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Regex format for a particular Match

I am trying to write a regex for the following format
PA-123456-067_TY
It's always PA, followed by a dash, 6 digits, another dash, then 3 digits, and ends with _TY
Apparently, when I write this regex to match the above format it shows the output correctly
^[^[PA]-]+-(([^-]+)-([^_]+))_([^.]+)
with all the Negation symbols ^
This does not work if I write the regex in the below format without negation symbols
[[PA]-]+-(([-]+)-([_]+))_([.]+)
Can someone explain to me why is this so?

The negation symbol means that the character cannot be anything within the specified class. Your regex is much more complicated than it needs to be and is therefore obfuscating what you really want.
You probably want something like this:
^PA-(\d+)-(\d+)_TY$
... which matches anything that starts with PA-, then includes two groups of numbers separated by a dash, then an underscore and the letters TY. If you want everything after the PA to be what you capture, but separated into the three groups, then it's a little more abstract:
^PA-(.+)-(.+)_(.+)$
This matches:
PA-
a capture group of any characters
a dash
another capture group of any characters
an underscore
all the remaining characters until end-of-line
Character classes [...] are saying match any single character in the list, so your first capture group (([^-]+)-([^_]+)) is looking for anything that isn't a dash any number of times followed by a dash (which is fine) followed by anything that isn't an underscore (again fine). Having the extra set of parentheses around that creates another capture group (probably group 1 as it's the first parentheses reached by the regex engine)... that part is OK but probably makes interpreting the answer less intuitive in this case.
In the re-write however, your first capture group (([-]+)-([_]+)) matches [-]+, which means "one or more dashes" followed by a dash, followed by any number of underscores followed by an underscore. Since your input does not have a dash immediately following PA-, the entire regex fails to find anything.
Putting the PA inside embedded character classes is also making things complicated. The first part of your first one is looking for, well, I'm not actually sure how [^[PA]-]+ is interpreted in practice but I suspect it's something like "not either a P or an A or a dash any number of times". The second one is looking for the opposite, I think. But you don't want any of that, you just want to start without anything other than the actual sequence of characters you care about, which is just PA-.
Update: As per the clarifications in the comments on the original question, knowing you want fixed-size groups of digits, it would look like this:
^PA-(\d{6})-(\d{3})_TY$
That captures PA-, then a 6-digit number, then a dash, then a 3-digit number, then _TY. The six digit number and 3 digit numbers will be in capture groups 1 and 2, respectively.
If the sizes of those numbers could ever change, then replace {x} with + to just capture numbers regardless of max length.

according to your comment this would be appropriate PA-\d{6}-\d{3}_TY
EDIT: if you want to match a line use it with anchors: ^PA-\d{6}-\d{3}_TY$

Regex for multiple instances of character

In Java, using a regular expression, how would I check a string to see if it had a correct amount of instances of a character.
For example take the string hello.world.hello:world:. How could this string be checked to see if it contained two instances of a . or two instances of a :?
I have tried
Pattern p = Pattern.compile("[:]{2}");
Matcher m = p.matcher(hello.world.hello:world:);
m.find();
but that failed.
Edit
First I would like to say thank you for all the answers. I noticed a lot of the answers said something along the lines of "This means: zero or more non-colons, followed by a single colon, followed by zero or more non-colons - matched exactly twice". So if you were checking for 3 : in a string such as Hello::World: how would you do it?

Well, using matches you could use:
"([^:]*:[^:]*){2}"
This means: "zero or more non-colons, followed by a single colon, followed by zero or more non-colons - matched exactly twice".
Using find is not as good, as there may be additional : and it will just ignore them.

You can use this regex based on two lookaheads assertions:
^(?=(?:[^.]*\.){2}[^.]*$)(?=(?:[^:]*:){2}[^:]*$)
(?=(?:[^.]*\.){2}[^.]*$) makes sure there are exactly 2 DOTS and (?=(?:[^:]*:){2}[^:]*$) asserts that there are exactly 2 colons in input string.
RegEx Demo

You can determine whether the string has exectly the given number of a certain character, say ':', by attempting to match it against a pattern of this form:
^(?:[^:]*[:]){2}[^:]*$
That says exactly two non-capturing groups consisting of any number (including zero) of characters other than ':' followed by one colon, with the second group followed by any number of additional characters other than ':'.

Set minimum and maximum characters in a regular expression

I've written a regular expression that matches any number of letters with any number of single spaces between the letters. I would like that regular expression to also enforce a minimum and maximum number of characters, but I'm not sure how to do that (or if it's possible).
My regular expression is:
[A-Za-z](\s?[A-Za-z])+
I realized it was only matching two sets of letters surrounding a single space, so I modified it slightly to fix that. The original question is still the same though.
Is there a way to enforce a minimum of three characters and a maximum of 30?

Yes
Just like + means one or more you can use {3,30} to match between 3 and 30
For example [a-z]{3,30} matches between 3 and 30 lowercase alphabet letters
From the documentation of the Pattern class
X{n,m} X, at least n but not more than m times
In your case, matching 3-30 letters followed by spaces could be accomplished with:
([a-zA-Z]\s){3,30}
If you require trailing whitespace, if you don't you can use: (2-29 times letter+space, then letter)
([a-zA-Z]\s){2,29}[a-zA-Z]
If you'd like whitespaces to count as characters you need to divide that number by 2 to get
([a-zA-Z]\s){1,14}[a-zA-Z]
You can add \s? to that last one if the trailing whitespace is optional. These were all tested on RegexPlanet
If you'd like the entire string altogether to be between 3 and 30 characters you can use lookaheads adding (?=^.{3,30}$) at the beginning of the RegExp and removing the other size limitations
All that said, in all honestly I'd probably just test the String's .length property. It's more readable.

This is what you are looking for
^[a-zA-Z](\s?[a-zA-Z]){2,29}$
^ is the start of string
$ is the end of string
(\s?[a-zA-Z]){2,29} would match (\s?[a-zA-Z]) 2 to 29 times..

Actually Benjamin's answer will lead to the complete solution to the OP's question.
Using lookaheads it is possible to restrict the total number of characters AND restrict the match to a set combination of letters and (optional) single spaces.
The regex that solves the entire problem would become
(?=^.{3,30}$)^([A-Za-z][\s]?)+$
This will match AAA, A A and also fail to match AA A since there are two consecutive spaces.
I tested this at http://regexpal.com/ and it does the trick.

You should use
[a-zA-Z ]{20}
[For allowed characters]{for limiting of the number of characters}

Regular Expression to match one or more digits 1-9, one '|', one or more '*" and zero or more ','

I'm new to regular expressions and I need to find a regular expression that matches one or more digits [1-9] only ONE '|' sign, one or more '*' sign and zero or more ',' sign.
The string should not contain any other characters.
This is what I have:
if(this.ruleString.matches("^[1-9|*,]*$"))
{
return true;
}
Is it correct?
Thanks,
Vinay

I think you should test separately for every type of symbols rather than write complex expression.
First, test that you don't have invalid symbols - "^[0-9|*,]$"
Then test for digits "[1-9]", it should match at least one.
Then test for "\\|", "\\*" and "\\," and check the number of matches.
If all test are passed then your string is valid.

Nope, try this:
"^[1-9]+\\|\\*+,*$"

Please give us at least 10 possible matching strings of what you are looking to accept, and 10 of what you want to reject, and tell us if either this have to keep some sequence or its order doesn't matter. So we can make a reliable regex.
By now, all I can offer is:
^[1-9]+\|{1}\*+,*$
This RegEx was tested against these sample strings, accepting them:
56421|*****,,,
2|*********,,,
1|*
7|*,
18|****
123456789|*
12|********,,
1516332|**,,,
111111|*
6|*****,,,,
And it was tested against these sample strings, rejecting them:
10|*,
2***525*|*****,,,
123456,15,22*66*****4|,,,*167
1|2*3,4,5,6*
,*|173,
|*,
||12211
12
1|,*
1233|54|***,,,,
I assume your given order is strict and all conditions apply at the same time.

It looks like the pattern you need is
n-n, one or more times seperated by commas
then a bar (|)
then n*n, one or more times seperated by commas.
Here is a regular expression for that.
([1-9]{1}[0-9]*\-[0-9]+){1}
(,[1-9]{1}[0-9]*\-[0-9]+)*
\|
([1-9]{1}[0-9]*\*[0-9]+){1}
(,[1-9]{1}[0-9]*\*[0-9]+)*
But it is so complex, and does not take into account the details, such as
for the case of n-m, you want
n less than m
(I guess).
And you likely want the same number of n-m before the bar, and x*y after the bar.
Depends whether you want to check the syntax completely or not.
(I hope you do want to.)
Since this is so complex, it should be done with a set of code instead of a single regular expression.

this regex should work
"^[1-9\\|\\*,-]*$"
Assert position at the beginning of the string «^»
Match a single character present in the list below «[1-9\|*,-]»
Between zero and unlimited times, as many times as possible, giving back as needed (greedy) «»
A character in the range between “1” and “9” «1-9»
A | character «\|»
A * character «*»
The character “,” «,»
The character “-” «-»
Assert position at the end of the string (or before the line break at the end of the string, if any) «$»

regex for specific digit prefix

I am trying to have the following regx rule, but couldn't find solution.
I am sorry if I didn't make it clear. I want for each rule different regx. I am using Java.
rule should fail for all digit inputs start with prefix '1900' or '1901'.
(190011 - fail, 190111 - fail, 41900 - success...)
rule should success for all digit inputs with the prefix '*'
different regex for each rule (I am not looking for the combination of both of them together)

Is this RE fitting the purpose ? :
'\A(\*|(?!190[01])).*'
\A means 'the beginning of string' . I think it's the same in Java's regexes
.
EDIT
\A : "from the very beginning of the string ....". In Python (which is what I know, in fact) this can be omitted if we use the function match() that always analyzes from the very beginning, instead of search() that search everywhere in a string. If you want the regex able to analyze lines from the very beginning of each line, this must be replaced by ^
(...|...) : ".... there must be one of the two following options : ....."
\* : "...the first option is one character only, a star; ..." . As a star is special character meaning 'zero, one or more times what is before' in regex's strings, it must be escaped to strictly mean 'a star' only.
(?!190[01]) : "... the second option isn't a pattern that must be found and possibly catched but a pattern that must be absent (still after the very beginning). ...". The two characters ?! are what says 'there must not be the following characters'. The pattern not to be found is 4 integer characters long, '1900' or '1901' .
(?!.......) is a negative lookahead assertion. All kinds of assertion begins with (? : the parenthese invalidates the habitual meaning of ? , that's why all assertions are always written with parentheses.
If \* have matched, one character have been consumed. On the contrary, if the assertion is verified, the corresponding 4 first characters of the string haven't been consumed: the regex motor has gone through the analysed string until the 4th character to verify them, and then it has come back to its initial position, that is to say, presently, at the very beginning of the string.
If you want the bi-optional part (...|...) not to be a capturing group, you will write ?: just after the first paren, then '\A(?:\*|(?!190[01])).*'
.* : After the beginning pattern (one star catched/matched, or an assertion verified) the regex motor goes and catch all the characters until the end of the line. If the string has newlines and you want the regex to catch all the characters until the end of the string, and not only of a line, you will specify that . must match the newlines too (in Python it is with re.MULTILINE), or you will replace .* with (.|\r|\n)*
I finally understand that you apparently want to catch strings composed of digits characters. If so the RE must be changed to '\A(?:\*|(?!190[01]))\d*' . This RE matches with empty strings. If you want no-match with empty strings, put \d+ in place of \d* . If you want that only strings with at least one digit, even after the star when it begins with a star, match, then do '\A(?:\*|(?!190[01]))(?=\d)\d*'

For the first rule, you should use a combo regex with two captures, one to capture the 1900/1901-prefixed case, and one the capture the rest. Then you can decide whether the string should succeed or fail by examining the two captures:
(190[01]\d+)|(\d+)
Or just a simple 190[01]\d+ and negate your logic.
Regex's are not really very good at excluding something.
You may exclude a prefix using negative look-behind, but it won't work in this case because the prefix is itself a stream of digits.
You seem to be trying to exclude 1-900/901 phone numbers in the US. If the number of digits is definite, you can use a negative look-behind to exclude this prefix while matching the remaining exact number digits.
For the second rule, simply:
\*\d+