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JS regex to match two possible combinations

I need to capture a certain combination of letters followed by a number (any amount), represented in a variable called input. The letters are strict, the numbers are not. The letters are either at the beginning of a string or followed immediately after a backslash.
So for example, I would need to non-case-sensitively capture:
ab12345678google
cd4321newyorkpost
anything\here\ab1357
something\too\cd2468
For these, I have a simple rule that works (well, two rules):
input.value.match(/^(ab|cd)[0-9]+/i) || input.value.match(/\\(ab|cd)[0-9]+/i)
However, it is also possible to a string called test to exist right before the set letters which I would also need to capture (either from the beginning or after a backslash again). So besides capturing just the given two letters, I would also need to capture these occurrences as well where the test before the letters is the strict factor, e.g.:
testcd4321newyorkpost
anything\here\testab1357
I'm quite sure it's possible to place an "optional" lookup of some sort in the match query without rewriting the rules for test separately, but as new as I am with regex I'm not sure what would I be looking here?

You may use this regex:
(?:^|\\)(?:test)?(?:ab|cd)\d+
Which is:
Match start or \
Match optional string test
Match ab or cd
Match 1+ digits

Why not just make the text test optional?
(?:test)?(ab|cd)[0-9]+
should work for any of your situations.

javascript regex : possible to have a range in the quantifier? [duplicate]

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}

Regex - must contain number and must not contain special character

I want to check by regex if:
String contains number
String does not contain special characters (!<>?=+#{}_$%)
Now it looks like:
^[^!<>?=+#{}_$%]+$
How should I edit this regex to check if there is number anywhere in the string (it must contain it)?

you can add [0-9]+ or \d+ into your regex, like this:
^[^!<>?=+#{}_$%]*[0-9]+[^!<>?=+#{}_$%]*$
or
^[^!<>?=+#{}_$%]*\d+[^!<>?=+#{}_$%]*$
different between [0-9] and \d see here

Just look ahead for the digit:
var re = /^(?=.*\d)[^!<>?=+#{}_$%]+$/;
console.log(re.test('bob'));
console.log(re.test('bob1'));
console.log(re.test('bob#'))
The (?=.*\d) part is the lookahead for a single digit somewhere in the input.

You only needed to add the number check, is that right? You can do it like so:
/^(?=.*\d)[^!<>?=+#{}_$%]+$/
We do a lookahead (like peeking at the following characters without moving where we are in the string) to check to see if there is at least one number anywhere in the string. Then we do our normal check to see if none of the characters are those symbols, moving through the string as we go.
Just as a note: If you want to match newlines (a.k.a. line breaks), then you can change the dot . into [\W\w]. This matches any character whatsoever. You can do this in a number of ways, but they're all pretty much as clunky as each other, so it's up to you.

JavaScript regular expressions to match no digits, whitespace and selected symbols

Thanks for taking a look.
My goal is to come up with a regexp that will match input that contains no digits, whitespace or the symbols !#£$%^&*()+= or any other symbol I may choose.
I am however struggling to grasp precisely how regular expressions work.
I started out with the simple pattern /\D/, which from my understanding will match the first non-digit character it can find. This would match the string 'James' which is correct but also 'James1' which I don't want.
So, my understanding is that if I want to ensure that a pattern is not found anywhere in a given string, I use the ^ and $ characters, as in /^\D$/. Now because this will only match a single character that is not a digit, I needed to use + to specify that 1 or more digits should not be founds in the entire string, giving me the expression /^\D+$/. Brilliant, it no longer matches 'James1'.
Question 1
Is my reasoning up to this point correct?
The next requirement was to ensure no whitespace is in the given string. \s will match a single whitespace and [^\s] will match the first non-whitespace character. So, from my understanding I just had to add this to what I have already to match strings that contain no digits and no whitespace. Again, because [^\s] will only match a single non-white space character, I used + to match one or more whitespace characters, giving the new regexp of /^\D+[^\s]+$/.
This is where I got lost, as the expression now matches 'James1' or even 'James Smith25'. What? Massively confused at this point.
Question 2
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Question 3
How would I go about writing the regular expression I'm trying to solve?
While I am keen to solve the problem I am more interested in figuring where my understanding of regular expressions is lacking, so any explanations would be helpful.

