-zA-Z]+,?\s[A-Z][-'a-zA-Z]{0,19}$/
This will catch names with apostrophes or dashes, allows a space after the comma between the names if they are separated by a comma, and allows for single-letter last names. But as the commenters have pointed out, this still fails to match a bunch of legitimate names and matches stuff that is definitely not name-like.
It also adds anchors ^ and $ to mean the entire string must match. If you are looking for a substring, you can remove those anchors and add in word boundary checks instead:
/\b[A-Z][-'a-zA-Z]+,?\s[A-Z][-'a-zA-Z]{0,19}\b/

You need {0,19} not just {19}. The latter means "exactly 19 chars".
"John Smith".match(/^[A-Z][a-z]{0,19}[\s,][A-Z][a-z]{0,19}$/)
Of course, this regexp doesn't match many totally valid names like "José Ortega y Gasset" or "Charles de Batz-Castelmore d'Artagnan".
Depending upon how long acceptable surnames can be, you can replace "{0,19}" with "{1,19}" or "{2,19}". The same applies to first names.

Here are some expressions that might help for more complex names
(^[A-Z][a-z]*$) - A typical First Name or Last Name like Thomas
(^[A-Z][a-z][A-Z][a-z]*$) - names like McDonald
(^[A-Z][a-z]*(-|\s)[A-Z][a-z]*$) - names like Tonya-Smith or Tonya Smith
(^[A-Z]('|’)[A-Z][a-z]*$) - names like Tim O’Reilly
(^[a-z]+\s[A-Z][a-z]*$) - names like von Gogh

/^[A-Z][a-z]+[\s|,][A-Z][a-z]{1,19}$/.test("John Smith") // true

The {19} means that the last name must have exactly 19 lowercase characters after the first uppercase character.

This should work for you
\b[A-Z]+.+[?^ ][A-Z].{1,19}|\b[A-Z]+.+[?^,][A-Z].{1,19}
This starts with the beginning of a word, checks that the first letter is caps, matches the first word up to a white space or comma, then checks to make sure the first letter of the next word is capitalized, and matches everything up to 19 characters after that. Also makes sure each name is 2 or more characters long.

regular expression for ends with some word

I want to build regular expression for series
cd1_inputchk,rd_inputchk,optinputchk where inputchk is common (ending characters)
please guide for the same

Very simply, it's:
/inputchk$/
On a per-word basis (only testing matching /inputchk$/.test(word) ? 'matches' : 'doesn\'t match';). The reason this works, is it matches "inputchk" that comes at the end of a string (hence the $)
As for a list of words, it starts becoming more complicated.
Are there spaces in the list?
Are they needed?
I'm going to assume no is the answer to both questions, and also assume that the list is comma-separated.
There are then a couple of ways you could proceed. You could use list.split() to get an array of each word, and teast each to see if they end in inputchk, or you could use a modified regular expression:
/[^,]*inputchk(?:,|$)/g
This one's much more complicated.
[^,] says to match non-, characters
* then says to match 0 or more of those non-, chars. (it will be greedy)
inputchk matches inputchk
(?:...) is a non-capturing parenthesis. It says to match the characters, but not store the match as part of the result.
, matches the , character
| says match one side or the other
$ says to match the end of the string
Hopefully all of this together will select the strings that you're looking for, but it's very easy to make a mistake, so I'd suggest doing some rigorous testing to make sure there aren't any edge-conditions that are being missed.

This one should work (dollar sign basically means "end of string"):
/inputchk$/

RegEx string for three letters and two numbers with pre- and post- spaces

Two quick questions:
What would be a RegEx string for three letters and two numbers with space before and after them (i.e. " LET 12 ")?
Would you happen to know any good RegEx resources/tools?

For a good resource, try this website and the program RegexBuddy. You may even be able to figure out the answer to your question yourself using these sites.
To start you off you want something like this:
/^[a-zA-Z]{3}\s+[0-9]{2}$/
But the exact details depend on your requirements. It's probably a better idea that you learn how to use regular expressions yourself and then write the regular expression instead of just copying the answers here. The small details make a big difference. Examples:
What is a "letter"? Just A-Z or also foreign letters? What about lower case?
What is a "number"? Just 0-9 or also foreign numerals? Only integers? Only positive integers? Can there be leading zeros?
Should there be a single space between the letters and numbers? Or any amount of any whitespace? Even none?
Do you want to search for this string in a larger text? Or match a line exactly?
etc..
The answers to these questions will change the regular expression. It would be much faster for you in the long run to learn how to create the regular expression than to completely specify your requirements and wait for other people to reply.
I forgot to mention that there will be a space before and after. How do I include that?
Again you need to consider the questions:
Do you mean just one space or any amount of spaces? Possibly not always a space but only sometimes?
Do you mean literally a space character or any whitespace characters?
My guess is:
/^\s+[a-zA-Z]{3}\s+[0-9]{2}\s+$/

