Probably dead simple for someone out there, but I want to build a regex that will search across this string:
foo[0][2]
and return
foo
0
2
in the array of matches
So far I have come up with is
\([a-z]*)(\[(\d)\])/g
which only gives back
0: "foo[0]"
1: "foo"
2: "[0]"
3: "0"
It's not repeating onto the next literal array in my string, so not hitting the [2] part.
I'm no regex ninja, and would appreciate some assistance with arriving at the correct expression I need.
Thanks
I think you're looking for this pattern: ([a-z]+)|(\[\d\])
console.log('foo[0][2]'.match(/([a-z]+)|(\[\d\])/g));If you don't want the [] parts, you can try this:
console.log('foo[0][2]'.match(/([a-z]+)|(\d)/g));If you use the pattern ([a-z]*)(\[(\d)\]) you could use a while loop to get the first and the third capturing group but note that using [a-z]* matches 0+ times and could also match the parts in [0][2] for example.
const regex = /([a-z]*)(\[(\d)\])/g;
const str = `foo[0][2]`;
let m;
let result = [];
while ((m = regex.exec(str)) !== null) {
// This is necessary to avoid infinite loops with zero-width matches
if (m.index === regex.lastIndex) {
regex.lastIndex++;
}
if (m[1] !== "") result.push(m[1]);
if (m[3] !== "") result.push(m[3]);
}
console.log(result);If you want to match that exact format, you could use 2 capturing groups to first match foo in group 1 and [0][2] in group 2
\b([a-z]+)((?:\[\d])+)
Explanation
\b Word boundary( Capture group 1
[a-z]+ Match 1+ times a char a-z) Close group( Capture group 2
(?:\[\d])+ Repeat 1+ times matching a single digit between [] (use \d+ to match 1 or more)) Close groupIf that pattern matches you could get all the digits from group 2 as the overall pattern already matches.
const pattern = /\b([a-z]+)((?:\[\d])+)/;
let parts = "foo[0][2]".match(pattern);
console.log([parts[1], ...parts[2].match(/\d/g)]);
