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Regex to validate words do not contain numbers or special characters

I am developing a java app, running on android. I am trying to pick all words which do not contain any embedded digits or symbols.
The best I have come up with is:
\b[a-zA-Z]+[a-zA-Z]*+\b
Test Data:
this is a test , an0ther gr8 WW##ee one, w1n 1test test1 end
This results in picking the following: this, is, a, test, WW##ee, one, end
I need to eliminate the WW##ee from the results.

You shouldn't use a word boundary meta-character \b since it matches the position right after WW which sees a hash # character. This position is a word boundary itself. So you should pick up a different way:
(?<![\S&&[^,]])[a-zA-Z]+(?![\S&&[^,]])
Using character class intersection feature of Java's regex you are able to define punctuation characters that are allowed to follow or precede a word character. Here it is a comma ,.

You could use look behind and look ahead to check there is no #.
\b(?<!\#)[a-zA-Z]+(?!\#)\b

My solution has evolved a bit as I have gotten additional help with this. So, this is now my best solution but still a bit lacking. I have not been able to accept "as-is" while rejecting "-this-" and a similar case of accept "and/or" while rejecting "/slash/". Also for simplicity I have made the input data single word per line.
^(?:[\p{P}\p{S}])?((?:[\p{L}\p{Pd}'])+)(?:[\p{P}\p{S}])$
as-is is picked valid
-this- is valid but I wish it weren't
and/or is not valid but I wish it would be picked
/slash/ "slash" is picked valid
(test) "test" is picked valid
[test] "test" is picked valid
<test> "test" is picked valid

password Regexp doesn't work when split up

I cannot get a regexp that checks if password has at least one digit to work.
This has been answered everywhere but all the answers stop working if split up.
For example in this Working Password Validation if I remove:
(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])
from
^(?=.*[0-9])(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])(?=\S+$).{8,}$
in order to check for the presence of a single digit, the whole thing stops working
I'm new with regular expressions, this seems to make sense but it doesn't, show me the light if you can.

I'm not really sure what you mean by the whole things stops working.
(?=.*[a-z])(?=.*[A-Z])(?=.*[##$%^&+=])
All the above does is mandate that:
A lower case letter must appear at least once
An upper-case letter must appear at least once
Any of ##$%^&+= must appear at least once
So, there is no reason that taking them out should break anything--they are essentially independent components.
There are myriad ways to check if a String contains a number. How you want to check really depends on your specific requirements. The method used in the presented regex does this through a positive-look ahead: ^(?=.*[0-9])
^ : begins with
.* : matches 0 or more non-newline characters
[0-9]: is a character class that matches the numbers [0,1,2,...,9]
?= is the positive look ahead, which in this case says to match iff there exists at least one number
Hope that helped. You can start off with Oracle's Tutorial on Regular Expressions. After you digest that, I'm sure you'll be able to find more advanced resources via Google.

Regex: How not to match a few letters

I have the following string: SEE ATTACHED ADDENDUM TO HUD-1194,520.07
Inside that string is HUD-1 and after that is 194,520.07. What I want is the 194,520.07 part.
I have written the following regular expression to pull that value out:
[^D\-1](?:-|\()?\$?(?:\d{1,3}[ ,]?)*(?:\.\d+)\)?
However, this pulls out: 94,520.07
I know it has something to do with this part: [^D\-1] "eating" to many of the 1's. Any ideas how I can stop it from "eating" 1's after the first one that appears in HUD-1?
UPDATED:
The reason for all the other stuff is I only want to match as well if the value after HUD-1 is a money amount. And the rest of that regex tries to determine all the different ways a money amount could be written

Why not something as simple as:
.*HUD\-1(.*+)
Ok, you need to be more restrictive I see based on your updated question. Try changing [^D\-1] to just (?:HUD\-1)?. For what it's worth, your currency RegEx is vary lax, allowing input like:
001 001 .31412341234123
You might consider not reinventing the wheel there, I'm sure you can find a currency RegEx quickly via Google. Otherwise, I'd also suggest anchoring your RegEx with a $ at the end of it.

this change will make the second match group of the regex include the full number you would like (everything after the first 1), and put the possible HUD-1 in a separate matching group, if present.
(HUD-1)?((?:-|\()?\$?(?:\d{1,3}[ ,]?)*(?:\.\d+)\)?)

Regular expression, excluding .. in suffix of email addy [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Using a regular expression to validate an email address
This is homework, I've been working on it for a while, I've done lots of reading and feel I have gotten pretty familiar with regex for a beginner.
I am trying to find a regular expression for validating/invalidating a list of emails. There are two addresses which are giving me problems, I can't get them both to validate the correct way at the same time. I've gone through a dozen different expressions that work for all the other emails on the list but I can't get those two at the same time.
First, the addresses.
me#example..com - invalid
someone.nothere#1.0.0.127 - valid
The part of my expression which validates the suffix
I originally started with
#.+\\.[[a-z]0-9]+
And had a second pattern for checking some more invalid addresses and checked the email against both patterns, one checked for validity the other invalidity but my professor said he wanted it all in on expression.
#[[\\w]+\\.[\\w]+]+
or
#[\\w]+\\.[\\w]+
I've tried it written many, many different ways but I'm pretty sure I was just using different syntax to express these two expressions.
I know what I want it to do, I want it to match a character class of "character+"."character+"+
The plus sign being at least one. It works for the invalid class when I only allow the character class to repeat one time(and obviously the ip doesn't get matched), but when I allow the character class to repeat itself it matches the second period even thought it isn't preceded by a character. I don't understand why.
I've even tried grouping everything with () and putting {1} after the escaped . and changing the \w to a-z and replacing + with {1,}; nothing seems to require the period to surrounded by characters.

