切片不是函数

问题描述 投票:0回答:1

尽管我将切片从数字转换为字符串,但它给了我一个错误,即切片不起作用。 问题出在哪里?

var weight = prompt("Enter your Weight");
var height = prompt("Enter your Height");
function bmi(weight, height) {
  var result = weight / Math.pow(height, 2);
  var count = result.toString.slice(0, 4);
  if (result < 18.5) {
    alert("Your result is " + count + ", so you are underweight.");
  } else if (result >= 18.5 && result <= 24.9) {
    alert("Your result is " + count + ", so you have a normal weight.");
  } else {
    alert("Your result is " + count + ", so you are overweight.");
  }
}

bmi(weight, height);

我将切片从数字转换为字符串,但给了我同样的错误,问题出在哪里?

function slice tostring
1个回答
0
投票

你实际上并没有打电话给

toString()

toString
解析为函数对象,然后您可以使用
()
调用该函数对象以获取字符串。您看到的错误告诉您函数对象本身没有
slice()
方法,这是正确的。

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