Question

我制作了这段代码，但它根本不起作用＆＃xA;我应该使用switch语句＆＃xA;你能告诉我哪里出错了＆＃xA;感谢战利品：）

＆＃xA;＆＃xA;

  var side == parseInt（prompt（'输入3到10之间的数字：'））;＆＃xA; var shape == ['triangle'， 'square'，'pentagon'，'hexagon'，'heptagon'，'octagon'，'nonagon'，'octagon'];＆＃xA; switch（shape）{＆＃xA; case == 3：＆＃xA ; shape = [0];＆＃xA; break;＆＃xA; case == 4：＆＃xA; shape == [1];＆＃xA; break;＆＃xA; case == 5：＆＃ xA; shape == [2];＆＃xA; break;＆＃xA; case == 6：＆＃xA; shape == [3];＆＃xA; break;＆＃xA; case == 7： ＆＃xA; shape == [4];＆＃xA; break;＆＃xA; case == 8：＆＃xA; shape == [5];＆＃xA; break;＆＃xA; case == 9：＆＃xA; shape == [6];＆＃xA; break;＆＃xA; case == 10：＆＃xA; shape == [7];＆＃xA; break;＆＃xA;} ＆＃xA;警告（'形状是'+形状）; ＆＃XA;

＆＃XA;

Answer 1

您正在使数组shape=[0];不引用具有值的数组。

Answer 2

你有很多语法错误：

case== 3:

应该只是

case 3:

和

shape==[1];

==测试质量，因此您基本上是在说shape is the same as an array containing the number 1。

Answer 3

当您使用==时，您正在使用=。在您的交换机中，您要检查用户输入（side），然后只需要选项无符号。最后，需要将形状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=。

==是比较检查，=是一个赋值运算符。

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 - 您可以使用更短更简洁的代码。您可以只指定形状，但请确保检查提示侧是否在范围内。

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);
}

除非您确定不再使用shape数组，否则最好将检查数组的结果分配给新变量（此处{{1否则你只是覆盖了数组。

另外，十边形是十字形：）

DEMO

切换状态javascript

5 个答案: