Develop Reference

Regexp to capture comma separated values

I have a string that can be a comma separated list of \w, such as:
abc123
abc123,def456,ghi789
I am trying to find a JavaScript regexp that will return ['abc123'] (first case) or ['abc123', 'def456', 'ghi789'] (without the comma).
I tried:
^(\w+,?)+$ -- Nope, as only the last repeating pattern will be matched, 789
^(?:(\w+),?)+$ -- Same story. I am using non-capturing bracket. However, the capturing just doesn't seem to happen for the repeated word
Is what I am trying to do even possible with regexp? I tried pretty much every combination of grouping, using capturing and non-capturing brackets, and still not managed to get this happening...

If you want to discard the whole input when there is something wrong, the simplest way is to validate, then split:
if (/^\w+(,\w+)*$/.test(input)) {
var values = input.split(',');
// Process the values here
}
If you want to allow empty value, change \w+ to \w*.
Trying to match and validate at the same time with single regex requires emulation of \G feature, which assert the position of the last match. Why is \G required? Since it prevents the engine from retrying the match at the next position and bypass your validation. Remember than ECMA Script regex doesn't have look-behind, so you can't differentiate between the position of an invalid character and the character(s) after it:
something,=bad,orisit,cor&rupt
^^ ^^
When you can't differentiate between the 2 positions, you can't rely on the engine to do a match-all operation alone. While it is possible to use a while loop with RegExp.exec and assert the position of last match yourself, why would you do so when there is a cleaner option?
If you want to savage whatever available, torazaburo's answer is a viable option.

Live demo
Try this regex :
'/([^,]+)/'
Alternatively, strings in javascript have a split method that can split a string based on a delimeter:
s.split(',')

Split on the comma first, then filter out results that do not match:
str.split(',').filter(function(s) { return /^\w+$/.test(s); })

This regex pattern separates numerical value in new line which contains special character such as .,,,# and so on.
var val = [1234,1213.1212, 1.3, 1.4]
var re = /[0-9]*[0-9]/gi;

var str = "abc123,def456, asda12, 1a2ass, yy8,ghi789";
var re = /[a-z]{3}\d{3}/g;
var list = str.match(re);
document.write("<BR> list.length: " + list.length);
for(var i=0; i < list.length; i++) {
document.write("<BR>list(" + i + "): " + list[i]);
}
This will get only "abc123" code style in the list and nothing else.

May be you can use split function
var st = "abc123,def456,ghi789";
var res = st.split(',');

Extract specific chars from a string using a regex

I need to split an email address and take out the first character and the first character after the '#'
I can do this as follows:
'bar#foo'.split('#').map(function(a){ return a.charAt(0); }).join('')
--> bf
Now I was wondering if it can be done using a regex match, something like this
'bar#foo'.match(/^(\w).*?#(\w)/).join('')
--> bar#fbf
Not really what I want, but I'm sure I miss something here! Any suggestions ?

Why use a regex for this? just use indexOf to get the char at any given position:
var addr = 'foo#bar';
console.log(addr[0], addr[addr.indexOf('#')+1])
To ensure your code works on all browsers, you might want to use charAt instead of []:
console.log(addr.charAt(0), addr.charAt(addr.indexOf('#')+1));
Either way, It'll work just fine, and This is undeniably the fastest approach
If you are going to persist, and choose a regex, then you should realize that the match method returns an array containing 3 strings, in your case:
/^(\w).*?#(\w)/
["the whole match",//start of string + first char + .*?# + first string after #
"groupw 1 \w",//first char
"group 2 \w"//first char after #
]
So addr.match(/^(\w).*?#(\w)/).slice(1).join('') is probably what you want.

If I understand correctly, you are quite close. Just don't join everything returned by match because the first element is the entire matched string.
'bar#foo'.match(/^(\w).*?#(\w)/).splice(1).join('')
--> bf

Using regex:
matched="",
'abc#xyz'.replace(/(?:^|#)(\w)/g, function($0, $1) { matched += $1; return $0; });
console.log(matched);
// ax

The regex match function returns an array of all matches, where the first one is the 'full text' of the match, followed by every sub-group. In your case, it returns this:
bar#f
b
f
To get rid of the first item (the full match), use slice:
'bar#foo'.match(/^(\w).*?#(\w)/).slice(1).join('\r')

Use String.prototype.replace with regular expression:
'bar#foo'.replace(/^(\w).*#(\w).*$/, '$1$2'); // "bf"

Or using RegEx
^([a-zA-Z0-9])[a-zA-Z0-9.!#$%&'*+\/=?^_`{|}~-]+#([a-zA-Z0-9-])[a-zA-Z0-9-]+(?:\.[a-zA-Z0-9-]+)*$
Fiddle

Remove everything after the first instance of one of several characters

Say I have a string like:
var str = "Good morningX Would you care for some tea?"
Where the X could be one of several characters, like a ., ?, or !.
How can I remove everything after that character?
If it could only be one type of character, I would use indexOf and substr, but it looks like I need a different method to find the position in this case. Perhaps a regular expression?
Clarification: I do not know what character X is. I'd like to cut the string off at the first occurrence of any one of the specified characters.
Ok, further clarification:
What I'm actually doing is scrubbing posts from a website. I'm taking the first bit from each post and stitching them together. By 'bit', I mean characters before the first piece of punctuation. I need to cut everything off after that punctuation. Does that make sense?

