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(" "));

regex encapsulation

I've got a question concerning regex.
I was wondering how one could replace an encapsulated text, something like {key:23} to something like <span class="highlightable">23</span, so that the entity will still remain encapsulated, but with something else.
I will do this in JS, but the regex is what is important, I have been searching for a while, probably searching for the wrong terms, I should probably learn more about regex, generally.
In any case, is there someone who knows how to perform this operation with simplicity?
Thanks!

It's important that you find {key:23} in your text first, and then replace it with your wanted syntax, this way you avoid replacing {key:'sometext'} with that syntax which is unwanted.
var str = "some random text {key:23} some random text {key:name}";
var n = str.replace(/\{key:[\d]+\}/gi, function myFunction(x){return x.replace(/\{key:/,'<span>').replace(/\}/, '</span>');});
this way only {key:AnyNumber} gets replaced, and {key:AnyThingOtherThanNumbers} don't get touched.

It seems you are new to regex. You need to learn more about character classes and capturing groups and backreferences.
The regex is somewhat basic in your case if you do not need any nested encapsulated text support.
Let's start:
The beginning is {key: - it will match the substring literally. Note that { can be a special character (denoting start of a limiting quantifier), thus, it is a good idea to escape it: {key:.
([^}]+) - This is a bit more interesting: the round brackets around are a capturing group that let us later back-reference the matched text. The [^}]+ means 1 or more characters (due to +) other than } (as [^}] is a negated character class where ^ means not)
} matches a } literally.
In the replacement string, we'll get the captured text using a backreference $1.
So, the entire regex will look like:
{key:([^}]+)}
See demo on regex101.com
Code snippet:
var re = /{key:([^}]+)}/g;
var str = '{key:23}';
var subst = '<span class="highlightable">$1</span>';
document.getElementById("res").innerHTML = str.replace(re, subst);
.highlightable
{
color: red;
}
<div id="res"/>
If you want to use a different behavior based on the value of key, then you'll need to adjust the regex to either match digits only (with \d+) or letters only (say, with [a-zA-Z] for English), or other shorthand classes, ranges (= character classes), or their combinations.

If your string is in var a, then:
var test = a.replace( /\{key:(\d+)\}/g, "<span class='highlightable'>$1</span>");

JavaScript string with regex search

I have a returned string consisting of multiple classes:
"xxx1sbu xxx2sdf xxx1sef xxx1sb1 xxx1su xxx1s1 dxxx1s xxx1sdfg xxx1sbf"
I need the regex to search the string and find the classes based on the following criteria:
1) class begins with 'xxx1s'
2) class is no longer than 7 characters (letters and numbers) excluding a following space ('xxx1sbu')
3) if a space follows the class ('xxx1sbu ') then it is also found so that it can be removed.
I then use the regex to replace the found classes to
var classesReplaced = classesString.replace(regex, "")
The string should then look as follows:
"xxx2sdf dxxx1s xxx1sdfg"
So far the best I have come up with is:
RegExp: /\bxxx1s([a-z1-9])([a-z1-9])(\s)|\bxxx1s([a-z1-9])([a-z1-9])\b/g
pattern: \bxxx1s([a-z1-9])([a-z1-9])(\s)|\bxxx1s([a-z1-9])([a-z1-9])\b
I also tried to use javascript to build the expression but it keeps stripping off the '\b':
var classId = 'xxx1s';
var regex2 = new RegExp('\b'+ classId +'([a-z1-9])([a-z1-9])(\s)|\b'+ classId +'([a-z1-9])([a-z1-9])\b','g');
Is there a better way to write this?? My understanding of regex isn't great!
Thank you in advance

I think this is the command and regex you need:
NewString = OldString.replace( /\bxxx1s\S{0,2}\b\s?/, "" );
Here's how that breaks down:
\b is the opening word border for the section you want to replace.
xxx1s is the base text you want to remove.
\S{0,2} (upper-case S) finds any non-white-space between 0 and 2 characters long; added to the item above, it makes sure that only items of 7 or fewer characters are found.
\b is the closing word border.
\s? (lower-case s) finds the trailing space; the ? makes it optional so the last item in the series won't be skipped.

Here is the regex you should use:
var regex = /((?:\s*(xxx1s)\w{1,2})(?=\W))|(xxx1sbu )/g;

Regex trying to match characters before and after symbol

I'm trying to match characters before and after a symbol, in a string.
string: budgets-closed
To match the characters before the sign -, I do: ^[a-z]+
And to match the other characters, I try: \-(\w+) but, the problem is that my result is: -closed instead of closed.
Any ideas, how to fix it?
Update
This is the piece of code, where I was trying to apply the regex http://jsfiddle.net/trDFh/1/
I repeat: It's not that I don't want to use split; it's just I was really curious, and wanted to see, how can it be done the regex way. Hacking into things spirit
Update2
Well, using substring is a solution as well: http://jsfiddle.net/trDFh/2/ and is the one I chosed to use, since the if in question, is actually an else if in a more complex if syntax, and the chosen solutions seems to be the most fitted for now.

Use exec():
var result=/([^-]+)-([^-]+)/.exec(string);
result is an array, with result[1] being the first captured string and result[2] being the second captured string.
Live demo: http://jsfiddle.net/Pqntk/

I think you'll have to match that. You can use grouping to get what you need, though.
var str = 'budgets-closed';
var matches = str.match( /([a-z]+)-([a-z]+)/ );
var before = matches[1];
var after = matches[2];
For that specific string, you could also use
var str = 'budgets-closed';
var before = str.match( /^\b[a-z]+/ )[0];
var after = str.match( /\b[a-z]+$/ )[0];
I'm sure there are better ways, but the above methods do work.

If the symbol is specifically -, then this should work:
\b([^-]+)-([^-]+)\b
You match a boundry, any "not -" characters, a - and then more "not -" characters until the next word boundry.
Also, there is no need to escape a hyphen, it only holds special properties when between two other characters inside a character class.
edit: And here is a jsfiddle that demonstrates it does work.

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,"");