multiple conditions within a while loop - Javascript

Question

I want to prompt the user for an input and if it isn't a valid option reprompt until a valid input is entered.

Right now it always reprompts not matter the input, correct or not.

Here is the code:

var userChoice = prompt("Do you choose rock, paper, scissors or rope?");
while (userChoice != "rock" || "paper" || "scissors" || "rope") {
    userChoice = prompt("Sorry invalid input, please enter either: rock, paper,scissors, or rope.");
}

Seems like it should be a simple thing to do, am I misunderstanding how to use while loops? maybe the operators are wrong? haven't been able to find anything useful, any help is appreciated!

@djechlin For hanging yourself after you get terminally bored of the game, instant loss. ;-) — Platinum Azure
– Platinum Azure, Commented Jan 4, 2016 at 21:39

smcd · Accepted Answer · 2016-01-04 21:39:36Z

5

You could also use an array and indexOf

var userChoice = prompt("rock, paper, scissors, rope?").toLowerCase();
while (["rock", "paper", "scissors", "rope"].indexOf(userChoice) === -1) {
    userChoice = prompt("Please choose: rock, paper, scissors, rope?").toLowerCase();
}

alert("You chose '" + userChoice + "'");

answered Jan 4, 2016 at 21:39

smcd

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3 Comments

T_Turner Over a year ago

First off thanks for the help, so let me try an interpret your code, I really am that new to this ha! The initial prompt asks for the choice (which will all display in lowercase), then in the while loop you created an array using the string "rock" etc, "rock" being 0, "paper" being 1 in that array? then indexof(userchoice) searches for the inputs I want aka the parameters of the array. the -1 means if it is not any of the correct answers it will reprompt and if correct will display the choice with the alert? That right?

T_Turner Over a year ago

and just looked up .indexOf so can see the -1 means unsuccessful search, are there any meaningful differences between .indexOf and str.search? They both appear to provide similar functions

smcd Over a year ago

To your first comment that sounds like a correct understanding of the code. Strings have an indexOf function which provides functionality comparable to that of array's indexOf (providing the offset of the found item, or -1 if no match).

willeM_ Van Onsem · Accepted Answer · 2016-01-04 21:39:19Z

4

The syntax "rock" || "paper" is valid, but the OR will bind later than the equality check, so basically you write:

while ((userChoice != "rock") || "paper" || "scissors" || "rope") {
    //...
}

Now since you use such or's, you basically do an effectiveness check and at least "paper" is effective. You can however get the behavior you aim, by using explicit checks and using the && operator instead of the ||:

while (userChoice !== "rock" && userChoice !== "paper" && userChoice !== "scissors" && userChoice !== "rope") {
    //...
}

answered Jan 4, 2016 at 21:39

willeM_ Van Onsem

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Comments

Hogan · Accepted Answer · 2016-01-04 21:36:57Z

2

While userChoice does not equal any of the choices... so AND

while (userChoice != "rock" && userChoice != "paper" && userChoice != "scissors" && userChoice != "rope") {

answered Jan 4, 2016 at 21:36

Hogan

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Comments

Nina Scholz · Accepted Answer · 2016-01-04 21:39:23Z

2

Maybe an array is smarter to use with Array.prototype.indexOf().

var userChoice;
while (!~["rock", "paper", "scissors", "rope"].indexOf(userChoice)) {
    userChoice = prompt("Sorry invalid input, please enter either: rock, paper,scissors, or rope.");
}

answered Jan 4, 2016 at 21:39

Nina Scholz

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Comments

Aprillion · Accepted Answer · 2016-01-04 22:19:56Z

2

for more complicated rules (such as ignoring case sensitivity), it is possible to use regular expressions, e.g. match:

while( ! userChoice.match(/^\s*(?:rock|paper|scissors|rope)\s*$/i)) { ... }

(see explanation of the regex on https://regex101.com/r/sV3tJ1/1#javascript)

^{note: thanks Willem Van Onsem for suggesting regex}

also avaliable are object literals:

var validChoices = {rock: true,
                    paper: true,
                    scissors: true,
                    rope: true},
    userChoice = ...
while ( ! validChoices.hasOwnProperty(userChoice)) {
  ...
}

or ES6 Set for the newest browsers as of 2016: new Set(["rock", "paper", ...])

edited Jan 4, 2016 at 22:19

answered Jan 4, 2016 at 21:45

Aprillion

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2 Comments

willeM_ Van Onsem Over a year ago

In case you want to check against a huge set of values, a regex will probably be more efficient. Nevertheless +1 for creativity ;)

Aprillion Over a year ago

@WillemVanOnsem actually, regex will be better even for a case insensitive solution accepting stray spaces in this particular problem, thanks for the suggestion

ruslaniv · Accepted Answer · 2019-02-21 08:37:41Z

I'll try a more generic answer:

The while loop runs while whatever in the brackets is TRUE.

The way || works, it will evaluate to TRUE if at least one of the operands is TRUE.

So in your case even if you select

userChoice = "rock";
userChoice != "rock" => FALSE

whatever comes after || will still evaluate to TRUE because

"rock" != "paper" => TRUE

and thus your whole condition in the brackets will evaluate to TRUE.

So in order to break out of the loop, you need to have the condition in the brackets evaluate to FALSE.

The way && works, it will evaluate to FALSE if at least one of the operands evaluates to FALSE.

And since

userChoice = "rock";
userChoice != "rock" => FALSE

the loop does not even check the rest of the conditions you might have and breaks out of the loop.

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