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Put terminal symbol(`$`) inside character group(`[]`)? [duplicate]

Is it possible to use anchors inside a character class? This doesn't work:
analyze-string('abcd', '[\s^]abcd[\s$]')
It looks like ^ and $ are treated as literal when inside a character class; however, escaping them (\^, \$) doesn't work either.
I'm trying to use this expression to create word boundaries (\b is not available in XSLT/XQuery), but I would prefer not to use groups ((^|\s)) -- since non-capturing groups aren't available, that means in some scenarios I may end up with a large amount of unneeded capture groups, and that creates a new task of finding the "real" capture groups in the set of unneeded ones.

I believe the answer is no, you can't include ^ and $ as anchors in a [], only as literal characters. (I've wished you could do that before too.)
However, you could concat a space on the front and back of the string, then just look for \s as word boundaries and never mind the anchors. E.g.
analyze-string(concat(' ', 'abcd xyz abcd', ' '), '\sabcd\s')
You may also want + after each \s, but that's a separate issue.

If you're using analyze-string as a function, then presumably you're using a 3.0 implementation of either XSLT or XQuery.
In that case, why do you say "non-capturing groups aren't available"? The XPath Functions and Operators 3.0 spec is explicit that "Non-capturing groups are also recognized. These are indicated by the syntax (?:xxxx)."

Using the caret after the first square bracket will negate the character class. It essentially gives you the opposite of what you're looking to do, meaning the character class will match any character that is not in the character class. Negated character classes also match (invisible) line break characters.
You could try doing a negative look-ahead possibly.
(?!\s)

Parsing units with javascript regex

Say I have a string which contains some units (which may or may not have prefixes) that I want to break into the individual units. For example the string may contain "Btu(th)" or "Btu(th).ft" or even "mBtu(th).ft" where mBtu(th) is the bastardised unit milli thermochemical BTU's (this is purely an example).
I currently have the following (simplified) regex however it fails for the case "mBtu(th).ft":
/(m|k)??(Btu\(th\)|ft|m)(?:\b|\s|$)/g
Currently this does not correctly detect the boundary between the end of 'Btu(th)' and the start of 'ft'. I understand javascript regex does not support look back so how do I accurately parse the string?
Additional notes
The regex presented above is greatly simplified around the prefixes and units groups. The prefixes could span multiple characters like 'Ki' and therefore character sets are not suitable.
The desire is for each group to catch the prefix match as group 1 and the unit as match two i.e for 'mBtu(th).ft' match one would be ['m','Btu(th)'] and match two would be ['','ft'].
The prefix match needs to be lazy so that the string 'm' would be matched as the unit metres rather than the prefix milli. Likewise the match for 'mm' would need to be the prefix milli and the unit metres.

I would try with:
/((m)|(k)|(Btu(\(th\))?)|(ft)|(m)|(?:\.))+/g
at least with example above, it matches all units merged into one string.
DEMO
EDIT
Another try (DEMO):
/(?:(m)|(k)|(Btu)|(th)|(ft)|[\.\(\)])/g
this one again match only one part, but if you use $1,$2,$3,$4, etc, (DEMO) you can extract other fragments. It ignores ., (, ), characters. The problem is to count proper matched groups, but it works to some degree.
Or if you accept multiple separate matches I think simple alternative is:
/(m|k|Btu|th|ft)/g

A word boundary will not separate two non-word characters. So, you don't actually want a word boundary since the parentheses and period are not valid word characters. Instead, you want the string to not be followed by a word character, so you can use this instead:
[mk]??(Btu\(th\)|ft|m)(?!\w)
Demo

I believe you're after something like this. If I understood you correctly that want to match any kind of element, possibly preceded by the m or k character and separated by parantheses or dots.
/[\s\.\(]*(m|k?)(\w+)[\s\.\)]*/g
https://regex101.com/r/eQ5nR4/2
If you don't care about being able to match the parentheses but just return the elements you can just do
/(m|k?)(\w+)/g
https://regex101.com/r/oC1eP5/1

Negate random regular expression

Is there a way to negate any regular expression? I'm using regular expressions to validate input on a form. I'm now trying to create a button that sanitizes my input. Is there a way so I can use the regular expression used for the validating also for stripping the invalid characters?
I'm using this regex for validation of illegal characters
<input data-val-regex-pattern="[^|<>:\?'\*\[\]\=%\$\+,;~&\{\}]*" type="text" />
When clicking on a button next to it, I'm calling this function:
$('#button').click(function () {
var inputElement = $(this).prev();
var regex = new RegExp(inputElement.attr('data-val-regex-pattern'), 'g');
var value = inputElement.val();
inputElement.val(value.replace(regex, ''));
});
At the moment the javascript is doing the exact opposite of what I'm trying to accomplish. I need to find a way to 'reverse' the regex.
Edit: I'm trying to reverse the regex in the javascript function. The regex in the data-val-regex-pattern-attribute is doing his job for validation.

