I have found very similar posts, but I can\'t quite get my regular expression right here.
I am trying to write a regular expression which returns a string which is b
I find regex to be tedious and time consuming given the syntax. Since you are already using javascript it is easier to do the following without regex:
const text = 'My cow always gives milk'
const start = `cow`;
const end = `milk`;
const middleText = text.split(start)[1].split(end)[0]
console.log(middleText) // prints "always gives"
If the data is on multiple lines then you may have to use the following,
/My cow ([\s\S]*)milk/gm
My cow always gives
milk
Regex 101 example
A lookahead (that (?=
part) does not consume any input. It is a zero-width assertion (as are boundary checks and lookbehinds).
You want a regular match here, to consume the cow
portion. To capture the portion in between, you use a capturing group (just put the portion of pattern you want to capture inside parenthesis):
cow(.*)milk
No lookaheads are needed at all.
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.
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).
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).
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.
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.
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);
Using the modern
String#matchAll
methodconst 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]));
I was able to get what I needed using Martinho Fernandes' solution below. The code is:
var test = "My cow always gives milk";
var testRE = test.match("cow(.*)milk");
alert(testRE[1]);
You'll notice that I am alerting the testRE variable as an array. This is because testRE is returning as an array, for some reason. The output from:
My cow always gives milk
Changes into:
always gives
Just use the following regular expression:
(?<=My cow\s).*?(?=\smilk)