使用 react-hotkeys-hook 只调用一次快捷键。

huangapple go评论71阅读模式
英文:

Call hotkey only once with react-hotkeys-hook

问题

我在我的应用程序中有两个快捷键,12。如果我按下 1 然后释放,再按下 2,那没问题。但如果我同时按下它们,当我同时释放它们时,它不会按预期递增。

我的期望是,如果我按下 1 和 2,然后同时释放它们,它们应该都递增 1。但它没有。我查看了 API 并尝试了一些不同的选项,但似乎总是在某些情况下递增 2。

import { useHotkeys } from 'react-hotkeys-hook';

export default function App() {
  const [cnt1, set1] = React.useState(0);
  const [cnt2, set2] = React.useState(0);

  useHotkeys('1', () => set1((v) => v + 1), { keyup: true });
  useHotkeys('2', () => set2((v) => v + 1), { keyup: true });

  return (
    <div>
      <div>{cnt1}</div>
      <div>{cnt2}</div>
    </div>
  );
}

一个运行示例:

https://stackblitz.com/edit/stackblitz-starters-6mbnsa?file=src%2FApp.tsx

英文:

I have 2 hotkeys in my app, 1 and 2. If I press 1 release and then 2, it's fine. But if I press both at the same time, it doesn't increment as expected when I release both at the same time.

My expectation is that if I press 1 and 2, and then release both, I should have both incrementing by 1. But it doesn't. I checked the API and tried some different options but it always seems to increment 2 in one of the cases.

import { useHotkeys } from &#39;react-hotkeys-hook&#39;;

export default function App() {
  const [cnt1, set1] = React.useState(0);
  const [cnt2, set2] = React.useState(0);

  useHotkeys(&#39;1&#39;, () =&gt; set1((v) =&gt; v + 1), { keyup: true });
  useHotkeys(&#39;2&#39;, () =&gt; set2((v) =&gt; v + 1), { keyup: true });

  return (
    &lt;div&gt;
      &lt;div&gt;{cnt1}&lt;/div&gt;
      &lt;div&gt;{cnt2}&lt;/div&gt;
    &lt;/div&gt;
  );
}

A running example:

https://stackblitz.com/edit/stackblitz-starters-6mbnsa?file=src%2FApp.tsx

答案1

得分: 1

我已经尝试了很多方法来防止这种情况发生,找到了一个不太好的解决方案,我想分享一下。

有一个isHotkeyPressed钩子,可以在第二个回调中使用,来检查另一个热键是否处于活动状态:

useHotkeys('2', () => {
    if (!isHotkeyPressed('1')) {
        set2((v) => v + 1);
    }
}, { keyup: true });

你可以将这两个热键钩子合并成一个(可以用逗号分隔多个键),以使另一个键上的isHotkeyPressed更加动态。

英文:

I've tried quite some ways to prevent this, found a single, not so great solution I thought I'd share.


There seems to be a isHotkeyPressed hook that can be used in the second callback to check if the other hotkey is active:

useHotkeys(&#39;2&#39;, () =&gt; {
    if (!isHotkeyPressed(&#39;1&#39;)) {
        set2((v) =&gt; v + 1);
    }
}, { keyup: true });

You could combine both hotkeys hooks into a single one (you can pass multiple keys comma separated) to make the isHotkeyPressed on the other key more dynamic.

答案2

得分: 0

我认为我在复制 @0stone0 的回答时犯了一个错误,并无意中使它起作用。我仍然困惑于这是如何工作的,但即使我有4个数字,它也可以工作。

请注意,只有在我放开的按键不被按下时它才会递增。

  useHotkeys(
    '1',
    () => {
      if (!isHotkeyPressed('1')) {
        set1((v) => v + 1);
      }
    },
    {
      keyup: true,
    }
  );
  useHotkeys(
    '2',
    () => {
      if (!isHotkeyPressed('2')) {
        set2((v) => v + 1);
      }
    },
    {
      keyup: true,
    }
  );
英文:

I think I made a mistake copying @0stone0 answer and inadvertently made it work. I'm still confused how this is working, but it even works if I have 4 numbers.

Notice it only increments if the key I'm letting go up is not pressed.

  useHotkeys(
    &#39;1&#39;,
    () =&gt; {
      if (!isHotkeyPressed(&#39;1&#39;)) {
        set1((v) =&gt; v + 1);
      }
    },
    {
      keyup: true,
    }
  );
  useHotkeys(
    &#39;2&#39;,
    () =&gt; {
      if (!isHotkeyPressed(&#39;2&#39;)) {
        set2((v) =&gt; v + 1);
      }
    },
    {
      keyup: true,
    }
  );

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  • 本文由 发表于 2023年7月4日 23:13:08
  • 转载请务必保留本文链接:https://go.coder-hub.com/76613950.html
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