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I need help getting the first n characters of a string up to when a number character starts

I'm working with a string where I need to extract the first n characters up to where numbers begin. What would be the best way to do this as sometimes the string starts with a number: 7EUSA8889er898 I would need to extract 7EUSA But other string examples would be SWFX74849948, I would need to extract SWFX from that string.
Not sure how to do this with regex my limited knowledge is blocking me at this point:
^(\w{4}) that just gets me the first four characters but I don't really have a stopping point as sometimes the string could be somelongstring292894830982 which would require me to get somelongstring

Using \w will match a word character which includes characters and digits and an underscore.
You could match an optional digit [0-9]? from the start of the string ^and then match 1+ times A-Za-z
^[0-9]?[A-Za-z]+
Regex demo
const regex = /^[0-9]?[A-Za-z]+/;
[
"7EUSA8889er898",
"somelongstring292894830982",
"SWFX74849948"
].forEach(s => console.log(s.match(regex)[0]));

Can use this regex code:
(^\d+?[a-zA-Z]+)|(^\d+|[a-zA-Z]+)
I try with exmaple and good worked:
1- somelongstring292894830982 -> somelongstring
2- 7sdfsdf5456 -> 7sdfsdf
3- 875werwer54556 -> 875werwer

If you want to create function where the RegExp is parametrized by n parameter, this would be
function getStr(str,n) {
var pattern = "\\d?\\w{0,"+n+"}";
var reg = new RegExp(pattern);
var result = reg.exec(str);
if(result[0]) return result[0].substr(0,n);
}

There are answers to this but here is another way to do it.
var string1 = '7EUSA8889er898';
var string2 = 'SWFX74849948';
var Extract = function (args) {
var C = args.split(''); // Split string in array
var NI = []; // Store indexes of all numbers
// Loop through list -> if char is a number add its index
C.map(function (I) { return /^\d+$/.test(I) === true ? NI.push(C.indexOf(I)) : ''; });
// Get the items between the first and second occurence of a number
return C.slice(NI[0] === 0 ? NI[0] + 1 : 0, NI[1]).join('');
};
console.log(Extract(string1));
console.log(Extract(string2));
Output
EUSA
SWFX7

Since it's hard to tell what you are trying to match, I'd go with a general regex
^\d?\D+(?=\d)

Matching an exact word from a string

I need a way to match a word against a string and not get false positives. Let me give an example of what I mean:
"/thing" should match the string "/a/thing"
"/thing" should match the string "/a/thing/that/is/here"
"/thing" should NOT match the string "/a/thing_foo"
Basically, it should match if the exact characters are there in the first string and the second, but not if there are run-ons in the second (such as an underscore like in thing_foo).
Right now, I'm doing this, which is not working.
let found = b.includes(a); // true
Hopefully my question is clear enough. Thanks for the help!

Boy did this turn in to a classic XY Problem.
If I had to guess, you want to know if a path contains a particular segment.
In that case, split the string on a positive lookahead for '/' and use Array.prototype.includes()
const paths = ["/a/thing", "/a/thing/that/is/here", "/a/thing_foo"]
const search = '/thing'
paths.forEach(path => {
const segments = path.split(/(?=\/)/)
console.log('segments', segments)
console.info(path, ':', segments.includes(search))
})
Using the positive lookahead expression /(?=\/)/ allows us to split the string on / whilst maintaining the / prefix in each segment.
Alternatively, if you're still super keen in using a straight regex solution, you'll want something like this
const paths = ["/a/thing", "/a/thing/that/is/here", "/a/thing_foo", "/a/thing-that/is/here"]
const search = '/thing'
const rx = new RegExp(search + '\\b') // note the escaped backslash
paths.forEach(path => {
console.info(path, ':', rx.test(path))
})
Note that this will return false positives if the search string is followed by a hyphen or tilde as those are considered to be word boundaries. You would need a more complex pattern and I think the first solution handles these cases better.

