使用条件的JavaScript switch-case

Question

I just started learning java script and I have really basic knowledge in C#.

Tried running that piece of code:

var number = parseInt(prompt("Enter a number:"));
    switch(number)
        {
            case number >= 1 && number <= 9: alert("The number " + number + " is a single digit number."); break;
            case number >= 10 && number <= 99: alert("The number " + number + " is a two digit number."); break;
            case number >= 100 && number <= 999: alert("The number " + number + " is a three digit number."); break;
            case number >= 1000 && number <= 9999: alert("The number " + number + " is a four digit number."); break;
            default: alert("Your number has 5 or more digits.");
        }

but it kept failing after the -

number <= 9

can you guys please give me a hand here?

Thanks a lot!

Answer 1

The switch statement does compare its argument with the values of the expressions in the cases - in your code, it does compare the number with those boolean results from the cases, so it only gets in the first case when your number is 1 (because 1 == true ) and falls to the default otherwise. You could simply compare them to true , like

switch(true)
    {
        case number >= 1 && number <= 9: alert("The number " + number + " is a single digit number."); break;
        case number >= 10 && number <= 99: alert("The number " + number + " is a two digit number."); break;
        case number >= 100 && number <= 999: alert("The number " + number + " is a three digit number."); break;
        case number >= 1000 && number <= 9999: alert("The number " + number + " is a four digit number."); break;
        default: alert("Your number has 5 or more digits.");
    }

but it would be cleaner to use an if-else construct:

if( number >= 1 && number <= 9)
    alert("The number " + number + " is a single digit number.");
else if( number >= 10 && number <= 99)
    alert("The number " + number + " is a two digit number.");
else if( number >= 100 && number <= 999)
     alert("The number " + number + " is a three digit number.");
else if( number >= 1000 && number <= 9999)
     alert("The number " + number + " is a four digit number.");
else
     alert("Your number has 5 or more digits.");

Btw, much shorter would be

var l = String(number).length;
alert( l<5
   ? "The number "+number+" is a "+[,"single","two","three","four"][number]+" digit number."
   : "Your number has 5 or more digits."
);

Answer 2

你需要切换（true），其余的保持不变（我猜）。

Answer 3

What all the other answers are missing so far is the fact that you don't need all these conditions.

if (number >= 10000) {
    ...
} else if (number >= 1000) {
    ...
} else if (number >= 100) {
    ...
} else if (number >= 10) {
    ...
} else if (number >= 1) {
    ...
} else {
    ...
}

Of course there are other ways, too. For this particular example, an easy way would be to use

var digits = (number >= 0) ? number.toString().length : 0;

For positive integers it will be the number of digits, for negative numbers it will simply be 0 . A switch/case statement over this number would be possible.

A switch(true) is terrible style. It simply isn't what a switch/case is intended for.

Answer 4

Do it like this way using if/else:

if( number >= 1 && number <= 9)
    alert("The number " + number + " is a single digit number.");
else if( number >= 10 && number <= 99)
    alert("The number " + number + " is a two digit number.");
else if( number >= 100 && number <= 999)
     alert("The number " + number + " is a three digit number.");
else if( number >= 1000 && number <= 9999)
     alert("The number " + number + " is a four digit number.");
else
     alert("Your number has 5 or more digits.");

Or Try using switch(true) as condition is to be compared with Boolean expression.