You need a negative look-ahead :
#\w+\.(?!\.)
See http://www.regular-expressions.info/lookaround.html
test in Perl :
Perl> $_ = 'someone.nothere#1.0.0.127'
someone.nothere#1.0.0.127
Perl> print "OK\n" if /\#\w+\.(?!\.)/
OK
1
Perl> $_ = 'me#example..com'
me#example..com
Perl> print "OK\n" if /\#\w+\.(?!\.)/
Perl>

#([\\w]+\\.)+[\\w]+
Matches at least one word character, followed by a '.'. This is repeated at least once, and is then followed by at least on more word character.

I think you want this:
#[\\w]+(\\.[\\w]+)+
This matches a "word" followed by one or more "." "word" sequences. (You can also do the grouping the other way around; e.g. see Dailin's answer.)
The problem with what you are doing before was that you were trying to embed a repeat inside a character class. That doesn't make sense, and there is no syntax that would support it. A character class defines a set of characters and matches against one character. Nothing more.

The official standard RFC 2822 describes the syntax that valid email addresses with this regular expression:
(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|"(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21\x23-\x5b\x5d-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])*")#(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\[(?:(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.){3}(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?|[a-z0-9-]*[a-z0-9]:(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21-\x5a\x53-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])+)\])
More practical implementation of RFC 2822 (if we omit the syntax using double quotes and square brackets), which will still match 99.99% of all email addresses in actual use today, is:
[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*#(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?

Regex for checking > 1 upper, lower, digit, and special char

^.(?=.{15,})(?=.\d)(?=.[a-z])(?=.[A-Z])(?=.[!##$%^&+=]).*$
This is the regex I am currently using which will evaluate on 1 of each: upper,lower,digit, and specials of my choosing. The question I have is how do I make it check for 2 of each of these? Also I ask because it is seemingly difficult to write a test case for this as I do not know if it is only evaluating the first set of criteria that it needs. This is for a password, however the requirement is that it needs to be in regex form based on the package we are utilizing.
EDIT
Well as it stands in my haste to validate the expression I forgot to validate my string length. Thanks to Ken and Gumbo on helping me with this.
This is the code I am executing:
I do apologize as regex is not my area.
The password I am using is the following string "$$QiouWER1245", the behavior I am experiencing at the moment is that it randomly chooses to pass or fail. Any thoughts on this?
Pattern pattern = Pattern.compile(regEx);
Matcher match = pattern.matcher(password);
while(match.find()){
System.out.println(match.group());
}
From what I see if it evaluates to true it will throw the value in password back to me else it is an empty string.

Personally, I think a password policy that forces use of all three character classes is not very helpful. You can get the same degree of randomness by letting people make longer passwords. Users will tend to get frustrated and write passwords down if they have to abide by too many password rules (which make the passwords too difficult to remember). I recommend counting bits of entropy and making sure they're greater than 60 (usually requires a 10-14 character password). Entropy per character would depend roughly on the number of characters, the range of character sets they use, and maybe how often they switch between character sets (I would guess that passwords like HEYthere are more predictable than heYThEre).
Another note: do you plan not to count the symbols to the right of the keyboard (period, comma, angle brackets, etc.)?
If you still have to find groups of two characters, why not just repeat each pattern? For example, make (?=.\d) into (?=.\d.*\d).
For your test cases, if you are worried that it would only check the first criteria, then write a test case that makes sure each of the following passwords fails (because one and only one of the criteria is not met in each case): Just for fun I reversed the order of expectation of each character set, though it probably won't make a difference unless someone removes/forgets the ?= at some future date.
!##TESTwithoutnumbers
TESTwithoutsymbols123
&*(testwithoutuppercase456
+_^TESTWITHOUTLOWERCASE3498
I should point out that technically none of these passwords should be acceptable because they use dictionary words, which have about 2 bits of entropy per character instead of something more like 6. However, I realize that it's difficult to write a (maintainable and efficient) regular expression to check for dictionary words.

Try this:
"^(?=(?:\\D*\\d){2})(?=(?:[^a-z]*[a-z]){2})(?=(?:[^A-Z]*[A-Z]){2})(?=(?:[^!##$%^&*+=]*[!##$%^&*+=]){2}).{15,}$"
Here non-capturing groups (?:…) are used to group the conditions and repeat them. I’ve also used the complements of each character class for optimization instead of the universal ..

If I understand your question correctly, you want at least 15 characters, and to require at least 2 uppercase characters, at least 2 lowercase characters, at least 2 digits, and at least 2 special characters. In that case you could it like this:
^.*(?=.{15,})(?=.*\d.*\d)(?=.*[a-z].*[a-z])(?=.*[A-Z].*[A-Z])(?=.*[!##$%^&*+=].*[!##$%^&*+=]).*$
BTW, your original regex had an extra backslash before the \d

I'm not sure that one big regex is the right way to go here. It already looks far too complicated and will be very difficult to change in the future.
My suggestion is to structure the code in the following way:
check that the string has 2 lower case characters
return failure if not found or continue
check that the string has 2 upper case characters
return failure if not found or continue
etc.
This will also allow you to pass out a return code or errors string specifying why the password was not accepted and the code will be much simpler.