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Need to replace a string from a long string in javascript

I have a long string
Full_str1 = 'ab#xyz.com;cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;c_ab#xyz.com;';
removable_str2 = 'ab#xyz.com;';
I need to have a replaced string which will have
resultant Final string should look like,
cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;c_ab#xyz.com;
I tried with
str3 = Full_str1.replace(new RegExp('(^|\\b)' +removable_str2, 'g'),"");
but it resulted in
cab#xyz.com;c-c.c_ab#xyz.com;

Here a soluce using two separated regex for each case :
the str to remove is at the start of the string
the str to remove is inside or at the end of the string
PS :
I couldn't perform it in one regex, because it would remove an extra ; in case of matching the string to remove inside of the global string.
const originalStr = 'ab#xyz.com;cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;ab#xyz.com;c_ab#xyz.com;';
const toRemove = 'ab#xyz.com;';
const epuredStr = originalStr
.replace(new RegExp(`^${toRemove}`, 'g'), '')
.replace(new RegExp(`;${toRemove}`, 'g'), ';');
console.log(epuredStr);

First, the dynamic part must be escaped, else, . will match any char but a line break char, and will match ab#xyz§com;, too.
Next, you need to match this only at the start of the string or after ;. So, you may use
var Full_str1 = 'ab#xyz.com;cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;c_ab#xyz.com;';
var removable_str2 = 'ab#xyz.com;';
var rx = new RegExp("(^|;)" + removable_str2.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&'), "g");
console.log(Full_str1.replace(rx, "$1"));
// => cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;c_ab#xyz.com;
Replace "g" with "gi" for case insensitive matching.
See the regex demo. Note that (^|;) matches and captures into Group 1 start of string location (empty string) or ; and $1 in the replacement pattern restores this char in the result.
NOTE: If the pattern is known beforehand and you only want to handle ab#xyz.com; pattern, use a regex literal without escaping, Full_str1.replace(/(^|;)ab#xyz\.com;/g, "$1").

i don't find any particular description why you haven't tried like this it will give you desired result cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;c_ab#xyz.com;
const full_str1 = 'ab#xyz.com;cab#xyz.com;c-ab#xyz.com;c.ab#xyz.com;c_ab#xyz.com;';
const removable_str2 = 'ab#xyz.com;';
const result= full_str1.replace(removable_str2 , "");
console.log(result);

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)

javascript match specific name plus / and characters after it

match a path for a specific word and a / and any characters that follow.
For example.
const str = 'cars/ford';
const isCars = str.match('cars');
What I want to do is make sure it matches cars and has a slash and characters after the / then return true or false.
The characters after cars/... will change so I can't match it excatly. Just need to match any characters along with the /

Would love to use regex not sure what it should be. Looking into how to achieve that via regex tutorials.
var str = "cars/ford";
var patt = new RegExp("^cars/"); //or var patt = /^cars\//
var res = patt.test(str); //true
console.log(res);
https://www.w3schools.com/js/js_regexp.asp
https://www.rexegg.com/regex-quickstart.html

You could use test() that returns true or false.
const str = "cars/ford";
const str2 = "cars/";
var isCars = (str)=>/^cars\/./i.test(str)
console.log(isCars(str));
console.log(isCars(str2));

Here is a quick regex to match "cars/" followed by any characters a-z.
(cars\/[a-z]+)
This will only match lowercase letters, so you can add the i flag to make it case insensitive.
/(cars\/[a-z]+)/i

It is a basic regular expression
var str = "cars/ford"
var result = str.match(/^cars\/(.*)$/)
console.log(result)
^ - start
cars - match exact characters
\/ - match /, the \ escapes it
(.*) - capture group, match anything
$ - end of line
Visualize it: RegExper

Regex to grab strings between square brackets

I have the following string: pass[1][2011-08-21][total_passes]
How would I extract the items between the square brackets into an array? I tried
match(/\[(.*?)\]/);
var s = 'pass[1][2011-08-21][total_passes]';
var result = s.match(/\[(.*?)\]/);
console.log(result);
but this only returns [1].
Not sure how to do this.. Thanks in advance.

