切换语句 javascript

Question

我制作了这段代码，但它根本不起作用我应该使用 switch 语句你能告诉我错误在哪里吗非常感谢 :)

var side == parseInt(prompt('Enter a number of side between 3 and 10: '));
var shape ==['triangle','square','pentagon','hexagon','heptagon','octagon','nonagon','octagon'];
switch (shape){
case== 3:
shape=[0];
break;
case== 4:
shape==[1];
break;
case== 5:
shape==[2];
break;
case== 6:
shape==[3];
break;
case== 7:
shape==[4];
break;
case== 8:
shape==[5];
break;
case== 9:
shape==[6];
break;
case== 10:
shape==[7];
break;
}
alert('The shape is' + shape);

Answer 1

您正在创建一个数组 shape=[0]; 未引用具有值的数组。

Answer 2

您有很多语法错误：

case== 3:

应该只是

case 3:

和

shape==[1];

== 测试质量，所以你基本上是在说 shape is the same as an array containing the number 1。

Answer 3

当您应该使用 = 时，您却使用了 ==。在您的开关中，您想检查用户输入（side），然后大小写只需要选项没有符号。最后形状需要重新分配给数组中的值。

var side = parseInt(prompt('Enter a number of side between 3 and 10: '));
var shape =['triangle','square','pentagon','hexagon','heptagon','octagon','nonagon','octagon'];
switch (side){
case 3:
shape=shape[0];
break;
case 4:
shape=shape[1];
break;
case 5:
shape=shape[2];
break;
case 6:
shape=shape[3];
break;
case 7:
shape=shape[4];
break;
case 8:
shape=shape[5];
break;
case 9:
shape=shape[6];
break;
case 10:
shape=shape[7];
break;
}
alert('The shape is ' + shape);

Answer 4

停止使用 == 代替 =。

==是比较校验，=是赋值运算符。

使用开关：

var side = parseInt(prompt('Enter a number of side between 3 and 10')),
shapes = ['triangle','square','pentagon','hexagon','heptagon','octagon','nonagon','octagon'],
shape;
switch (side){
case 3:
    shape = shapes[0];
    break;
case 4:
    shape = shapes[1];
    break;
case 5:
    shape = shapes[2];
    break;
case 6:
    shape = shapes[3];
    break;
case 7:
    shape = shapes[4];
    break;
case 8:
    shape = shapes[5];
    break;
case 9:
    shape = shapes[6];
    break;
case 10:
    shape = shapes[7];
    break;
}
alert('The shape is ' + shape);

更好的方法：

var side = parseInt(prompt('Enter a number of side between 3 and 10')),
shapes =    ['triangle','square','pentagon','hexagon','heptagon','octagon','nonagon','octagon'];

alert('The shape is' + shapes[side - 3]);

Answer 5

您不需要 switch 语句 - 您可以拥有更短更简洁的代码。您可以只分配形状，但请确保检查提示的边是否在范围内。

var shape = ['triangle','square','pentagon','hexagon','heptagon','octagon','nonagon','decagon'];

var side = parseInt(prompt('Enter a number of side between 3 and 10: '));

var userShape = shape[side - 3];

if (side < 3 || side > 10) {
    alert('Not a recognised shape')
} else {
    alert(userShape);
}

除非您确定不会再次使用 shape 数组，否则最好将检查数组的结果分配给一个新变量（此处为 userShape），否则您只是覆盖数组。

另外，十边形是十边形:)

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