To find the invalid characters, just take the ^ off from your regex. The carret is the negative of everything that is inside the brackets.
data-val-regex-pattern="[|<>:\?'\*\[\]\=%\$\+,;~&\{\}]*"
This will return the undesired characters so you can replace them.
Also, as you want to take off a lot of non-word characters, you could try a simpler regex. If you want only word characters and spaces, you could use something like this:
data-val-regex-pattern="[\W\S]*"

Your reges is as so:
[^|<>:\?'\*\[\]\=%\$\+,;~&\{\}]*
That means, it matches any non-invalid character multiple times.
Then you replace this for empty, so you leave only the bad characters.
Try this instead, without the negation (hat moved somewhere else):
[|^<>:\?'\*\[\]\=%\$\+,;~&\{\}]*

The following answer is to the general question of negating a regular expression. In your specific case you just need to negate a character group, or more precisely remove the negation of a character group - which is detailed in other answers.
Regular languages – those consisting of all strings entirely by matched some RE – are in fact closed under negation: there is another RE which matches exactly those strings the original RE does not. It is however not trivial to construct, which perhaps explains why RE implementations often do not offer a negation operator.
However the Javascript regexp language has extensions that make it more expressive than regular languages; in particular there is the construct of negative lookahead.
If R1 is a regexp then
^(?!.*(R1))
matches precisely the strings that does not contain a match for R1.
And
^(?!R1$)
matches precisely the strings where the whole string is not a match for R1.
Ie. negation.
For rewriting any substring not matching a given regexp, the above is insufficient. One would have to do something like
((?!R1).)*
Which would catch any substring not containing a subsubstring that matches R1. - But consideration of the edge cases show that this does not quite do what we are after. For example ((?!ab).)* matches "b" in "ab", because "ab" is not a substring of "b".
One can cheat, and make your regexp like;
(.*)(R1|$)
And rewrite to T1$2
Where T1 is the target string you want to rewrite to.
This should rewrite any portion of the string not matching R1 to T1. However I would be very careful about any edge cases for this. So much so that it might be better to write the regexp from scratch rather than trying a general approach.

Javascript regular expression (unbroken repetitions of a pattern)

Let's say that I have a given string in javascript - e.g., var s = "{{1}}SomeText{{2}}SomeText"; It may be very long (e.g., 25,000+ chars).
NOTE: I'm using "SomeText" here as a placeholder to refer to any number of characters of plain text. In other words, "SomeText" could be any plain text string which doesn't include {{1}} or {{2}}. So the above example could be var s = "{{1}}Hi there. This is a string with one { curly bracket{{2}}Oh, very nice to meet you. I also have one } curly bracket!"; And that would be perfectly valid.
The rules for it are simple:
It does not need to have any instances of {{2}}. However, if it does, then after that instance we cannot encounter another {{2}} unless we find a {{1}} first.
Valid examples:
"{{2}}SomeText"
"{{1}}SomeText{{2}}SomeText"
"{{1}}SomeText{{1}}SomeText{{2}}SomeText"
"{{1}}SomeText{{1}}SomeText{{2}}SomeText{{1}}SomeText"
"{{1}}SomeText{{1}}SomeText{{2}}SomeText{{1}}SomeText{{1}}SomeText"
"{{1}}SomeText{{1}}SomeText{{2}}SomeText{{1}}SomeText{{1}}SomeText{{2}}SomeText"
etc...
Invalid examples:
"{{2}}SomeText{{2}}SomeText"
"{{1}}SomeText{{2}}SomeText{{2}}SomeText"
"{{1}}SomeText{{2}}SomeText{{2}}SomeText{{1}}SomeText"
etc...
This seems like a relatively easy problem to solve - and indeed I could easily solve it without regular expressions, but I'm keen to learn how to do something like this with regular expressions. Unfortunately, I'm not even sure if "conditionals and lookaheads" is a correct description of the issue in this case.
NOTE: If a workable solution is presented that doesn't involve "conditionals and lookaheads" then I will edit the title.

