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replace words that starts with "(uuid: )"

I would like to replace text using javascript/regex
"TV "my-samsung" (UUID: a1c3bbc1d27c5be8:8baabe2fa7f5d9ca) is already switched off."
with
TV 'my-samsung' is already switched off.
by removing text (UUID: ) and replace " with '
Looks like regex can be used
\([\s\S]*?\)
https://regex101.com/r/xXDncn/1
or have also tried using replace method in JS
str = str.replace("(UUID", "");

You can use
const str = '" "Tv "my-samsung" (UUID: a1c3bbc1d27c5be8:8baabe2fa7f5d9ca) is already switched-off""';
console.log(
str.replace(/\s*\(UUID:[^()]*\)/g, '').replace(/^[\s"]+|[\s"]+$/g, '').replaceAll('"', "'")
)
See the first regex demo. It matches
\s* - zero or more whitespaces
\(UUID: - (UUID: string
[^()]* - zero or more chars other than ( and )
\) - a ) char.
The g flag makes it replace all occurrences.
The second regex removes trailing and leading whitespace and double quotation marks:
^[\s"]+ - one or more whitespaces and double quotes at the start of string
| - or
[\s"]+$ - one or more whitespaces and double quotes at the end of string.
The .replaceAll('"', "'") is necessary to replace all " with ' chars.
It is not a good idea to merge these two operations into one as the replacements are different. Here is how it could be done, just for learning purposes:
const str = '" "Tv "my-samsung" (UUID: a1c3bbc1d27c5be8:8baabe2fa7f5d9ca) is already switched-off""';
console.log(
str.replace(/^[\s"]+|[\s"]+$|\s*\(UUID:[^()]*\)|(")/g, (x,y) => y ? "'" : "")
)
That is, " is captured into Group 1, the replacement is now a callable, where x is the whole match and y is the Group 1 contents. If Group 1 matched, the replacement is ', else, the replacement is an empty string (to remove the match found).

you can try this
str.replace(/\(.*?\)/, "")
str.replace(/\(.*?\)/, "with")
--- update ---
const str = `"TV "my-samsung" (UUID: a1c3bbc1d27c5be8:8baabe2fa7f5d9ca) is already switched off."`;
const a = str.replace(/"(.*?)\(.*\)(.*)"/, (a, b, c) => {
return b.replace(/"/g, "'") + c
});
console.log(a); //TV 'my-samsung' is already switched off.

Format and replace a string with a regular expression

I have a number that's at least 7 digits long.
Typical examples: 0000123, 00001234, 000012345
I want to transform them so that they become respectively:
01:23, 12:34, 23:45
Which mean replacing the whole string by the last 4 characters and putting a colon in the middle.
I can get the last 4 digits with (\d{4})$
And I can get 2 groups with this: (\d{2})(\d{2})$
With the last option, on a string 0000123 $1:$2 match gives me 00001:23
where I want 01:23
I replace the string like so:
newVal = val.replace(/regex/, '$1:$2');

You need to match the beginning digits with \d* (or with just .* if there can be anything):
var val = "0001235";
var newVal = val.replace(/^\d*(\d{2})(\d{2})$/, '$1:$2');
console.log(newVal);
Pattern details:
^ - start of string
\d* - 0+ digits (or .* will match any 0+ chars other than line break chars)
(\d{2}) - Group 1 capturing 2 digits
(\d{2}) - Group 2 capturing 2 digits
$ - end of string.

As Alex K. said, no need for a regular expression, just extract the parts you need with substr:
val = val.substr(-4, 2) + ":" + val.substr(-2);
Note that when the starting index is negative, it's from the end of the string.
Example:
function update(val) {
return val.substr(-4, 2) + ":" + val.substr(-2);
}
function test(val) {
console.log(val + " => " + update(val));
}
test("0000123");
test("0001234");
test("000012345");

You could throw the first characters away and the replace only the last matched parts.
console.log('00000001234'.replace(/^(.*)(\d{2})(\d{2})$/, '$2:$3'));

Use this regex: ^(\d+?)(\d{2})(\d{2})$:
var newVal = "0000123".replace(/^(\d+?)(\d{2})(\d{2})$/, '$2:$3');
console.log(newVal);

How to get first 2 words?