Just replace everything within the [ and ] with your delimiters. Escape if necessary.
var str = "Good morning! Would you care for some tea?";
var beginning = str.split(/[.?!]/)[0];
// "Good morning"

Try this, If the X have this ',' character , then try below
var s = 'Good morning, would you care for some tea?';
s = s.substring(0, s.indexOf(','));
document.write(s);
Demo : http://jsfiddle.net/L4hna/490/
and if the X have '!' , then try below
var s = 'Good morning! would you care for some tea?';
s = s.substring(0, s.indexOf('!'));
document.write(s);
Demo : http://jsfiddle.net/L4hna/491/
Try this way for your requirement string.
Both are will return Good Morning

The below code will do as you expect:
var s = "Good morningX Would you care for some tea?";
s = s.substring(X, n != -1 ? n : s.length);
document.write(s);

http://jsfiddle.net/JEFnY/
The regex would be
str.replace(/(.*?)([\.\?\!])(.*)/i, '$1$2');
The first capturing group is a lazy expression to match everything before the next capturing group.
The second capturing group only looks for the characters that you specify - which in this case are .!?, all escaped.
The last capturing group is discarded. Hence the substitution string is $1$2, or the first two capturing groups together.

getting contents of string between digits

have a regex problem :(
what i would like to do is to find out the contents between two or more numbers.
var string = "90+*-+80-+/*70"
im trying to edit the symbols in between so it only shows up the last symbol and not the ones before it. so trying to get the above variable to be turned into 90+80*70. although this is just an example i have no idea how to do this. the length of the numbers, how many "sets" of numbers and the length of the symbols in between could be anything.
many thanks,
Steve,

The trick is in matching '90+-+' and '80-+/' seperately, and selecting only the number and the last constant.
The expression for finding the a number followed by 1 or more non-numbers would be
\d+[^\d]+
To select the number and the last non-number, add parens:
(\d+)[^\d]*([^\d])
Finally add a /g to repeat the procedure for each match, and replace it with the 2 matched groups for each match:
js> '90+*-+80-+/*70'.replace(/(\d+)[^\d]*([^\d])/g, '$1$2');
90+80*70
js>

Or you can use lookahead assertion and simply remove all non-numerical characters which are not last: "90+*-+80-+/*70".replace(/[^0-9]+(?=[^0-9])/g,'');

You can use a regular expression to match the non-digits and a callback function to process the match and decide what to replace:
var test = "90+*-+80-+/*70";
var out = test.replace(/[^\d]+/g, function(str) {
return(str.substr(-1));
})
alert(out);
See it work here: http://jsfiddle.net/jfriend00/Tncya/
This works by using a regular expression to match sequences of non-digits and then replacing that sequence of non-digits with the last character in the matched sequence.

i would use this tutorial, first, then review this for javascript-specific regex questions.

This should do it -
var string = "90+*-+80-+/*70"
var result = '';
var arr = string.split(/(\d+)/)
for (i = 0; i < arr.length; i++) {
if (!isNaN(arr[i])) result = result + arr[i];
else result = result + arr[i].slice(arr[i].length - 1, arr[i].length);
}
alert(result);
Working demo - http://jsfiddle.net/ipr101/SA2pR/

Similar to #Arnout Engelen
var string = "90+*-+80-+/*70";
string = string.replace(/(\d+)[^\d]*([^\d])(?=\d+)/g, '$1$2');
This was my first thinking of how the RegEx should perform, it also looks ahead to make sure the non-digit pattern is followed by another digit, which is what the question asked for (between two numbers)
Similar to #jfriend00
var string = "90+*-+80-+/*70";
string = string.replace( /(\d+?)([^\d]+?)(?=\d+)/g
, function(){
return arguments[1] + arguments[2].substr(-1);
});
Instead of only matching on non-digits, it matches on non-digits between two numbers, which is what the question asked
Why would this be any better?
If your equation was embedded in a paragraph or string of text. Like:
This is a test where I want to clean up something like 90+*-+80-+/*70 and don't want to scrap the whole paragraph.
Result (Expected) :
This is a test where I want to clean up something like 90+80*70 and don't want to scrap the whole paragraph.
Why would this not be any better?
There is more pattern matching, which makes it theoretically slower (negligible)
It would fail if your paragraph had embedded numbers. Like:
This is a paragraph where Sally bought 4 eggs from the supermarket, but only 3 of them made it back in one piece.
Result (Unexpected):
This is a paragraph where Sally bought 4 3 of them made it back in one piece.