It's probably easier to invert the condition. Try to match any text that contains two consecutive instances of {{2}}, and if it doesn't match that, it's good.
Using this strategy, your pattern can be as simple as:
/{\{2}}([^{]*){\{2}}/
Demonstration
This will match a literal {{2}}, followed by zero or more characters other than {, followed by a literal {{2}}.
Notice that the second { needs to be escaped, otherwise, the regex engine will consider the {2} as to be a quantifier on the previous { (i.e. {{2} matches exactly two { characters).
Just in case you need to allow characters like {, and between the two {{2}}, you can use a pattern like this:
/{\{2}}((?!{\{1}}).)*{\{2}}/
Demonstration
This will match a literal {{2}}, followed by zero or more of any character, so long as those characters create a sequence like {{1}}, followed by a literal {{2}}.

(({{1}}SomeText)+({{2}}SomeText)?)*
Broken down:
({{1}}SomeText)+ - 1 to many {{1}} instances (greedy match)
({{2}}SomeText)? - followed by an optional {{2}} instance
Then the whole thing is wrapped in ()* such that the sequence can appear 0 to many times in a row.
No conditionals or lookaheads needed.

You said you can have one instance of {2} first, right?
^(.(?!{2}))(.{2})?(?!{2})((.(?!{2})){1}(.(?!{2}))({2})?)$
Note if {2} is one letter replace all dots with [^{2}]

RegEx string for three letters and two numbers with pre- and post- spaces

Two quick questions:
What would be a RegEx string for three letters and two numbers with space before and after them (i.e. " LET 12 ")?
Would you happen to know any good RegEx resources/tools?

For a good resource, try this website and the program RegexBuddy. You may even be able to figure out the answer to your question yourself using these sites.
To start you off you want something like this:
/^[a-zA-Z]{3}\s+[0-9]{2}$/
But the exact details depend on your requirements. It's probably a better idea that you learn how to use regular expressions yourself and then write the regular expression instead of just copying the answers here. The small details make a big difference. Examples:
What is a "letter"? Just A-Z or also foreign letters? What about lower case?
What is a "number"? Just 0-9 or also foreign numerals? Only integers? Only positive integers? Can there be leading zeros?
Should there be a single space between the letters and numbers? Or any amount of any whitespace? Even none?
Do you want to search for this string in a larger text? Or match a line exactly?
etc..
The answers to these questions will change the regular expression. It would be much faster for you in the long run to learn how to create the regular expression than to completely specify your requirements and wait for other people to reply.
I forgot to mention that there will be a space before and after. How do I include that?
Again you need to consider the questions:
Do you mean just one space or any amount of spaces? Possibly not always a space but only sometimes?
Do you mean literally a space character or any whitespace characters?
My guess is:
/^\s+[a-zA-Z]{3}\s+[0-9]{2}\s+$/

/[a-z]{3} [0-9]{2}/i will match 3 letters followed by a whitespace character, and then 2 numbers. [a-z] is a character class containing the letters a through z, and the {3} means that you want exactly 3 members of that class. The space character matches a literal space (alternately, you could use \s, which is a "shorthand" character class that matches any whitespace character). The i at the end is a pattern modifier specifying that your pattern is case-insenstive.
If you want the entire string to only be that, you need to anchor it with ^ and $:
/^[a-z]{3} [0-9]{2}$/i
Regular expression resources:
http://www.regular-expressions.info - great tutorial with a lot of information
http://rexv.org/ - online regular expression tester that supports a variety of engines.

^([A-Za-z]{3}) ([0-9]{2})$ assuming one space between the letters/numbers, as in your example. This will capture the letters and numbers separately.
I use http://gskinner.com/RegExr/ - it allows you to build a regex and test it with your own text.

As you can probably tell from the wide variety of answers, RegEx is a complex subject with a wide variety of opinions and preferences, and often more than one way of doing things. Here's my preferred solution.
^[a-zA-Z]{3}\s*\d{2}$
I used [a-zA-Z] instead of \w because \w sometimes includes underscores.
The \s* is to allow zero or more spaces.
I try to use character classes wherever possible, which is why I went with \d.

\w{3}\s{1}\d{2}
And I like this site.
EDIT:[a-zA-Z]{3}\s{1}\d{2} - The \w supports numeric characters too.

try this regularexpression
[^"\r\n]{3,}