Matching if/else statement with regular expression

Question

I am currently writing a regular expression which will allow me to use a basic conditional if/else system written in JavaScript. (I know there is probably easier ways to do this!)

This is the content I need to work with:

{?if language = german}Hallo!{?else}Hello!{?endif} {?if language = german}{#greeting}{?endif}

I have written /{\?if ([^{}]+)}(.+)(?:{\?else})?(.+)?{\?endif}/g which unfortunately only matches part of this. I have been using RegExr for testing so far.

My expected output, using the output expression 1: $1\n2: $2\n3: $3\n, would be:

1: language = german 2: Hallo! 3: Hello! 1: language = german 2: {#greeting} 3:

Unfortunately, I'm getting this instead:

1: language = german 2: Hallo!{?else}Hello! 3: 1: language = german 2: {#greeting} 3:

What am I doing wrong here? I'm relatively new to writing regular expressions, so I assume there is a way, and I'm just not doing it correctly.

I agree that there are easier ways to do this! But, +1 for a good clear question. — Smandoli
– Smandoli, Commented Aug 29, 2014 at 21:30
Try using non-greedy quantifiers. Since else is optional, the greedy quantifier is going past it. — Barmar
– Barmar, Commented Aug 29, 2014 at 21:30

hwnd · Accepted Answer · 2014-08-29 21:45:15Z

You first want to make the + operators non-greedy. Then instead of making the third capturing group optional, simply place that group inside of the Non-capturing group leaving the Non-capturing group optional.

/{\?if ([^{}]+)}(.+?)(?:{\?else}(.+?))?{\?endif}/g

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