Not quite; ^ and $ are actually "anchors" - they mean "start" and "end", it's actually a little more complicated, but you can consider them to mean the start and end of a line for now - look up the various modifiers on regular expressions if you're interested in learning more about this. Unfortunately ^ has an overloaded meaning; if used inside square brackets it means "not", which is the meaning you are already acquainted with. It's very important that you understand the difference between these two meanings and that the definition in your head actually applies only to character range matching!
Contributing further to your confusion is that \d means "a numerical digit" and \D means "not a numerical digit". Similarly \s means "a whitespace (space/tab/newline/etc.) character" and \S means "not a whitespace character."
It's worth noting that \d is effectively a shortcut for [0-9] (note that - has a special meaning inside square brackets), and \D is a shortcut for [^0-9].
The reason it's matching strings that contain spaces is that you've asked for "1+ non-numerical digits followed by 1+ non-space characters" - so it'll match lots of strings! I think that perhaps you don't understand that regular expressions match bits of strings, you're not adding constraints as you go, but rather building up bots of matchers that will match bits of corresponding strings.
/^[^\d\s!#£$%^&*()+=]+$/ is the answer you're looking for - I'd look at it like this:
i. [] - match a range of characters
ii. []+ - match one or more of that range of characters
iii. [^\d\s]+ - match one or more characters that do not match \d (numerical digit) or \s (whitespace)
iv. [^\d\s!#£$%^&*()+=]+ - here's a bunch of other characters I don't want you to match
v. ^[^\d\s!#£$%^&*()+=]+$ - now there are anchors applied, so this matcher has to apply to the whole line otherwise it fails to match
A useful website to explore regexs is http://regexr.com/3b9h7 - which I supply with my suggested solution as an example. Edit: Pruthvi Raj's link to debuggerx is awesome!

Is my reasoning up to this point correct?
Almost. /\D/ matches any character other than a digit, but not just the first one (if you use g option).
and [^\s] will match the first non-whitespace character
Almost, [^\s] will match any non-whitespace character, not just the first one (if you use g option).
/^\D+[^\s]+$/ matching strings that contain spaces?
Yes, it does, because \D matches a space (space is not a digit).
Why is /^\D+[^\s]+$/ matching strings that contain spaces?
Because \D+ in /^\D+[^\s]+$/can match spaces.
Conclusion:
Use
^[^\d\s!#£$%^&*()+=]+$
It will match strings that have no digits and spaces, and the symbols you do not allow.
Mind that to match a literal -, ] or [ with a character class, you either need to escape them, or use at the start or end of the expression. To play it safe, escape them.

Just insert every character you don't want to include in a negated character class as follows:
^[^\s\d!#£$%^&*()+=]*$
DEMO
Debuggex Demo
^ - start of the string
[^...] - matches one character that is not in `...`
\s - matches a whitespace (space, newline,tab)
\d - matches a digit from 0 to 9
* - a quantifier that repeats immediately preceeding element by 0 or more times
so the regex matches any string that has
1. string that has a beginning
2. containing 0 or more number of characters that is not whitesapce, digit, and all the symbols included in the character class ( In this example !#£$%^&*()+=) i.e., characters that are not included in the character class `[...]`
3.that has ending
NOTE:
If the symbols you don't want it to have also includes - , a hyphen, don't put it in between some other characters because it is a metacharacter in character class, put it at last of character class

UK Currency Regular Expression for javascript

I'm after a regular expression that matches a UK Currency (ie. £13.00, £9,999.99 and £12,333,333.02), but does not allow negative (-£2.17) or zero values (£0.00 or 0).
I've tried to create one myself, but I've got in a right muddle!
Any help greatfully received.
Thanks!

This'll do it (well mostly...)
/^£?[1-9]{1,3}(,\d{3})*(\.\d{2})?$/
Leverages the ^ and $ to make sure no negative or other character is in the string, and assumes that commas will be used. The pound symbol, and pence are optional.
edit: realised you said non-zero so replaced the first \d with [1-9]
Update: it's been pointed out the above won't match £0.01. The below improvement will but now there's a level of complexity where it may quite possibly be better to test /[1-9]/ first and then the above - haven't benchmarked it.
/^£?(([1-9]{1,3}(,\d{3})*(\.\d{2})?)|(0\.[1-9]\d)|(0\.0[1-9]))$/
Brief explanation:
Match beginning of string followed by optional "£"
Then match either:
a >£1 amount with potential for comma separated groupings and optional pence
OR a <£1 >=£0.10 amount
OR a <=£0.09 amount
Then match end of line
The more fractions of pence (zero in the above) you require adding to the regex the less efficient it becomes.

Under Unix/Linux, it's not always possible to type in the '£' sign in a JavaScript file, so I tend to use its hexadecimal representation, thus:
/^\xA3?\d{1,3}?([,]\d{3}|\d)*?([.]\d{1,2})?$/
This seems to take care of all combinations of UK currency amounts representation that I have come across.

/^\xA3?\d{1,}(?:\,?\d+)*(?:.\d{1,2})?$/;
Explanation:
^ Matches the beginning of the string, or the beginning of a line.
xA3 Matches a "£" character (char code 163)
? Quantifier for match between 0 and 1 of the preceding token.
\d Matches any digit character (0-9).
{1,} Match 1 or more of the preceding token.
(?: Groups multiple tokens together without creating a capture group.
\, Matches a "," character (char code 44).
{1,2} Match between 1 and 2 of the preceding token.
$ Matches the end of the string, or the end of a line if the multiline flag (

You could just make two passes:
/^£\d{1,3}(,\d{3})*(\.\d{2})?$/
to validate the format, and
/[1-9]/
to ensure that at least one digit is non-zero.
This is less efficient than doing it in one pass, of course (thanks, annakata, for the benchmark information), but for a first implementation, just "saying what you want" can significantly reduce developing time.