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Accepted
Dan
Asked:2022-02-26 14:53:17 +0000 UTC2022-02-26 14:53:17 +0000 UTC

如何优化给定的 JavaScript 代码？

  • 772

写了代码，一切正常，但在我看来，它可以更好地优化。

const [working, setWorking] = useState(true)
  const workTime = () => {
    const simpleNotWorking = [6, 7]
    const weekendNotWorking = [6, 7, 8, 9, 10]
    const currentDate = new Date()
    const currentTime = currentDate.getHours()
    const currentDay = currentDate.getDay()

    if (currentDay >= 1 && currentDay <= 5) {
      if (simpleNotWorking.includes(currentTime))
        setWorking(false)
    }
    if (currentDay === 6 || currentDay === 7) {
      if (weekendNotWorking.includes(currentTime)) {
        setWorking(false)
      }
    }
  }

底线是，有一家餐厅，工作日从早上 8 点工作到第二天早上 6 点（工作日只有 2 小时从早上 6 点到早上 8 点不工作）周末他们从早上 11 点到早上 6 点工作)

我制作了 2 个数组，我在其中指出了餐厅不工作的位置。我还通过 getDay (1-7) 获取星期几，并通过通常的检查

如何缩短代码？

javascript

2 个回答

  1. Best Answer

    唯一想到的：

      const [working, setWorking] = useState(true)
  const workTime = () => {
    const notWorking = [[6, 7, 8, 9, 10], [6, 7]]
    const currentDate = new Date()
    const currentTime = currentDate.getHours()
    const currentDay = currentDate.getDay()
    
    if (notWorking[ +(currentDay >= 1 && currentDay <= 5) ].includes(currentTime))
        setWorking(false)
  }

    PS getDay 周日不返回 0 吗？

    • 2

  2. 你可以看起来像这样。

     const [working, setWorking] = useState(true)
  const workTime = () => {
    const currentDate = new Date()
    const currentTime = currentDate.getHours()
    const currentDay = currentDate.getDay();
    ([6, 7].includes(currentTime) || ([0,6].includes(currentDay) && [8, 9, 10].includes(currentTime))) && setWorking(false); // смотрим на время [6, 7] закрыто в любой день, а  на выходных смотрим дополнительно на [8, 9, 10] 
  }

    • 0

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