Problem and Solution

This page carries the full answer

Problem: A replace call looked broken, but the real issue was that the string behaved like a URI and needed the right encoding and decoding strategy.

Original question: replace() Method don't replace anything by user3944364.

Full imported answer

Updated:

var str   = encodeURI(cup[number][9]);
str = decodeURI( str.replace( encodeURI('amazon®ion'), encodeURI('amazon&region') );

encodeURI() & decodeURI()

• Since it's a url, this is the favorable function
• With the use of encodeURI('amazon®ion') you don't have to seccond guess the encoded character(s)
• & in amazon&region needs to be in the form of &, like so: amazon&region
• use decodeURI() to use the new URI

JsFiddle: http://jsfiddle.net/cbsL9ed7/2/

Want an even better and more flexible solution?

Why not make it a function then?

function updateURI(uri, oldValue, newValue) {
    var str = encodeURI( uri );
    str = decodeURI( str.replace( encodeURI(oldValue), encodeURI(newValue) ) );
    return str;
}


var newURI = updateURI(cup[number][9], 'amazon®ion', 'amazon&region');
document.write( newURI )

jsFiddle: http://jsfiddle.net/cbsL9ed7/4/

Running Demo

Working code in-house

This demo is embedded here so the page carries the working result, not just a link away from the portfolio.

Source and Credit

Also posted on Stack Overflow

This page keeps the full solution in-house, while still pointing back to the original Stack Overflow post for proof and attribution.

Original answer: Stack Overflow answer #29685434 by Omar Juvera.

License note: this answer was originally published on Stack Overflow under CC BY-SA 3.0. Keeping a small source note here is the cleanest way to preserve attribution while the portfolio page remains the main destination.