I'd recommend using regular expressions...
e.g. The following regular expression /\/thing$/ - matches anything that ends with /thing.
console.log(/\/thing$/.test('/a/thing')) // true
console.log(/\/thing$/.test('/a/thing_foo')) // false
Update: To use a variable...
var search = '/thing'
console.log(new RegExp(search + '$').test('/a/thing')) // true
console.log(new RegExp(search + '$').test('/a/thing_foo')) // false

Simply with following regex you can do it
var a = "/a/thing";
var b = "/a/thing/that/is/here";
var c = "/a/thing_foo";
var pattern = new RegExp(/(:?(thing)(([^_])|$))/);
pattern.test(a) // true
pattern.test(b) // true
pattern.test(c) // false

Regex - match the better part of a word in a search string

I am using Javascript and currently looking for a way to match as many of my pattern's letters as possible, maintaining the original order..
For example a search pattern queued should return the march Queue/queue against the any of the following search strings:
queueTable
scheduledQueueTable
qScheduledQueueTable
As of now I've reached as far as this:
var myregex = new RegExp("([queued])", "i");
var result = myregex.exec('queueTable');
but it doesn't seem to work correctly as it highlights the single characters q,u,e,u,e and e at the end of the word Table.
Any ideas?

Generate the regex with optional non-capturing group part where regex pattern can be generate using Array#reduceRight method.
var myregex = new RegExp("queued"
.split('')
.reduceRight(function(str, s) {
return '(?:' + s + str + ')?';
}, ''), "i");
var result = myregex.exec('queueTable');
console.log(result)
The method generates regex : /(?:q(?:u(?:e(?:u(?:e(?:d?)?)?)?)?)?)?/
UPDATE : If you want to get the first longest match then use g modifier in regex and find out the largest using Array#reduce method.
var myregex = new RegExp(
"queued".split('')
.reduceRight(function(str, s) {
return '(?:' + s + str + ')?';
}, ''), "ig");
var result = 'qscheduledQueueTable'
.match(myregex)
.reduce(function(a, b) {
return a.length > b.length ? a : b;
});
console.log(result);

I think the logic would have to be something like:
Match as many of these letters as possible, in this order.
The only real answer that comes to mind is to get the match to continue if possible, but allow it to bail out. In this case...
myregex = /q(?:u(?:e(?:u(?:e(?:d|)|)|)|)|)/;
You can generate this, of course:
function matchAsMuchAsPossible(word) { // name me something sensible please!
return new RegExp(
word.split("").join("(?:")
+ (new Array(word.length).join("|)"))
);
}

You are using square brackets - which mean that it will match a single instance of any character listed inside.
There are a few ways of interpreting your intentions:
You want to match the word queue with an optional 'd' at the end:
var myregex = new RegExp("queued?", "i");
var result = myregex.exec('queueTable');
Note this can be shorter try this:
'queueTable'.match(/queued?/i);
I also removed the brackets as these were not adding anything here.
This link provides some good examples that may help you further: https://www.w3schools.com/js/js_regexp.asp

When you use [] in a regular expression, it means you want to match any of the characters inside the brackets.
Example: if I use [abc] it means "match a single character, and this character can be 'a', 'b' or 'c'"
So in your code [queued] means "match a single character, and this character can be 'q', 'u', 'e' or 'd'" - note that 'u' and 'e' appear twice so they are redundant in this case. That's why this expression matches just one single character.
If you want to match the whole string "queued", just remove the brackets. But in this case it won't match, because queueTable doesn't have 'd'. If you want 'd' to be optional, you can use queued? as already explained in previous answers.

