javascript - switch...case...

Question

export const setID = (v) => {
  console.log('执行setID',v);
  let l = v.length;
  switch(l)
  {
    case l < 6 :
      console.log('qq');
      break;
    default:
      console.log('11111');
  }

}

v is the incoming string. The first console of this code is executed normally, and the second console is not executed under any circumstances. Now I am sure that there is something wrong with my switch...case.... Please tell me what the problem is

Answer 1

Change it to the following

const setID = (v) => {
  console.log('执行setID',v);
  let l = v.length;
  switch(l < 6)
  {
    case true :
      console.log('qq');
      break;
    default:
      console.log('11111');
  }

}

But you can definitely use an if statement:

if (l < 6) {
    console.log("qq");
} else {
    console.log("1111");
}

According to the original writing, it should be whether l and l < 6 are equal. l is an integer, indicating the length of the string. l<6 is a Boolean value. Integers and Boolean values ​​are not congruent. , so the default statement will always be used;
The misunderstanding of the original writing method: it is not that the case statement will be executed if it is true, but the content in the switch expression l and the content after the case statement l< Matches only when 6 are congruent; assuming v="111", then l=3 l<6 is true, but 3!==true, so the default statement is used.

Answer 2

Are you sure your l is less than 6?

switch(n)
{
case 1:
  执行代码块 1
  break;
case 2:
  执行代码块 2
  break;
default:
  n 与 case 1 和 case 2 不同时执行的代码
}

Answer 3

export const setID = (v) => {
  console.log('执行setID',v);
  let l = v.length;
  switch(true)
  {
    case l < 6 :
      console.log('qq');
      break;
    default:
      console.log('11111');
  }

}