Question

I am new to regular expression and this may be a very easy question (hopefully).

I am trying to use one solution for 3 kinds of string

• "45%", expected result: "45"
• "45", expected result: "45"
• "", expected result: ""

What I am trying (let the string be str):

str.match(/(.*)(?!%*)/i)[1]

This is in my head would sound like "match any instance of anything up until ‘%’ if it is found, or else just match anything"

In firebug’s head, it seems to sound more like "just match anything and completely disregard the negative lookahead". Also to make it lazy – (.*)? – doesn’t seem to help.

Let’s forget for a second that in this specific situation I am only matching numbers, so a /\d*/ would do. I am trying to understand a general rule so that I can apply it whenever.

Anybody would be so kind to help me out?

Answer 1

How about the simpler

str.match(/[^%]*/i)[0]

Which means, match zero-or-more character, which is not a %.

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Edit: If need to parse until </a>, then you could parse a sequence pf characters, followed by </a>, then then discard the </a>, which means you should use positive look-ahead instead of negative.

str.match(/.*?(?=<\/a>|$)/i)[0]

This means: match zero-or-more character lazily, until reaching a </a> or end of string.

Note that *? is a single operator, (.*)? is not the same as .*?.

(And don’t parse HTML with a single regex, as usual.)