Javascript Email Validation function

This function is fantastic, but there is one problem:

It accepts an invalid email address, e.g. try@haveago.

is accepted as a valid address. The address should not finish with a dot unless I am mistaken.

<script language="javascript"><!--

function verifyEmail(form) {
checkEmail = form.email.value

if ((checkEmail.indexOf('@') < 0) || ((checkEmail.charAt(checkEmail.length-4) != '.') && (checkEmail.charAt(checkEmail.length-3) != '.')))
{alert("You have entered an invalid email address. Please try again.");
form.email.select();
return false;
}

else {
form.method="get";
form.target="_self";
form.action="myscript.cgi";
form.submit();
}

}
//--></script>

Leuk hoor

There are certain points to be noted

1) What about domain names ending with .de , .se ,.in

2)The e-mail address like "bikash" is valid one. Try any unix box.

3)The e-mail address need not always contain "." . Again try any Unix box.

Scott Kreischer' code works for .in,.de etc

function validate()
{
validRegExp = /^[^@]+@[^@]+.[a-z]{2,}$/i;
strEmail = document.form1.email.value;

if (strEmail.search(validRegExp) == -1)
{
alert(" A valid e-mail address is required.\r\Please check you have entered your details correctly.\r\r\ © Paul Holmes 2001-2003");
return false;
}

hi there...
when i enter an e-mail like this: dino@@yahoo.com .it's still accept this e-mail... can u make it better?

To validate email format, I've had luck with the following regular expression:

/^\w(\.?[\w-])*@\w(\.?[\w-])*\.[a-z]{2,6}$/i;

However, some browsers (Mac IE?) will result in an alert if there is an underscore. Any ideas why?

THe email validation should check for the postition of @ and the (.) and the last 3 characters after dot.a maximum length of email not just @ and .

this is the best I found and it works!


function isEmail(string) {
if (string.search(/^\w+((-\w+)|(\.\w+))*\@[A-Za-z0-9]+((\.|-)[A-Za-z0-9]+)*\.[A-Za-z0-9]+$/) != -1)
return true;
else
return false;
}




For the explanation:

http://tech.irt.org/articles/js049/index.htm

Scott's stuff is better. But thanks anyways !

i need the function of e-mail to be state out for me

Can any body provide me a solution of running a c++ code using Java.Is there any command.

Same as William's example above, but true and false are returned implicitly.

function isEmail(string) {
return (string.search(/^\w+((-\w+)|(\.\w+))*\@[A-Za-z0-9]+((\.|-)[A-Za-z0-9]+)*\.[A-Za-z0-9]+$/) != -1);
}

The beauty of this function is that it is extraordinarly simple.

Although there are some flaws. For example it will return true with the following string: 'asd@.com' or 'asd.asd@.com', which are obviously not valid e-mail addresses.

So i made a little change, keeping it simple anyway:

function isValidEmail(str) {
return (str.lastIndexOf(".") > 2) && (str.indexOf("@") > 0) && (str.lastIndexOf(".") > (str.indexOf("@")+1));
}

Cheers

<Added>

Here is an updated version.
Now it checks if exists more than one '@', like asd@@asd.com

function isValidEmail(str) {
return (str.lastIndexOf(".") > 2) && (str.indexOf("@") > 0) && (str.lastIndexOf(".") > (str.indexOf("@")+1)) && (str.indexOf("@") == str.lastIndexOf("@"));
}

This is regarding Posted by : bestondoa at 09:15 on Thursday, December 07, 2006

have you done simple unit tests for your function? In my opinion you might not have tested it with some cases.
Definitely it is not a valid function to check email addresses, for instance if the email value is '.@......' then your function assumes it is a valid email.

Do you think this is a valid email address?

Thanks,

This is regarding the function provided in the article by Jeff Anderson:

For your convineince here is the function:

function isValidEmail(str) {
return (str.indexOf(".") > 2) && (str.indexOf("@") > 0);

}

if the input is '#@@@...' then obviously the function reports as this is a valid email, is that?

I wonder why people get into rush to write something that is erroneous.


Thanks,
Goit

To goit:

See my post above? Did you get the logic?

Well, next time you find an error, try and help the community by submitting a possible solution.
I did that, solved a couple of errors, but obviously there are more as you correctly pointed out.

Nevertheless your post lacks a solution. If everyone did as you, we wouldn't probably be here reading other people's suggestions/solutions.

Thanks,
Bestondoa