Hello All,

I am very new to using Regular expression. 

I wanted to create a regular expression for the following rule for validating the password.

1. Passwords must be atleast 8 non blank characters in length 

2. contain  at least 1 capital letters,1 lower case letters, and 1 numeric characters and at least one of the following special characters # ! $ % - _ = + < >

 It will be really very nice of you if you could help me with this...

^(?=.*?\d)(?=.*?[a-z])(?=.*?[A-Z])(?=.*?[-#!$%_=+<>]).{8,}$

prometheuzz, due to a Microsoft client-side regex engine lookahead bug, that pattern will fail to properly match in JavaScript code etc.  I've found that the workaround seems to be to add the length check to the front as a lookahead instead:

^(?=.{8,})(?=.*?\d)(?=.*?[a-z])(?=.*?[A-Z])(?=.*?[-#!$%_=+<>]).*$

Note that I added the .*$ at the end to simply return the match if necessary, for just string test purposes than can be left off, also the ?'s and the comma after the 8 in the pattern are also optional.

The bug is discussed here:

http://blog.stevenlevithan.com/archives/regex-lookahead-bug

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Thanks ddrudik, didn't know that!

No worries, Microsoft should have fixed it by now but has failed to for some reason.

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Thankx a lot to both of you for helping me out !!!

I have a similar password strength problem: My client would like to be able to require 2 of 3 conditions. That is,

* one or more alpha character

* one or more numeric characters

* one or more special characters (!@#$%^&*.?-_+=~`;:)

If any 2 of those are true, then the password is sufficiently complex.

And yes, I'm a regex noob... my eyes still glaze over at expressions > 5 characters. But I'm blown away at how powerful they are.

Thanks in advance,

== Ross-a-roni ==

Ross, it's always best to start a new thread rather than reuse someone else's thread, but here's that as I understand your requirements:

(?:(?=.*([a-zA-Z]))(?=.*(\d))|(?=.*(?1))(?=.*([!@#$%^&*.?_+=~`;:-]))|(?=.*(?2))(?=.*(?3))).*

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Wow. My eyes can hardly take it all in. :-)

Thank you for whipping that up. I'll give it a go and report back.

== Ross ==