想请教一下关于获取计时器内部数据的问题 - V2EX
aglsv

想请教一下关于获取计时器内部数据的问题

  •  
  •   aglsv Aug 1, 2021 2339 views
    This topic created in 1771 days ago, the information mentioned may be changed or developed.
    function a (){ let a = 0 a = 1 let loader = setInterval(function () { abc () clearInterval(loader) }, 1000) function abc (){ a +=1 } return a } 

    我的预想是 return 的 a 为 2,但是现在 return 的结果依旧是 1,请问这个该怎么解决啊

    9 replies    2021-08-03 15:24:33 +08:00
    lianyue
        1
    lianyue  
       Aug 1, 2021
    返回的时候 setInterval 里面的函数并没被执行
    aglsv
        2
    aglsv  
    OP
       Aug 1, 2021
    @lianyue 那这个该怎么解决
    lianyue
        3
    lianyue  
       Aug 1, 2021
    async
    await
    promise

    自己了解
    lianyue
        4
    lianyue  
       Aug 1, 2021
    rabbbit
        5
    rabbbit  
       Aug 1, 2021
    const foo = () => {
      let bar = 1
      return new Promise(resolve => {
       setTimeout(() => resolve(++bar), 3000)
     })
    }

    (async() => console.log(await foo()))()
    aglsv
        6
    aglsv  
    OP
       Aug 2, 2021
    @rabbbit 如果没法子使用 async 的话请问这个应该如何解决
    aglsv
        7
    aglsv  
    OP
       Aug 2, 2021
    @lianyue 主要是没法子使用 async
    rabbbit
        8
    rabbbit  
       Aug 2, 2021
    const foo = () => {
    let bar = 1;
    return new Promise((resolve) => {
      setTimeout(() => resolve(++bar), 3000);
    });
    };

    (() => {
      foo().then((bar) => {
       console.log(bar)
     })
    })();


    const foo = (callback) => {
    let bar = 1;
    setTimeout(() => {
      bar++;
      callback?.(bar)
    }, 3000);
    };

    (() => {
      foo((bar) => console.log(bar))
    })();
    aglsv
        9
    aglsv  
    OP
       Aug 3, 2021
    @rabbbit 我想问一下第一种,比如 a 有一个```b+=bar```,然后把 b 给 return 出去,这时候因为 foo 里有个异步操作使用了 promise,这样的话就需要在 a 方法里面执行.then 来获取 bar,这样的话会是先```return b```,然后才执行```b+=bar```。
    因为这个 b 要被上层方法使用,不能再把 a 方法也写成 promise,这样的话有办法解决吗
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