Question

var speed = prompt("Do you know how to type?");
speed = speed.toLowerCase();
if (speed === "yes" ) {
    var howFast = prompt("what is your wpm?(the answer must be a number between 1 and 200");
    switch(howFast) {
        case (howFast <= 10):
        console.log("you are a snail! practice and type at least 20 wpm, then try this again.");
        break;
        case (howFast <= 30):
        console.log("you are still pretty slow, but you're getting there!");
        break;
        case (howFast <= 50):
        console.log("you are getting there, keep trying");
        break;
        case (howFast <= 90):
        console.log("WoW! Excellent job! Your tenacity has paid off");
        break;
        case (howFast > 90):
        console.log("you are a megaracer! congratulations!");
        break;
        default:
        console.log("DOES NOT COMPUTE... You're either superfast or playing around!");



        }

    } else { alert("learn how to type and comeback.");}

我正在尝试在javascript中编写一个简单的switch语句来询问用户的打字速度。令我沮丧的是，当这段代码执行最终警报时，我回来的情况总是默认情况。请告诉我我做错了什么！

Answer 1

只是改变：

switch(howFast) {
..

到

switch(true) {
..

它应该工作。
演示:: jsFiddle

Answer 2

它有点像黑客但你可以在JS case语句中做这样的表达式：

var wpm = parseInt(howFast); // convert string to number first

switch(true)
{
    case wpm >= 0 && wpm <= 10:
    console.log('snail');
    break;

    case wpm > 10 && wpm <= 30:
    console.log('slowpoke');
    break;

    // etc.
}

Answer 3

我有这个工作：

var speed = prompt("Do you know how to type?");
speed = speed.toLowerCase();
if (speed === "yes" ) {
    var howFast = prompt("what is your wpm?(the answer must be a number between 1 and 200");
    switch(true) {
        case (parseInt(howFast) <= 10):
        console.log("you are a snail! practice and type at least 20 wpm, then try this again.");
        break;
        case (parseInt(howFast) <= 30):
        console.log("you are still pretty slow, but you're getting there!");
        break;
        case (parseInt(howFast) <= 50):
        console.log("you are getting there, keep trying");
        break;
        case (parseInt(howFast) <= 90):
        console.log("WoW! Excellent job! Your tenacity has paid off");
        break;
        case (parseInt(howFast) > 90):
        console.log("you are a megaracer! congratulations!");
        break;
        default:
            console.log("DOES NOT COMPUTE... You're either superfast or playing around!");  
    }
    } else { alert("learn how to type and comeback.");}

jsFiddle：http://jsfiddle.net/kR4cy/6/

希望有所帮助

Answer 4

在这种情况下，没有适用于每种情况的表达式，因此，使用if-else块更有意义而不是切换。

Answer 5

根据我的经验，你的开关盒中不能有操作员;你必须有一个明确的价值。在这种情况下，即使切换看起来更好，也应使用IF ELSE块。

编辑：我还从类似的问题中找到了this的答案。

Answer 6

在比较之前，您需要将提示响应转换为整数，然后您需要将切换更改为一组IF。

<script>
    var speed = prompt("Do you know how to type?");
    var howFast;
    var logMessage;

    speed = speed.toLowerCase();

    if (speed === "yes" ) {
        howFast = parseInt(prompt("what is your wpm?..."));
        logMessage = "DOES NOT COMPUTE... You're either superfast or playing around!";

        if (howFast <= 10)
            logMessage = "you are a snail! practice and type at least 20 wpm, then try this again.";

        if (howFast <= 30)
            logMessage = "you are still pretty slow, but you're getting there!";

        if (howFast <= 50)
            logMessage = "you are getting there, keep trying";

        if (howFast <= 90)
            logMessage = "WoW! Excellent job! Your tenacity has paid off";

        if (howFast > 90)
            logMessage = "you are a megaracer! congratulations!";

        console.log(logMessage);

    } else {

        alert("learn how to type and comeback.");

    }
</script>

简单的开关声明

6 个答案: