I am trying to write a regular expression which returns a string which is between parentheses. For example: I want to get the string which resides between the strings "(" and ")"
I expect five hundred dollars ($500).
would return
$500
Found Regular Expression to get a string between two strings in Javascript
But I'm new with regex. I don't know how to use '(', ')' in regexp
10 Answers
You need to create a set of escaped (with \) parentheses (that match the parentheses) and a group of regular parentheses that create your capturing group:
var regExp = /\(([^)]+)\)/;
var matches = regExp.exec("I expect five hundred dollars ($500).");
//matches[1] contains the value between the parentheses
console.log(matches[1]);Breakdown:
\(: match an opening parentheses(: begin capturing group[^)]+: match one or more non)characters): end capturing group\): match closing parentheses
Here is a visual explanation on RegExplained
Try string manipulation:
var txt = "I expect five hundred dollars ($500). and new brackets ($600)";
var newTxt = txt.split('(');
for (var i = 1; i < newTxt.length; i++) {
console.log(newTxt[i].split(')')[0]);
}
or regex (which is somewhat slow compare to the above)
var txt = "I expect five hundred dollars ($500). and new brackets ($600)";
var regExp = /\(([^)]+)\)/g;
var matches = txt.match(regExp);
for (var i = 0; i < matches.length; i++) {
var str = matches[i];
console.log(str.substring(1, str.length - 1));
}
Simple solution
Notice: this solution can be used for strings having only single "(" and ")" like string in this question.
("I expect five hundred dollars ($500).").match(/\((.*)\)/).pop();
To match a substring inside parentheses excluding any inner parentheses you may use
\(([^()]*)\)
pattern. See the regex demo.
In JavaScript, use it like
var rx = /\(([^()]*)\)/g;
Pattern details
\(- a(char([^()]*)- Capturing group 1: a negated character class matching any 0 or more chars other than(and)\)- a)char.
To get the whole match, grab Group 0 value, if you need the text inside parentheses, grab Group 1 value.
Most up-to-date JavaScript code demo (using matchAll):
const strs = ["I expect five hundred dollars ($500).", "I expect.. :( five hundred dollars ($500)."];
const rx = /\(([^()]*)\)/g;
strs.forEach(x => {
const matches = [...x.matchAll(rx)];
console.log( Array.from(matches, m => m[0]) ); // All full match values
console.log( Array.from(matches, m => m[1]) ); // All Group 1 values
});Legacy JavaScript code demo (ES5 compliant):
var strs = ["I expect five hundred dollars ($500).", "I expect.. :( five hundred dollars ($500)."];
var rx = /\(([^()]*)\)/g;
for (var i=0;i<strs.length;i++) {
console.log(strs[i]);
// Grab Group 1 values:
var res=[], m;
while(m=rx.exec(strs[i])) {
res.push(m[1]);
}
console.log("Group 1: ", res);
// Grab whole values
console.log("Whole matches: ", strs[i].match(rx));
}Ported Mr_Green's answer to a functional programming style to avoid use of temporary global variables.
var matches = string2.split('[')
.filter(function(v){ return v.indexOf(']') > -1})
.map( function(value) {
return value.split(']')[0]
})
Alternative:
var str = "I expect five hundred dollars ($500) ($1).";
str.match(/\(.*?\)/g).map(x => x.replace(/[()]/g, ""));
→ (2) ["$500", "$1"]
It is possible to replace brackets with square or curly brackets if you need
For just digits after a currency sign : \(.+\s*\d+\s*\) should work
Or \(.+\) for anything inside brackets
let str = "Before brackets (Inside brackets) After brackets".replace(/.*\(|\).*/g, '');
console.log(str) // Inside bracketsvar str = "I expect five hundred dollars ($500) ($1).";
var rex = /\$\d+(?=\))/;
alert(rex.exec(str));
Will match the first number starting with a $ and followed by ')'. ')' will not be part of the match. The code alerts with the first match.
var str = "I expect five hundred dollars ($500) ($1).";
var rex = /\$\d+(?=\))/g;
var matches = str.match(rex);
for (var i = 0; i < matches.length; i++)
{
alert(matches[i]);
}
This code alerts with all the matches.
References:
search for "?=n"
search for "x(?=y)"
Simple:
(?<value>(?<=\().*(?=\)))
I hope I've helped.