Regex to get string between curly braces

Unfortunately, despite having tried to learn regex at least one time a year for as many years as I can remember, I always forget as I use them so infrequently. This year my new year's resolution is to not try and learn regex again - So this year to save me from tears I'll give it to Stack Overflow. (Last Christmas remix).
I want to pass in a string in this format {getThis}, and be returned the string getThis. Could anyone be of assistance in helping to stick to my new year's resolution?
Related questions on Stack Overflow:
How can one turn regular quotes (i.e. ', ") into LaTeX/TeX quotes (i.e. `', ``'')
Regex: To pull out a sub-string between two tags in a string
Regex to replace all \n in a String, but no those inside [code] [/code] tag

Try
/{(.*?)}/
That means, match any character between { and }, but don't be greedy - match the shortest string which ends with } (the ? stops * being greedy). The parentheses let you extract the matched portion.
Another way would be
/{([^}]*)}/
This matches any character except a } char (another way of not being greedy)

/\{([^}]+)\}/
/ - delimiter
\{ - opening literal brace escaped because it is a special character used for quantifiers eg {2,3}
( - start capturing
[^}] - character class consisting of
^ - not
} - a closing brace (no escaping necessary because special characters in a character class are different)
+ - one or more of the character class
) - end capturing
\} - the closing literal brace
/ - delimiter

If your string will always be of that format, a regex is overkill:
>>> var g='{getThis}';
>>> g.substring(1,g.length-1)
"getThis"
substring(1 means to start one character in (just past the first {) and ,g.length-1) means to take characters until (but not including) the character at the string length minus one. This works because the position is zero-based, i.e. g.length-1 is the last position.
For readers other than the original poster: If it has to be a regex, use /{([^}]*)}/ if you want to allow empty strings, or /{([^}]+)}/ if you want to only match when there is at least one character between the curly braces. Breakdown:
/: start the regex pattern
{: a literal curly brace
(: start capturing
[: start defining a class of characters to capture
^}: "anything other than }"
]: OK, that's our whole class definition
*: any number of characters matching that class we just defined
): done capturing
}: a literal curly brace must immediately follow what we captured
/: end the regex pattern

Try this:
/[^{\}]+(?=})/g
For example
Welcome to RegExr v2.1 by #{gskinner.com}, #{ssd.sd} hosted by Media Temple!
will return gskinner.com, ssd.sd.

Try this
let path = "/{id}/{name}/{age}";
const paramsPattern = /[^{}]+(?=})/g;
let extractParams = path.match(paramsPattern);
console.log("extractParams", extractParams) // prints all the names between {} = ["id", "name", "age"]

Here's a simple solution using javascript replace
var st = '{getThis}';
st = st.replace(/\{|\}/gi,''); // "getThis"
As the accepted answer above points out the original problem is easily solved with substring, but using replace can solve the more complicated use cases
If you have a string like "randomstring999[fieldname]"
You use a slightly different pattern to get fieldname
var nameAttr = "randomstring999[fieldname]";
var justName = nameAttr.replace(/.*\[|\]/gi,''); // "fieldname"

This one works in Textmate and it matches everything in a CSS file between the curly brackets.
\{(\s*?.*?)*?\}
selector {.
.
matches here
including white space.
.
.}
If you want to further be able to return the content, then wrap it all in one more set of parentheses like so:
\{((\s*?.*?)*?)\}
and you can access the contents via $1.
This also works for functions, but I haven't tested it with nested curly brackets.

You want to use regex lookahead and lookbehind. This will give you only what is inside the curly braces:
(?<=\{)(.*?)(?=\})

i have looked into the other answers, and a vital logic seems to be missing from them . ie, select everything between two CONSECUTIVE brackets,but NOT the brackets
so, here is my answer
\{([^{}]+)\}

Regex for getting arrays of string with curly braces enclosed occurs in string, rather than just finding first occurrence.
/\{([^}]+)\}/gm

var re = /{(.*)}/;
var m = "{helloworld}".match(re);
if (m != null)
console.log(m[0].replace(re, '$1'));
The simpler .replace(/.*{(.*)}.*/, '$1') unfortunately returns the entire string if the regex does not match. The above code snippet can more easily detect a match.

Try this one, according to http://www.regextester.com it works for js normaly.
([^{]*?)(?=\})

This one matches everything even if it finds multiple closing curly braces in the middle:
\{([\s\S]*)\}
Example:
{
"foo": {
"bar": 1,
"baz": 1,
}
}

You can use this regex recursion to match everythin between, even another {} (like a JSON text) :
\{([^()]|())*\}

Even this helps me while trying to solve someone's problem,
Split the contents inside curly braces ({}) having a pattern like,
{'day': 1, 'count': 100}.
For example:
#include <iostream>
#include <regex>
#include<string>
using namespace std;
int main()
{
//string to be searched
string s = "{'day': 1, 'count': 100}, {'day': 2, 'count': 100}";
// regex expression for pattern to be searched
regex e ("\\{[a-z':, 0-9]+\\}");
regex_token_iterator<string::iterator> rend;
regex_token_iterator<string::iterator> a ( s.begin(), s.end(), e );
while (a!=rend) cout << " [" << *a++ << "]";
cout << endl;
return 0;
}
Output:
[{'day': 1, 'count': 100}] [{'day': 2, 'count': 100}]

Your can use String.slice() method.
let str = "{something}";
str = str.slice(1,-1) // something