/[a-z]{3} [0-9]{2}/i will match 3 letters followed by a whitespace character, and then 2 numbers. [a-z] is a character class containing the letters a through z, and the {3} means that you want exactly 3 members of that class. The space character matches a literal space (alternately, you could use \s, which is a "shorthand" character class that matches any whitespace character). The i at the end is a pattern modifier specifying that your pattern is case-insenstive.
If you want the entire string to only be that, you need to anchor it with ^ and $:
/^[a-z]{3} [0-9]{2}$/i
Regular expression resources:
http://www.regular-expressions.info - great tutorial with a lot of information
http://rexv.org/ - online regular expression tester that supports a variety of engines.

^([A-Za-z]{3}) ([0-9]{2})$ assuming one space between the letters/numbers, as in your example. This will capture the letters and numbers separately.
I use http://gskinner.com/RegExr/ - it allows you to build a regex and test it with your own text.

As you can probably tell from the wide variety of answers, RegEx is a complex subject with a wide variety of opinions and preferences, and often more than one way of doing things. Here's my preferred solution.
^[a-zA-Z]{3}\s*\d{2}$
I used [a-zA-Z] instead of \w because \w sometimes includes underscores.
The \s* is to allow zero or more spaces.
I try to use character classes wherever possible, which is why I went with \d.

\w{3}\s{1}\d{2}
And I like this site.
EDIT:[a-zA-Z]{3}\s{1}\d{2} - The \w supports numeric characters too.

try this regularexpression
[^"\r\n]{3,}

Javascript regex

I was trying to do a regex for someone else when I ran into this problem. The requirement was that the regex should return results from a set of strings that has, let's say, "apple" in it. For example, consider the following strings:
"I have an apple"
"You have two Apples"
"I give you one more orange"
The result set should have the first two strings.
The regex(es) I tried are:
/[aA]pple/ and /[^a-zA-Z0-9][aA]pple/
The problem with the first one is that words like "aapple", "bapple", etc (ok, so they are meaningless, but still...) test positive with it, and the problem with the second one is that when a string actually starts with the word "apple", "Apples and oranges", for example, it tests negative. Can someone explain why the second regex behaves this way and what the correct regex would be?

/(^.*?\bapples?\b.*$)/i
Edit: The above will match the entire string containing the word "apples", which I thought is what you were asking for. If you are just trying to see if the string contains the word, the following will work.
/\bapples?\b/i
The regex(es) I tried are:
/[aA]pple/ and /[^a-zA-Z0-9][aA]pple/
The first one just checks for the existence of the following characters, in order: a-p-p-l-e, regardless of what context they are used in. The \b, or word-boundary character, matches any spot where a non-word character and a word character meet, ala \W\w.
The second one is trying to match other characters before the occurrance of a-p-p-l-e, and is essentially the same as the first, except it requires other characters in front of it.
The one I answered with works like following. From the beginning of the string, matches any characters (if they exist) non-greedily until it encounters a word boundary. If the string starts with apple, the beginning of a string is a word-boundary, so it still matches. It then matches the letters a-p-p-l-e, and s if it exists, followed by another word boundary. It then matches all characters to the end of the string. The /i at the end means it's case-insensitive, so 'Apple', 'APPLE', and 'apple' are all valid.
If you have the time, I would highly recommend walking through the tutorial at http://regular-expressions.info. It really goes in-depth and talks about how the regular expression engines match different expressions, it helped me a ton.

To build on #tj111, the reason your second regex fails is that [^a-zA-Z0-9] requires that a character matches; that is, there is some character in that position, and its value is not contained in the set [a-zA-Z0-9]. Markers like \b are called "zero-width assertions". \b, in particular, matches against boundaries between characters or at the beginning or end of a string. Because it is not matching against any character, its "width" is zero.
In sum, [^a-zA-Z0-9] requires a character that does not take a particular value be present, while \b requires only that a boundary be present.
Edit: #tj111 has added most of this to his response. I'm in too late, again :)

This works for apple and apples and its case-insensitive spellings:
var strings = ["I have an apple", "You have two Apples", "I give you one more orange"];
var result = [];
var pattern = /\bapples?\b/i;
for (var i=0; i<strings.length; i++) {
if (pattern.test(strings[i])) {
result.push(strings[i]);
}
}

Your second regex requires a nonalphanumeric character before the first a in apple. "apple" doesn't satisfy this. As others note, "\b" matches not a character, but a word boundary position.

/\bapple/i
\b is a word boundary.
To explain why your attempts do not work, the first one does not check that it is the beginning of the word, so it can have something before it. The second regex you gave says that something must be before the word "apple", but it can't be alphanumeric.