Let data.title be ABC XYZ PQRS - www.aaa.tld.
Output needs to be like this ABC+XYZ
i've tried this:
var t = data.title.split(' ').join('+');
t = t.replace(/(([^\s]+\s\s*){1})(.*)/,"Unknown");
$("#log").text(t);

Here is one way to do it, no regex though, it only grabs the first two words and must have a space between those words.
First we split into and array, then we slice that array from the 0 index to 2(exclusive) or 1, and finally we join them with a '+':
var x = 'ABC XYZ PQRS';
var y = x.split(' ').slice(0,2).join('+');
// y = "ABC+XYZ"
Working Fiddle

Try using .match() with RegExp /([\w+]+)/g; concatenate first match, + character, second match
var matches = "ABC XYZ PQRS - www.aaa.tld".match(/([\w+]+)/g);
console.log(matches[0] + "+" + matches[1])

This is my general function for first n words. Haven't tested it extensively but it is fast even on long strings because it doesn't use a global regex or split every word. You can fine tune the regex for dealing with punctuation. I'm considering a hyphen as a delimiter but you can move that to the word portion instead if you prefer.
function regFirstWords(s, n) {
// ?: non-capturing subsequent sp+word.Change {} if you want to require n instead of allowing fewer
var a = s.match(new RegExp('[\\w\\.]+' + '(?:[\\s-]*[\\w\\.]+){0,' + (n - 1) + '}'));
return (a === undefined || a === null) ? '' : a[0];
}
To satisfy the OP's request to replace with '+'
regFirstWords('ABC XYZ PQRS - www.aaa.tld',2).replace(/\s/g,'+')

JS regex return string from url

I have the following URL structure:
https://api.bestschool.com/student/1102003120009/tests/json
I want to cut the student ID from the URL. So far I've came up with this:
/(student\/.*[^\/]*)/
which returns
student/1102003120009/tests/json
I only want the ID.

Your regex (student\/.*[^\/]*) matches and captures into Group 1 a literal sequence student/, then matches any characters other than a newline, 0 or more occurrences (.*) - that can match the whole line at once! - and then 0 or more characters other than /. It does not work because of .*. Also, a capturing group should be moved to the [^\/]* pattern.
You can use the following regex and grab Group 1 value:
student\/([^\/]*)
See regex demo
The regex matches student/ literally, and then matches and captures into Group 1 zero or more symbols other than /.
Alternatively, if you want to avoid using capturing, and assuming that the ID is always numeric and is followed by /tests/, you can use the following regex:
\d+(?=\/tests\/)
The \d+ matches 1 or more digits, and (?=\/tests\/) checks if right after the digits there is a /tests/ character sequence.
var re = /student\/([^\/]*)/;
var str = 'https://api.bestschool.com/student/1102003120009/tests/json';
var m = str.match(re);
if (m !== null) {
document.getElementById("r").innerHTML = "First method : " + m[1] + "<br/>";
}
var m2 = str.match(/\d+(?=\/tests\/)/);
if (m2 !== null) {
document.getElementById("r").innerHTML += "Second method: " + m2;
}
<div id="r"/>

JS string replace only replacing every other occurence

I have the following JS:
"a a a a".replace(/(^|\s)a(\s|$)/g, '$1')
I expect the result to be '', but am instead getting 'a a'. Can anyone explain to me what I am doing wrong?
Clarification: What I am trying to do is remove all occurrences of 'a' that are surronded by whitespace (i.e. a whole token)

