String split with RegExp returns empty strings at either end

Question

// pattern -- <some name with an optional 1/2/3 at end separated by a _>
var re = new RegExp("^\(.*\)_\([1-3]\)$")
"".split(re) 
// returns [""] --- OK
"abc_d_4".split(re) 
// returns ["abc_d_4"] --- OK
"abc_d_3".split(re)
// returns ["", "abc_d", "3", ""] --- WHOA! hoped-for was ["abc_d", "3"]
// similarly, "_3".split(re), "abc_d_3_4_3".split(re)

why the extra empty strings at either end in the last case, and how to avoid that? I would definitely appreciate an explanation.

I can see similar questions have been asked before on SO but not this case (or pl point me there)

@sln if you write that up, I can accept that as most concise answer (apologies OceansOnPluto :-( — Dinesh
– Dinesh, Commented Jul 2, 2015 at 1:41
Not sure why you escape ( and ). Although the string ends up being the same as "^(.*)_([1-3])$", it's better if you don't escape them at all "^(.*)_([1-3])$". Anyway, it's better to use RegExp literal instead of calling RegExp constructor when you have a fixed pattern. — nhahtdh
– nhahtdh, Commented Jul 2, 2015 at 3:34

user557597 · Accepted Answer · 2015-07-02 02:15:52Z

1

Moved from comment (per request).

Try to split on /_(?=[1-3]$)/

answered Jul 2, 2015 at 2:15

user557597

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Comments

whatoncewaslost · Accepted Answer · 2015-07-02 01:17:10Z

1

Use .match instead.

"abc_d_3".match(re)

That should return this:

["abc_d_3", "abc_d", "3"]

answered Jul 2, 2015 at 1:17

whatoncewaslost

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1 Comment

Dinesh Over a year ago

I also liked the comment by @sln, and am actually going to use that. It is more concise and the code describes exactly what I am trying to do.

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