We Keep Coding

REGEX - after bracket get data until end bracket

I have a string like the following:
SOME TEXT (BI1) SOME MORE TEXT (BI17) SOME FINAL TEXT (BI1234)
Question
I am trying to make a regex to get just the information between the curly brackets, for example the end string would look like:
BI1 BI17 BI1234
I have found this example on stackoverflow which will get the first value BI1, but will ignore the rest after.
Get text between two rounded brackets
this is the REGEX I created from the above link: /\(([^)]+)\)/g but it includes the brackets, I want to remove these.
I am using this website to attempt to solve this query which has a testing window to see if the regex entered works:
http://www.regexr.com
Additional Information
there can be any amount of numbers also, which is why I have given 3 different examples.
this is a continous string, not on seperate lines
thanks for any help on this matter.

While this isn't possible using just regexes, you can do it with string#split and the following regex:
\).*?\(|^.*?\(|\).*?$
Yielding code that looks a bit like this:
function getBracketed(str) {
return str.split(/\).*?\(|^.*?\(|\).*?$/).filter(Boolean);
}
(You need to filter out the empty strings that'll appear at the beginning and end if you do it this way - hence the extra operation).
Regex demo on Regex101
Code demo on Repl.it

If you need to keep all inside parentheses and remove everything else, you might use
var str = "SOME TEXT (BI1) SOME MORE TEXT (BI17) SOME FINAL TEXT (BI1234)";
var result = str.replace(/.*?\(([^()]*)\)/g, " $1").trim();
console.log(result);
If you need to get only the BI+digits pattern inside parentheses, use
/.*?\((BI\d+)\)/g
Details:
.*? - match any 0+ chars other than linebreak symbols
\( - match a (
(BI\d+) - Group 1 capturing BI + 1 or more digits (\d+) (or [^()]* - zero or more chars other than ( and ))
\) - a closing ).
To get all the values as array (say, for later joining), use
var str = "SOME TEXT (BI1) SOME MORE TEXT (BI17) SOME FINAL TEXT (BI1234)";
var re = /\((BI\d+)\)/g;
var res =str.match(re).map(function(s) {return s.substring(1, s.length-1);})
console.log(res);
console.log(res.join(" "));

How can I remove all white spaces?

I am using the following:
var xpto = "AB23. 3434-.34212 23 42."
I'm removing the "." "-" And ""
xpto.replace(/\./g,"").replace(/\-/g,"").replace("/\s/g","")
How can I remove all white spaces?

Your last replace is using a string, not a regular expression. You also don't seem to have kept the result:
xpto = xpto.replace(/\./g,"").replace(/\-/g,"").replace(/\s/g,"");
// ^ No quotes here -------------------------------^--^
// \--- Remember result
You can also shorten that and just call replace once, using a character class ([...]):
xpto = xpto.replace(/[-.\s]/g,"");
(Note that when using the - character literally in a character class, you must make it the first character after the opening [ or the last character before the closing ], or put a backslash in front of it. If it appears between two other characters ([a-z], for instance), it means "any character in the range".)

You can remove white spaces using replace function
xpto.replace(/\s/g,'');

Your error comes from the quotes around your last regex, however I might also point out that you are calling replace way more than needed:
xpto = xpto.replace(/[\s.-]/g,"");
This will strip out spaces, dots and hyphens.

You done it right, but forgot the quotation marks "" at /\s/g. Also, you want to change the string xpto, to the replaced xpto, so you can now do something with it.
Javascript
var xpto = "AB23. 3434-.34212 23 42."
xpto = xpto.replace(/\./g,"").replace(/\-/g,"").replace(/\s/g,"");
Output
AB233434342122342
JSFiddle demo

Javascript regex expression to replace multiple strings?

I've a string done like this: "http://something.org/dom/My_happy_dog_%28is%29cool!"
How can I remove all the initial domain, the multiple underscore and the percentage stuff?
For now I'm just doing some multiple replace, like
str = str.replace("http://something.org/dom/","");
str = str.replace("_%28"," ");
and go on, but it's really ugly.. any help?
Thanks!
EDIT:
the exact input would be "My happy dog is cool!" so I would like to get rid of the initial address and remove the underscores and percentage and put the spaces in the right place!
The problem is that trying to put a regex on Chrome "something goes wrong". Is it a problem of Chrome or my regex?

