Match all consecutive numbers of length n [duplicate] - regex

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How to use regex to find all overlapping matches
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Where n=4 in my example.
I'm very new to Regex and have searched for 20 minutes now. There are some helpful websites out there that simplify things but I can't work out how to proceed with this.
I wish to extract every combination of 4 consecutive digits from this:
12345
to get:
1234 - possible with ^\d{4}/g - Starts at the beginning
2345 - possible with \d{4}$/g - Starts at the end
But I can't get both! The input could be any length.

Answer 1

Your expression isn't working as expected because those two sub-strings are overlapping.
Aside from zero-length assertions, any characters in the input string will be consumed in the matching process, which results in the overlapping matches not being found.
You could work around this by using a lookahead and a capturing group to retrieve the overlapping matches. This works because lookahead assertions (as well as lookbehind assertions) are classified as zero-length assertions, which means that they don't consume the matches; thereby allowing you to find any overlapping matches.
(?=(\d{4}))
Here is a quick snippet demonstrating this:
var regex = /(?=(\d{4}))/g;
var input = '12345678';
var match;
while ((match = regex.exec(input)) !== null) {
if (match.index === regex.lastIndex) {
regex.lastIndex++;
}
console.log(match[1]);
}

Answer 2

You can use a lookahead with a capturing group:
(?=(\d{4}))
See demo

Answer 3

Use a look ahead assertion with all the possibilities
(?=(0123|1234|2345|3456|4567|5678|6789))
(?=
( # (1 start)
0123
| 1234
| 2345
| 3456
| 4567
| 5678
| 6789
) # (1 end)
)
Output
** Grp 0 - ( pos 0 , len 0 ) EMPTY
** Grp 1 - ( pos 0 , len 4 )
1234
------------------
** Grp 0 - ( pos 1 , len 0 ) EMPTY
** Grp 1 - ( pos 1 , len 4 )
2345