Javascript regular expression to extract characters from mid string with optional end character

Question

I would like to extract characters from mid string with optional end character. If the optional end character is not found, extract until end of string. The first characters are S= and the last optional character is &.

Example #1:

"rilaS=testingabc"

should extract:

"testingabc"

Example #2:

"rilaS=testing123&thistest"

should extract:

"testing123"

This is what I have so far (Javascript):

var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=(.*)[\&]{0,1}");
var newStr = tmpStr[1];
alert(newStr);

But it does not detect that the end should be the ampersand (if found). Thank you before hand.

Answer (By ggorlen)

var Str = "rilaS=testing123&thistest";
var tmpStr = Str.match("S=([^&]*)");
var newStr = tmpStr[1];
alert(newStr);

Answer 1

You may use /S=([^&]*)/ to grab from an S= to end of line or & :

 ["rilaS=testingabc", "rilaS=testing123&thistest"].forEach(s => console.log(s.match(/S=([^&]*)/)[1]) ); 

Answer 2

Just in case you are wondering why your original regex didn't work: the problem is that the (.*) pattern is greedy - meaning it will happily slurp up anything, including &, and not leave it for for later items to match. This is why you want the "not &" - it will match up to, but not including the &.