I'd suggest:
var str = "http://something.org/dom/My_happy_dog_%28is%29cool!";
str.substring(str.lastIndexOf('/')+1).replace(/(_)|(%\d{2,})/g,' ');
JS Fiddle demo.
The reason I took this approach is that RegEx is fairly expensive, and is often tricky to fine tune to the point where edge-cases become less troublesome; so I opted to use simple string manipulation to reduce the RegEx work.
Effectively the above creates a substring of the given str variable, from the index point of the lastIndexOf('/') (which does exactly what you'd expect) and adding 1 to that so the substring is from the point after the / not before it.
The regex: (_) matches the underscores, the | just serves as an or operator and the (%\d{2,}) serves to match digit characters that occur twice in succession and follow a % sign.
The parentheses surrounding each part of the regex around the |, serve to identify matching groups, which are used to identify what parts should be replaced by the ' ' (single-space) string in the second of the arguments passed to replace().
References:
lastIndexOf().
replace().
substring().

You can use unescape to decode the percentages:
str = unescape("http://something.org/dom/My_happy_dog_%28is%29cool!")
str = str.replace("http://something.org/dom/","");

Maybe you could use a regular expression to pull out what you need, rather than getting rid of what you don't want. What is it you are trying to keep?
You can also chain them together as in:
str.replace("http://something.org/dom/", "").replace("something else", "");

You haven't defined the problem very exactly. To get rid of all stretches of characters ending in %<digit><digit> you'd say
var re = /.*%\d\d/g;
var str = str.replace(re, "");

ok, if you want to replace all that stuff I think that you would need something like this:
/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g
test
var string = "http://something.org/dom/My_happy_dog_%28is%29cool!";
string = string.replace(/(http:\/\/.*\.[a-z]{3}\/.*\/)|(\%[a-z0-9][a-z0-9])|_/g,"");

replacing spaces in a string with hyphens

I have a string and I need to fix it in order to append it to a query.
Say I have the string "A Basket For Every Occasion" and I want it to be "A-Basket-For-Every-Occasion"
I need to find a space and replace it with a hyphen. Then, I need to check if there is another space in the string. If not, return the fixed string. If so, run the same process again.
Sounds like a recursive function to me but I am not sure how to set it up. Any help would be greatly appreciated.

You can use a regex replacement like this:
var str = "A Basket For Every Occasion";
str = str.replace(/\s/g, "-");
The "g" flag in the regex will cause all spaces to get replaced.
You may want to collapse multiple spaces to a single hyphen so you don't end up with multiple dashes in a row. That would look like this:
var str = "A Basket For Every Occasion";
str = str.replace(/\s+/g, "-");

Use replace and find for whitespaces \s globally (flag g)
var a = "asd asd sad".replace(/\s/g,"-");
a becomes
"asd-asd-sad"

Try
value = value.split(' ').join('-');
I used this to get rid of my spaces. Instead of the hyphen I made it empty and works great. Also it is all JS. .split(limiter) will delete the limiter and puts the string pieces in an array (with no limiter elements) then you can join the array with the hyphens.

Javascript Regex: replacing the last dot for a comma

I have the following code:
var x = "100.007"
x = String(parseFloat(x).toFixed(2));
return x
=> 100.01
This works awesomely just how I want it to work. I just want a tiny addition, which is something like:
var x = "100,007"
x.replace(",", ".")
x.replace
x = String(parseFloat(x).toFixed(2));
x.replace(".", ",")
return x
=> 100,01
However, this code will replace the first occurrence of the ",", where I want to catch the last one. Any help would be appreciated.

You can do it with a regular expression:
x = x.replace(/,([^,]*)$/, ".$1");
That regular expression matches a comma followed by any amount of text not including a comma. The replacement string is just a period followed by whatever it was that came after the original last comma. Other commas preceding it in the string won't be affected.
Now, if you're really converting numbers formatted in "European style" (for lack of a better term), you're also going to need to worry about the "." characters in places where a "U.S. style" number would have commas. I think you would probably just want to get rid of them:
x = x.replace(/\./g, '');
When you use the ".replace()" function on a string, you should understand that it returns the modified string. It does not modify the original string, however, so a statement like:
x.replace(/something/, "something else");
has no effect on the value of "x".

You can use a regexp. You want to replace the last ',', so the basic idea is to replace the ',' for which there's no ',' after.
x.replace(/,([^,]*)$/, ".$1");
Will return what you want :-).

You could do it using the lastIndexOf() function to find the last occurrence of the , and replace it.
The alternative is to use a regular expression with the end of line marker:
myOldString.replace(/,([^,]*)$/, ".$1");

You can use lastIndexOf to find the last occurence of ,. Then you can use slice to put the part before and after the , together with a . inbetween.

You don't need to worry about whether or not it's the last ".", because there is only one. JavaScript doesn't store numbers internally with comma or dot-delimited sets.