I have 2 kinds of strings coming back from an API. They look as follows:
let string1 = "\n1. foo\n2. bar\n3. foobar"
let string2 = "\n\n1. foo.\n\n2. bar.\n\n3. foobar."
To be clear, string 1 will always have 3 items coming back, string 2 has an unknown number of items coming back. But very similar patterns.
What I want to do is pull out the text only from each item into an array.
So from string1 I want ["foo", "bar", "foobar"]
From string2 I want ["foo.", "bar.", "foobar."]
I'm awful at regex but I somehow stumbled myself into an expression that accomplishes this for both string types, however, it uses regex's lookbehind which I'm trying to avoid as it isn't supported in all browsers:
let regex = /(?<=\. )(.*[a-zA-Z])/g;
let resultArray = str.match(regex);
Would someone be able to help me refactor this regex into something that doesn't use lookbehind?
Some notes about the pattern (?<=\. )(.*[a-zA-Z]) that you tried:
The first part asserts a dot and space to the left, but does not take any newlines into account so it could possibly also match on other positions
It does not take the digits into account of matching a list item
The second part of your pattern matches till the last occurrence of a character A-Za-z which does not match ending dot in the examples in string2
All your example strings start with a newline, a number, a dot and 1 or more spaces.
You can get the matches without using a lookbehind, and make the pattern more specific by starting the match with the list item format.
Then capture the rest of the line after it in a capture group.
\n\d+\.[^\S\r\n]+(.+)
Explanation
\n Match a newline\d+\. Match 1+ digits and a dot[^\S\r\n]+ Match 1 or more spaces without newlines(.+) Capture group 1, match 1 or more charsSee a regex demo.
const regex = /\n\d+\.[^\S\r\n]+(.+)/g;
[
"\n1. foo\n2. bar\n3. foobar",
"\n\n1. foo.\n\n2. bar.\n\n3. foobar."
].forEach(s => {
console.log(Array.from(s.matchAll(regex), m => m[1]))
});If the string should also match without a leading newline, you can use an anchor ^ to assert the start of the string and use the multiline flag /m
const regex = /^\d+\.[^\S\r\n]+(.+)/gm;
[
"1. foo\n2. bar\n3. foobar",
"1. foo.\n\n2. bar.\n\n3. foobar."
].forEach(s => {
console.log(Array.from(s.matchAll(regex), m => m[1]))
});OP's code, which uses a String.match, is actually better than the proposed solutions. It only needed a minor tweak to make it work, which is what the question asked:
string1.match(/([A-z].+)/g)
// TEST
[
"\n1. foo\n2. bar\n3. foobar",
"\n\n1. foo.\n\n2. bar.\n\n3. foobar."
].forEach(p => {
console.log( p.match(/([A-z].+)/g) )
});EDIT: see @Oleg-Barabanov 's solution, it's technically a bit quicker.
string.replace(/\n[0-9.]*/g, "").split(" ").slice(1)
\n for the new line0-9 for digits (also could use \d. for the dot after the numberg to replace all.split(" ") to chop it up wherever there's a space (\s)Demo:
let string1 = "\n1. foo\n2. bar\n3. foobar"
let string2 = "\n\n1. foo.\n\n2. bar.\n\n3. foobar."
const parse = (str) => str.replace(/\n[0-9.]*/g, "").split(" ").slice(1)
console.log(parse(string1))
console.log(parse(string2))