Why this switch statement isn't work perfectly?-CodePudding

const number = prompt("Enter your number");
const txt = "You result is : ";

switch (number) {
  case (number >= 80 && number <= 100):
    document.write(`${txt} A `);
    break;
  case (number >= 70 && number <= 80):
    document.write(`${txt} A gread`);
    break;
  case (number >= 60 && number <= 70):
    document.write(`${txt} B gread`);
    break;
  case (number >= 50 && number <= 60):
    document.write(`${txt} C gread`);
    break;
  case (number >= 33 && number <= 50):
    document.write(`${txt} D gread`);
    break;
  case (number >= 0 && number <= 33):
    document.write(`${txt}  Field !`);
    break;
  case (number > 100 || number < 0):
    document.write(`It's not a valid number. Please input any valid number.`);
    break;
  default:
    document.write(`Not input any number. Please input any number .`);
    break;
}

CodePudding user response：

With a switch-statement, you evaluate the expression right after switch. In your case that would be number.

So, your cases are evaluated with the number. This is what will happen:

Suppose, we enter '88' in the prompt.
The first thing you are evaluating number against is (number >= 80 && number <= 100). The result of the latter is true.
now the value of number, 88 is compared to true, like this: 88===true.
this results in a false, so the next case is evaluated.
the same happens for the following cases, right until the point you hit the default-case.

CodePudding user response：

There are two reasons why this code will not work:

The first reason is because the prompt method return a string as you can see by typing

  console.log('Type of number:', typeof number);
  // Expected output: Type of number: string

You can resolve this by parsing the output of the prompt method as follows

  const number = parseInt(prompt("Enter your number"));

The second reason is because in a switch statement statement, the evaluated value of the switch expression is compared the the evaluated values of the cases.

It means that the value of number (number) is compared to the expression (number >= 80 && number <= 100) (comparison expression) and so on with the others.

So unless one of yoru case expression yield a number the switch will try to evaluate the value of number to the expression (which is always true) meaning that the switch is comparing 45 === true, not matching any of the cases.

Just use if/else instead:

if (number >= 80 && number <= 100)
    document.write(`${txt} A `);
else if (number >= 70 && number <= 80)
    document.write(`${txt} A gread`);
else if (number >= 60 && number <= 70)
    document.write(`${txt} B gread`);
else if (number >= 50 && number <= 60)
    document.write(`${txt} C gread`);
else if (number >= 33 && number <= 50)
    document.write(`${txt} D gread`);
else if (number >= 0 && number <= 33)
    document.write(`${txt}  Field !`);
else if (number > 100 || number < 0)
    document.write(`It's not a valid number. Please input any valid number.`);
else
    document.write(`Not input any number. Please input any number .`);