Alternative to nested ternary operator in JS

Your alternatives here are basically:

1. That if/else you don't want to do
2. A switch combined with if/else

I tried to come up with a reasonable lookup map option, but it got unreasonable fairly quickly.

I'd go for #1, it's not that big:

if (res.distance == 0) {    word = 'a';} else if (res.distance == 1 && res.difference > 3) {    word = 'b';} else if (res.distance == 2 && res.difference > 5 && String(res.key).length > 5) {    word = 'c';} else {    word = 'd';}

If all the braces and vertical size bother you, without them it's almost as concise as the conditional operator version:

if (res.distance == 0) word = 'a';else if (res.distance == 1 && res.difference > 3) word = 'b';else if (res.distance == 2 && res.difference > 5 && String(res.key).length > 5) word = 'c';else word = 'd';

(I'm not advocating that, I never advocate leaving off braces or putting the statement following an if on the same line, but others have different style perspectives.)

#2 is, to my mind, more clunky but that's probably more a style comment than anything else:

word = 'd';switch (res.distance) {    case 0:        word = 'a';        break;    case 1:        if (res.difference > 3) {            word = 'b';        }        break;    case 2:        if (res.difference > 5 && String(res.key).length > 5) {            word = 'c';        }        break;}

And finally, and I am not advocating this, you can take advantage of the fact that JavaScript's switch is unusual in the B-syntax language family: The case statements can be expressions, and are matched against the switch value in source code order:

switch (true) {    case res.distance == 0:        word = 'a';        break;    case res.distance == 1 && res.difference > 3:        word = 'b';        break;    case res.distance == 2 && res.difference > 5 && String(res.key).length > 5:        word = 'c';        break;    default:        word = 'd';        break;}

How ugly is that? :-)

To my taste, a carefully structured nested ternary beats all those messy ifs and switches:

const isFoo = res.distance === 0;const isBar = res.distance === 1 && res.difference > 3;const isBaz = res.distance === 2 && res.difference > 5 && String(res.key).length > 5;const word =  isFoo ? 'a' :  isBar ? 'b' :  isBaz ? 'c' :          'd' ;

You could write an immediately invoked function expression to make it a little more readable:

const word = (() =>  {  if (res.distance === 0) return 'a';  if (res.distance === 1 && res.difference > 3) return 'b';  if (res.distance === 2 && res.difference > 5 && String(res.key).length > 5) return 'c';  return 'd';})();

Link to repl