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javascript regex insert new element into expression

I am passing a URL to a block of code in which I need to insert a new element into the regex. Pretty sure the regex is valid and the code seems right but no matter what I can't seem to execute the match for regex!
//** Incoming url's
//** url e.g. api/223344
//** api/11aa/page/2017
//** Need to match to the following
//** dir/api/12ab/page/1999
//** Hence the need to add dir at the front
var url = req.url;
//** pass in: /^\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var re = myregex.toString();
//** Insert dir into regex: /^dir\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var regVar = re.substr(0, 2) + 'dir' + re.substr(2);
var matchedData = url.match(regVar);
matchedData === null ? console.log('NO') : console.log('Yay');
I hope I am just missing the obvious but can anyone see why I can't match and always returns NO?
Thanks

Let's break down your regex
^\/api\/ this matches the beginning of a string, and it looks to match exactly the string "/api"
([a-zA-Z0-9-_~ %]+) this is a capturing group: this one specifically will capture anything inside those brackets, with the + indicating to capture 1 or more, so for example, this section will match abAB25-_ %
(?:\/page\/([a-zA-Z0-9-_~ %]+)) this groups multiple tokens together as well, but does not create a capturing group like above (the ?: makes it non-captuing). You are first matching a string exactly like "/page/" followed by a group exactly like mentioned in the paragraph above (that matches a-z, A-Z, 0-9, etc.
?$ is at the end, and the ? means capture 0 or more of the precending group, and the $ matches the end of the string
This regex will match this string, for example: /api/abAB25-_ %/page/abAB25-_ %
You may be able to take advantage of capturing groups, however, and use something like this instead to get similar results: ^\/api\/([a-zA-Z0-9-_~ %]+)\/page\/\1?$. Here, we are using \1 to reference that first capturing group and match exactly the same tokens it is matching. EDIT: actually, this probably won't work, since the text after /api/ and the text after /page/ will most likely be different, carrying on...
Afterwards, you are are adding "dir" to the beginning of your search, so you can now match someting like this: dir/api/abAB25-_ %/page/abAB25-_ %
You have also now converted the regex to a string, so like Crayon Violent pointed out in their comment, this will break your expected funtionality. You can fix this by using .source on your regex: var matchedData = url.match(regVar.source); https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/source
Now you can properly match a string like this: dir/api/11aa/page/2017 see this example: https://repl.it/Mj8h

As mentioned by Crayon Violent in the comments, it seems you're passing a String rather than a regular expression in the .match() function. maybe try the following:
url.match(new RegExp(regVar, "i"));
to convert the string to a regular expression. The "i" is for ignore case; don't know that's what you want. Learn more here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

Matching and returning a regex between two values

I am trying to get the values from a string using regex, the value is that of the text between tt=" and "&
So, for example, "tt="Value"&" I would only want to get the word "Value" out of this.
So far I have this: /tt=.*&/ which gives me "tt=Value"&, Then, to get the value I am thinking to split the match on = and remove the 2 characters from the end. I feel though, that this would be an awful way to do this and would like to see if it could be done in the regex?

You're on the right track for matching the entire context inside of the string, but you want to use a capturing group to match/capture the value between the quotes instead of splitting on = and having to remove the two quote chars.
var r = 'tt="Value"&'.match(/tt="([^"]*)"/)[1];
if (r)
console.log(r); //=> "Value"

I know this isn't really the answer you are looking for since it doesn't involve regex but it's the way I usually do it.
strvariable = strvariable.Remove(0,strvariable.IndexOf("=") + 2);
strvariable = strvariable.Remove(strvariable.IndexOf("\""), strvariable.Length - strvariable.IndexOf("\""));
this would give you the result you were looking for which is Value in this instance.

Remove everything after the first instance of one of several characters

Say I have a string like:
var str = "Good morningX Would you care for some tea?"
Where the X could be one of several characters, like a ., ?, or !.
How can I remove everything after that character?
If it could only be one type of character, I would use indexOf and substr, but it looks like I need a different method to find the position in this case. Perhaps a regular expression?
Clarification: I do not know what character X is. I'd like to cut the string off at the first occurrence of any one of the specified characters.
Ok, further clarification:
What I'm actually doing is scrubbing posts from a website. I'm taking the first bit from each post and stitching them together. By 'bit', I mean characters before the first piece of punctuation. I need to cut everything off after that punctuation. Does that make sense?

Just replace everything within the [ and ] with your delimiters. Escape if necessary.
var str = "Good morning! Would you care for some tea?";
var beginning = str.split(/[.?!]/)[0];
// "Good morning"

Try this, If the X have this ',' character , then try below
var s = 'Good morning, would you care for some tea?';
s = s.substring(0, s.indexOf(','));
document.write(s);
Demo : http://jsfiddle.net/L4hna/490/
and if the X have '!' , then try below
var s = 'Good morning! would you care for some tea?';
s = s.substring(0, s.indexOf('!'));
document.write(s);
Demo : http://jsfiddle.net/L4hna/491/
Try this way for your requirement string.
Both are will return Good Morning

The below code will do as you expect:
var s = "Good morningX Would you care for some tea?";
s = s.substring(X, n != -1 ? n : s.length);
document.write(s);

http://jsfiddle.net/JEFnY/
The regex would be
str.replace(/(.*?)([\.\?\!])(.*)/i, '$1$2');
The first capturing group is a lazy expression to match everything before the next capturing group.
The second capturing group only looks for the characters that you specify - which in this case are .!?, all escaped.
The last capturing group is discarded. Hence the substitution string is $1$2, or the first two capturing groups together.

what regular expression would I use to retrieve the digits that follow this string?

I'm using jQuery to retrieve the class attribute and I need to get the a substring of the digits that follow the substring "position"
for example
"position7 selected" I need to retrieve "7"
for example
"navitem position14 selected" I need to retrieve "14"
I started writing:
$(this).attr('class').match(/(\d+)$/))
But I am getting lost with the regular expression, any help much appreciated.
I actually really really like regular expressions but I'm still learning!
update due to first answer: there might be another group of digits which I need to ignore
For example "navitem2 position14 selected" I need to retrieve "14"

"navitem position14 selected".match(/position(\d+)/)[1]
The call returns ["position14", "14"], so the [1] element is 14.
You're on the right track with the (\d+), it will match a contiguous group of digits. This just says only match that group when it directly follows the literal string "position".

Your attempt has two problems. The first is you anchor it to the end of the string with $. So instead of giving you the digits after position it will give you digits at the endof the string. At the same time you don't mention position at all. What you are looking for is this:
var matches = str.match(/position(\d+)/);
Now matches will be
["position14", "14"]

You can replace everything that is not a number
str.replace(/\D/g,"");
If other numbers are present in the string you can still use replace
str.replace(/^.*position(\d+).*$/,"$1");
However, you may want to go for one of the other regex expressions in the other answers.

var r = "aaa 1 bb 2 position 113 dd".match(/position\s*(\d+)/)
if (r instanceof Array )
return r.pop()

Javascript string validation using the regex object

I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}

By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.

If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.