javascript - (How) Can i use 'or' in a switch statement? -

is following code right:

var user = prompt("is programming awesome?").touppercase(); switch (user) {     case 'hell yeah' || 'yes' || 'yeah' || 'yep' || 'yea':         console.log("that's i'd expected!");     break;     case 'kinda'||'cant say':         console.log("oh, you'd when you'll pro!");     break;     case 'hell no'||'no'||'nah'||'nope'||'na':         console.log("you can't that!");     break;     default:         console.log("wacha tryna mate?");     break; } 

i tried it. didn't work wanted to. worked if input 'hell no' or 'hell yeah', later strings ignored!

i approach using dictionary (object) containing answers can check function in switch statement. it's lot neater having multiple case checks:

var user = prompt("is programming awesome?").touppercase(); // 'hell yeah';  // 'dictionary' js object has key/value pairs // in instance have keys map case checks // making , each has corresponding array value, filled // or checks wanted make. // pass 'dict' in second parameter when call checkanswer var dict = {     yes: ['hell yeah', 'yes', 'yeah', 'yep', 'yea'],     maybe: ['kinda', 'cant say'],     no: ['hell no', 'no', 'nah', 'nope', 'na'] };  // checkanswer returns 'yes' user prompt 'hell yeah' // loops on 'dict' object , if finds instance of // user input in 1 of arrays returns key // in instance 'yes' 'hell yeah' (indexof returns 0) // otherwise returns false (for 'default' case) function checkanswer(user, dict) {     (var p in dict) {         if (dict[p].indexof(user) > -1) return p;     }     return false; }  // use checkanswer return key // user input found in 1 of dictionary arrays. // single string switch expects switch (checkanswer(user, dict)) {     case 'yes':         console.log("that's i'd expected!");         break;     case 'maybe':         console.log("oh, you'd when you'll pro!");         break;     case 'no':         console.log("you can't that!");         break;     default:         console.log("wacha tryna mate?");         break; } 

demo