It's because this regex /(^|\s)a(\s|$)/g match the previous char and the next char to each a
in string "a a a a" the regex matches :
"a " , then the string to check become "a a a"$ (but now the start of the string is not the beginning and there is not space before)
" a " (the third a) , then become "a"$ (that not match because no space before)
Edit:
Little bit tricky but working (without regex):
var a = "a a a a";
// Handle beginning case 'a '
var startI = a.indexOf("a ");
if (startI === 0){
var off = a.charAt(startI + 2) !== "a" ? 2 : 1; // test if "a" come next to keep the space before
a = a.slice(startI + off);
}
// Handle middle case ' a '
var iOf = -1;
while ((iOf = a.indexOf(" a ")) > -1){
var off = a.charAt(iOf + 3) !== "a" ? 3 : 2; // same here
a = a.slice(0, iOf) + a.slice(iOf+off, a.length);
}
// Handle end case ' a'
var endI = a.indexOf(" a");
if (endI === a.length - 2){
a = a.slice(0, endI);
}
a; // ""

First "a " matches.
Then it will try to match against "a a a", which will skip first a, and then match "a ".
Then it will try to match against "a", which will not match.
First match will be replaced to beginning of line. => "^"
Then we have "a" that didn't match => "a"
Second match will be replaced to " " => " "
Then we have "a" that didn't match => "a"
The result will be "a a".
To get your desired result you can do this:
"a a a a".replace(/(?:\s+a(?=\s))+\s+|^a\s+(?=[^a]|$|a\S)|^a|\s*a$/g, '')

As others have tried to point out, the issue is that the regex consumes the surrounding spaces as part of the match. Here's a [hopefully] more straight forward explanation of why that regex doesn't work as you expect:
First let's breakdown the regex, it says match the a space or start of string, followed by an 'a' followed by a space or the end of the string.
Now let's apply it to the string. I've added character indexes beneath the string to make things easier to talk about:
a a a a
0123456
The regex looks at the 0 index char, and finds an 'a' at that location, followed by a space at index 2. This is a match because it is the start of the string, followed by an a followed by a space. The length of our match is 2 (the 'a' and the space), so we consume two characters and start our next search at index 2.
Character 2 ('a') is neither a space nor the start of the string, and therefore it doesn't match the start of our regular expression, so we consume that character (without replacing it) and move on to the next.
Character 3 is a space, followed by an 'a' followed by another space, which is a match for our regex. We replace it with an empty string, consume the length of the match (3 characters - " a ") and move on to index 6.
Character 6 ('a') is neither a space nor the start of the string, and therefore it doesn't match the start of our regular expression, so we consume that character (without replacing it) and move on to the next.
Now we're at the end of the string, so we're done.
The reason why the regex #caeth suggested (/(^|\s+)a(?=\s|$)/g) works is because of the ?= quantifier. From the MDN Regexp Documentation:
Matches x only if x is followed by y. For example, /Jack(?=Sprat)/ matches "Jack" only if it is followed by "Sprat". /Jack(?=Sprat|Frost)/ matches "Jack" only if it is followed by "Sprat" or "Frost". However, neither "Sprat" nor "Frost" is part of the match results.
So, in this case, the ?= quantifier checks to see if the following character is a space, without actually consuming that character.

(^|\s)a(?=\s|$)
Try this.Replace by $1.See demo.
https://regex101.com/r/gQ3kS4/3

Use this instead:
"a a a a".replace(/(^|\s*)a(\s|$)/g, '$1')
With "* this you replace all the "a" occurrences
Greetings

Or you can just split the string up, filter it and glue it back:
"a ba sl lf a df a a df r a".split(/\s+/).filter(function (x) { return x != "a" }).join(" ")
>>> "ba sl lf df df r"
"a a a a".split(/\s+/).filter(function (x) { return x != "a" }).join(" ")
>>> ""
Or in ECMAScript 6:
"a ba sl lf a df a a df r a".split(/\s+/).filter(x => x != "a").join(" ")
>>> "ba sl lf df df r"
"a a a a".split(/\s+/).filter(x => x != "a").join(" ")
>>> ""
I assume that there is no leading and trailing spaces. You can change the filter to x && x != 'a' if you want to remove the assumption.