If you are using JavaScript, the solution provided by cletus, (?<=\[)(.*?)(?=\]) won't work because JavaScript doesn't support the lookbehind operator.

Edit: actually, now (ES2018) it's possible to use the lookbehind operator. Just add / to define the regex string, like this:

var regex = /(?<=\[)(.*?)(?=\])/;

Old answer:

Solution:

var regex = /\[(.*?)\]/;
var strToMatch = "This is a test string [more or less]";
var matched = regex.exec(strToMatch);

It will return:

["[more or less]", "more or less"]

So, what you need is the second value. Use:

var matched = regex.exec(strToMatch)[1];

To return:

"more or less"

(?<=period_1_)(.*)(?=.ssa)

This extracts the part between "period_1_" and ".ssa".

https://regexr.com/3jtbm

A very useful solution to this issue which doesn't require from you to do the indexOf is using Apache Commons libraries.

 StringUtils.substringBetween(s, "(", ")");

This method will allow you even handle even if there multiple occurrences of the closing string which wont be easy by looking for indexOf closing string.

You can download this library from here: https://mvnrepository.com/artifact/org.apache.commons/commons-lang3/3.4

Try this regular expression:

@([^.]*)\.

The expression [^.]* will match any number of any character other than the dot. And the plain dot needs to be escaped as it’s a special character.

Lazy Quantifier Needed

Resurrecting this question because the regex in the accepted answer doesn't seem quite correct to me. Why? Because

(?<=This is)(.*)(?=sentence)

will match my first sentence. This is my second in This is my first sentence. This is my second sentence.

See demo.

You need a lazy quantifier between the two lookarounds. Adding a ? makes the star lazy.

This matches what you want:

(?<=This is).*?(?=sentence)

See demo. I removed the capture group, which was not needed.

DOTALL Mode to Match Across Line Breaks

Note that in the demo the "dot matches line breaks mode" (a.k.a.) dot-all is set (see how to turn on DOTALL in various languages). In many regex flavors, you can set it with the online modifier (?s), turning the expression into:

(?s)(?<=This is).*?(?=sentence)

Reference

• The Many Degrees of Regex Greed
• Repetition with Star and Plus

How about

yourStringVariable.match(/\[(.*)\]/)[1]

Regular expression to get a string between two strings in JavaScript

The most complete solution that will work in the vast majority of cases is using a capturing group with a lazy dot matching pattern. However, a dot . in JavaScript regex does not match line break characters, so, what will work in 100% cases is a [^] or [\s\S]/[\d\D]/[\w\W] constructs.

ECMAScript 2018 and newer compatible solution

In JavaScript environments supporting ECMAScript 2018, s modifier allows . to match any char including line break chars, and the regex engine supports lookbehinds of variable length. So, you may use a regex like

var result = s.match(/(?<=cow\s+).*?(?=\s+milk)/gs); // Returns multiple matches if any
// Or
var result = s.match(/(?<=cow\s*).*?(?=\s*milk)/gs); // Same but whitespaces are optional

In both cases, the current position is checked for cow with any 1/0 or more whitespaces after cow, then any 0+ chars as few as possible are matched and consumed (=added to the match value), and then milk is checked for (with any 1/0 or more whitespaces before this substring).

Scenario 1: Single-line input

This and all other scenarios below are supported by all JavaScript environments. See usage examples at the bottom of the answer.

cow (.*?) milk

cow is found first, then a space, then any 0+ chars other than line break chars, as few as possible as *? is a lazy quantifier, are captured into Group 1 and then a space with milk must follow (and those are matched and consumed, too).

Scenario 2: Multiline input

cow ([\s\S]*?) milk

Here, cow and a space are matched first, then any 0+ chars as few as possible are matched and captured into Group 1, and then a space with milk are matched.

Scenario 3: Overlapping matches

If you have a string like >>>15 text>>>67 text2>>> and you need to get 2 matches in-between >>>+number+whitespace and >>>, you can't use />>>\d+\s(.*?)>>>/g as this will only find 1 match due to the fact the >>> before 67 is already consumed upon finding the first match. You may use a positive lookahead to check for the text presence without actually "gobbling" it (i.e. appending to the match):

/>>>\d+\s(.*?)(?=>>>)/g

See the online regex demo yielding text1 and text2 as Group 1 contents found.

Also see How to get all possible overlapping matches for a string.

Performance considerations

Lazy dot matching pattern (.*?) inside regex patterns may slow down script execution if very long input is given. In many cases, unroll-the-loop technique helps to a greater extent. Trying to grab all between cow and milk from "Their\ncow\ngives\nmore\nmilk", we see that we just need to match all lines that do not start with milk, thus, instead of cow\n([\s\S]*?)\nmilk we can use:

/cow\n(.*(?:\n(?!milk$).*)*)\nmilk/gm

See the regex demo (if there can be \r\n, use /cow\r?\n(.*(?:\r?\n(?!milk$).*)*)\r?\nmilk/gm). With this small test string, the performance gain is negligible, but with very large text, you will feel the difference (especially if the lines are long and line breaks are not very numerous).

Sample regex usage in JavaScript:

//Single/First match expected: use no global modifier and access match[1]
console.log("My cow always gives milk".match(/cow (.*?) milk/)[1]);
// Multiple matches: get multiple matches with a global modifier and
// trim the results if length of leading/trailing delimiters is known
var s = "My cow always gives milk, thier cow also gives milk";
console.log(s.match(/cow (.*?) milk/g).map(function(x) {return x.substr(4,x.length-9);}));
//or use RegExp#exec inside a loop to collect all the Group 1 contents
var result = [], m, rx = /cow (.*?) milk/g;
while ((m=rx.exec(s)) !== null) {
  result.push(m[1]);
}
console.log(result);

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Using the modern String#matchAll method

const s = "My cow always gives milk, thier cow also gives milk";
const matches = s.matchAll(/cow (.*?) milk/g);
console.log(Array.from(matches, x => x[1]));

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