You are almost there, you just need a global match (note the /g flag):
match(/\[(.*?)\]/g);
Example: http://jsfiddle.net/kobi/Rbdj4/
If you want something that only captures the group (from MDN):
var s = "pass[1][2011-08-21][total_passes]";
var matches = [];
var pattern = /\[(.*?)\]/g;
var match;
while ((match = pattern.exec(s)) != null)
{
matches.push(match[1]);
}
Example: http://jsfiddle.net/kobi/6a7XN/
Another option (which I usually prefer), is abusing the replace callback:
var matches = [];
s.replace(/\[(.*?)\]/g, function(g0,g1){matches.push(g1);})
Example: http://jsfiddle.net/kobi/6CEzP/

var s = 'pass[1][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r ; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]
example proving the edge case of unbalanced [];
var s = 'pass[1]]][2011-08-21][total_passes]';
r = s.match(/\[([^\]]*)\]/g);
r; //# => [ '[1]', '[2011-08-21]', '[total_passes]' ]

add the global flag to your regex , and iterate the array returned .
match(/\[(.*?)\]/g)

I'm not sure if you can get this directly into an array. But the following code should work to find all occurences and then process them:
var string = "pass[1][2011-08-21][total_passes]";
var regex = /\[([^\]]*)\]/g;
while (match = regex.exec(string)) {
alert(match[1]);
}
Please note: i really think you need the character class [^\]] here. Otherwise in my test the expression would match the hole string because ] is also matches by .*.

'pass[1][2011-08-21][total_passes]'.match(/\[.+?\]/g); // ["[1]","[2011-08-21]","[total_passes]"]
Explanation
\[ # match the opening [
Note: \ before [ tells that do NOT consider as a grouping symbol.
.+? # Accept one or more character but NOT greedy
\] # match the closing ] and again do NOT consider as a grouping symbol
/g # do NOT stop after the first match. Do it for the whole input string.
You can play with other combinations of the regular expression
https://regex101.com/r/IYDkNi/1

[C#]
string str1 = " pass[1][2011-08-21][total_passes]";
string matching = #"\[(.*?)\]";
Regex reg = new Regex(matching);
MatchCollection matches = reg.Matches(str1);
you can use foreach for matched strings.

Replace a substring with javascript

Need to replace a substring in URL (technically just a string) with javascript.
The string like
http://blah-blah.com/search?par_one=test&par_two=anothertest&SearchableText=TO_REPLACE
or
http://blah-blah.com/search?par_one=test&SearchableText=TO_REPLACE&par_two=anothertest
means, the word to replace can be either at the most end of the URL or in the middle of it.
I am trying to cover these with the following:
var newWord = NEW_SEARCH_TERM;
var str = 'http://blah-blah.com/search?par_one=test&SearchableText=TO_REPLACE&par_two=anothertest';
var regex = /^\S+SearchableText=(.*)&?\S*$/;
str = str.replace(regex, newWord);
But no matter what I do I get str = NEW_SEARCH_TERM. Moreover the regular expression when I try it in RegExhibit, selects the word to replace and everything that follows it that is not what I want.
How can I write a universal expression to cover both cases and make the correct string be saved in the variable?

str.replace(/SearchableText=[^&]*/, 'SearchableText=' + newWord)

The \S+ and \S* in your regex match all non-whitespace characters.
You probably want to remove them and the anchors.

http://jsfiddle.net/mplungjan/ZGbsY/
ClyFish did it while I was fiddling
var url1="http://blah-blah.com/search?par_one=test&par_two=anothertest&SearchableText=TO_REPLACE";
var url2 ="http://blah-blah.com/search?par_one=test&SearchableText=TO_REPLACE&par_two=anothertest"
var newWord = "foo";
function replaceSearch(str,newWord) {
var regex = /SearchableText=[^&]*/;
return str.replace(regex, "SearchableText="+newWord);
}
document.write(replaceSearch(url1,newWord))
document.write('<hr>');
document.write(replaceSearch(url2,newWord))