Regexp for "IF textTrue ELSE textFalse"

Question

I have two strings:

"~if~ text_if_true ~end~"
"~if~ text_if_true ~else~ text_if_false ~end~"

All I want is a regexp with give following output:

for 1. =>

group 1 = text_if_true

for 2. =>

group 1 = text_if_true, group 2 = text_if_false

I tried:

~if~(.*?)(?:~else~)(.*?)~end~

Work fine on II but not on I because the ~else~ is needed

if I use

~if~(.*?)(?:~else~)?(.*?)~end~   

With a ? behind (?:~else~) (for 0 or 1 matches) the text_if_true and the text_if_false is in the first group.

Is there a easy way to solve the problem?

const regex = /~if~(.*?)(?:~else~)(.*?)~end~/gm;
const regex2 = /~if~(.*?)(?:~else~)?(.*?)~end~/gm;

const str = `~if~ text_if_true ~else~ text_if_false ~end~
~if~ text_if_true ~end~`;
let m;


console.log('without ? quantifier')
while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

console.log('with ? quantifier')
while ((m = regex2.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

Answer 1

Instead of making just the word ~else~ optional, you should make the whole of ~else~ (.*?) optional (I have also added some spaces so that the groups don't have leading or trailing spaces):

if~ (.*?)(?: ~else~ (.*?))? ~end

Demo