Try something like the following :
var myregex = /queued?\B/g;
var result = myregex.exec('queueTable');
console.log(result);

Regexp, capture between parentheses, javascript

I have regexp that extracts values between parentheses.
It's working most of the time but not when it ends with a parentheses
var val = 'STR("ABC(t)")';
var regExp = /\(([^)]+)\)/;.
var matches = regExp.exec(val);
​
console.log(matches[1]); //"ABC(t"
What I want is "ABC(t)".
Any ideas how I can modify my regexp to Achive this?
Update
The value is always inside the parentheses.
Some examples:
'ASD("123")'; => '123'
'ASD(123)'; => '123'
'ASD(aa(10)asda(459))'; => 'aa(10)asda(459)'
So first there is some text (always text). Then there is a (, and it always ends with a ). I want the value between.

You may use greedy dot matching inside Group 1 pattern: /\((.+)\)/. It will match the first (, then any 1+ chars other than linebreak symbols and then the last ) in the line.
var vals = ['STR("ABC(t)")', 'ASD("123")', 'ASD(123)', 'ASD(aa(10)asda(459))'];
var regExp = /\((.+)\)/;
for (var val of vals) {
var matches = regExp.exec(val);
console.log(val, "=>", matches[1]);
}
Answering the comment: If the texts to extract must be inside nested balanced parentheses, either a small parsing code, or XRegExp#matchRecursive can help. Since there are lots of parsing codes around on SO, I will provide XRegExp example:
var str = 'some text (num(10a ) ss) STR("ABC(t)")';
var res = XRegExp.matchRecursive(str, '\\(', '\\)', 'g');
console.log(res);
<script src="https://cdnjs.cloudflare.com/ajax/libs/xregexp/2.0.0/xregexp-all-min.js"></script>

Modify formatting-pattern to replace placeholders in string

I write currently a simple formatting function to replace some placeholders in a string.
var format = function (a, c) {
return a.replace(/{ *([^} ]+) *}/g, function (b, a) {
b = c;
a.replace(/[^.|\[\]]+/g, function (a) {
b = b[a];
});
return b;
});
};
The syntax uses currently curly-bracket notation {key}, I try now to modify the RegExp-pattern to work with one percent instead %key.
var pattern = /{ *([^} ]+) *}/g;
I tried to just replace the parentheses {} with a percent %, but this still doesn't work properly.
var pattern = /% *([^% ]+) */g;
The original pattern works with the following conditions as expected:
var data = {
obj: {
foo: 'Foo',
bar: 'Bar'
},
arr: ['Foo', 'Bar']
};
var objTest = '{obj.foo}, is not equal to {obj.bar}.'
format(objTest, data) // => 'Foo, is not equal to Bar.'
var arrTest = '{arr[0]}, is not equal to {arr[1]}.'
format(arrTest, data) // => 'Foo, is not equal to Bar.'
If we use my modified pattern it seems that the last character after each placeholder-replacement will be removed:
'%obj.foo, is not equal to %obj.bar.' // => 'undefined is not equal to Bar'
'%arr[0], is not equal to %arr[1]' // => 'undefined is not equal to Bar'
Any ideas how to modify the pattern to make it possible to use it with percentage % instead of curly-brackets {}?

You can use this pattern:
var regex = /%([^\b]+)/g;
which means a % sign followed by a complete word. This excludes whitespace characters, underscores, etc.
If you instead want to be able to use those characters as well, you can write:
var regex = /%(\S+)/g;
which is the equivalent of:
var regex = /%([^\s]+)/g;

The reason for this is that your modified regex pattern does not know when to stop matching. In the previous one it terminated at the following } symbol.
This is not an easy thing to fix as there is a myriad of possibilities that could be seen to end your format, here you have a piece of punctuation at the end of each replacement string, i.e. a full stop %arr[1]. or comma %obj.foo,.
So to make this work in your case toy could replace } in the original pattern with [\.,] i.e.
/% *([^% ]+) *[\.,]/g
This will work, but now your replacement pattern needs to always be terminated with either a full stop or comma which i suspect is not exactly what you want. Better to terminate with a know character such as % which would make your matching pattern /% *([^% ]+) *%/g and your format %obj.foo% and you can output a % by doubling up i.e. %obj.bar%%%