正则表达式-匹配捕获组中的第一个和最后一个字符

Question

I want to capture the first and last character within a capturing group.

My current RegEx is -

([\w\.]+)@([\w]+)\.com

For example, if there is an email address -

xyz@test.com

This is the output -

Full match  0-12    `xyz@test.com`
Group 1.    0-3 `xyz`
Group 2.    4-8 `test`

The email address can have alphanumeric and period values.

If I want to curtail the Group 1 such that it starts and ends with only alphanumeric values, how to do that?

I want to modify this capturing group -

 ([\w\.]+)

The required output is -

xyz.@test.com Invalid
.xyz@test.com Invalid
xy.z@test.com Valid

Answer 1

To tell engine match English alphanumeric characters at the start position and one before @ you need to do this:

^([a-zA-Z0-9][\.a-zA-Z0-9]*[a-zA-Z0-9])@([a-zA-Z0-9]+)\.com$

Note: \\w includes _ that you may not desire.

But this doesn't allow usernames with one character long. So you have to modify it a little:

^([a-zA-Z0-9]+(?:\.+[a-zA-Z0-9]+)*)@([a-zA-Z0-9]+)\.com$

Also this shouldn't be considered a good email validator. But as it seems you narrow down matching to .com TLD so I assume this is a very specific requirement otherwise it limits domain name to alphanumerics and doesn't allow many more characters that would be valid in an email address according to RFC 822 . This would be enough for capturing an email address from user input:

^[^\s@]+@[^\s@]+$

Answer 2

This works:

^([0-9a-zA-Z][a-zA-Z0-9_\.]*)(?<!\.)@([a-zA-Z0-9_]+)\.com$

Demo

Basically, it tries to match alphanumeric characters at the start, then [a-zA-Z0-9_\\.] for 0 or more times. Before it reaches @ , it will look behind to check if there is a dot (if it is not an alphanumeric, it's gotta be a dot).

Answer 3

试试这个正则表达式-（^ [\\ w] [\\ w \\。\\ w] + [\\ w]）@（[[ww ++）\\。com