Question

Since I started writing this question, I think I figured out the answers to every question I had, but I thought I’d post anyway, as it might be useful to others and more clarification might be helpful.

I was trying to use a regular expression with lookahead with the javascript function split. For some reason it was not splitting the string even though it finds a match when I call match. I originally thought the problem was from using lookahead in my regular expression. Here is a simplified example:

Doesn’t work:

"aaaaBaaaa".split("(?=B).");

Works:

"aaaaBaaaa".match("(?=B).");

It appears the problem was that in the split example, the passed string wasn’t being interpreted as a regular expression. Using forward slashes instead of quotes seems to fix the problem.

"aaaaBaaaa".split(/(?=B)./);

I confirmed my theory with the following silly looking example:

"aaaaaaaa(?=B).aaaaaaa".split("(?=B).");

Does anyone else think it’s strange that the match function assumes you have a regular expression while the split function does not?

Answer 1

String.split accepts either a string or regular expression as its first parameter. The String.match method only accepts a regular expression.

I’d imagine that String.match will try and work with whatever is passed; so if you pass a string it will interpret it as a regular expression. The String.split method doesn’t have the luxury of doing this because it can accept regular expressions AND strings; in this case it would be foolish to second-guess.

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Edit: (From: “JavaScript: The Definitive Guide”)

String.match requires a regular expression to work with. The passed argument needs to be a RegExp object that specifies the pattern to be matched. If this argument is not a RegExp, it is first converted to one by passing it to the RegExp() constructor.