Guide on creating a delayed button display in a react component

Question

Guide on creating a delayed button display in a react component

This question is exactly as the title suggests. I'm looking for solutions in either CSS or JavaScript, so any help is welcome.

  animation-delay: 5s;
  animation-duration: 1ms;
  animation-name: slidein;
  opacity: 100;

}

@keyframes slidein {
  from {
      opacity:0;
  }

  to {
 opacity: 100;
  }
}

css reactjs

Answer 1

Answer №1

Here is a step-by-step guide on how to achieve this:

Start by creating a variable called "notVisible" and set it to true
Create a button and set its hidden property to the value of the "notVisible" variable
Next, utilize a new promise that will resolve after 5 seconds. Inside this promise, change the value of the "notVisible" variable to false which will make the button visible

await new Promise(resolve => setTimeout(resolve, 5000)).then(() => {setNotVisible(false)});

Answer 2

Here is a step-by-step guide on how to achieve this:

Start by creating a variable called "notVisible" and set it to true
Create a button and set its hidden property to the value of the "notVisible" variable
Next, utilize a new promise that will resolve after 5 seconds. Inside this promise, change the value of the "notVisible" variable to false which will make the button visible

await new Promise(resolve => setTimeout(resolve, 5000)).then(() => {setNotVisible(false)});

Answer 3

Answer №2

If I were faced with the same issue, here is my approach:

Declare a variable to keep track of the button's state

const [isButtonVisible, setButtonVisible] = useState(false)

Set up a callback function in setTimeout that will execute after 5000ms.
This callback function will change the state to true after 5000ms.
Your button can have a conditional expression to add a CSS class if isButtonVisible is true.

<button 
    style={{display: 'None'}} 
    className={isButtonVisible ? '' : 'MyClass'}
>
    Click Me
</button>

In this case, MyClass represents the CSS class containing properties to make the button visible.

I trust this explanation is helpful :)

Answer 4

If I were faced with the same issue, here is my approach:

Declare a variable to keep track of the button's state

const [isButtonVisible, setButtonVisible] = useState(false)

Set up a callback function in setTimeout that will execute after 5000ms.
This callback function will change the state to true after 5000ms.
Your button can have a conditional expression to add a CSS class if isButtonVisible is true.

<button 
    style={{display: 'None'}} 
    className={isButtonVisible ? '' : 'MyClass'}
>
    Click Me
</button>

In this case, MyClass represents the CSS class containing properties to make the button visible.

I trust this explanation is helpful :)

Answer 5

Answer №3

let [show, setShow] = useState(false);

To update visibility after the page has loaded:

useEffect(() => {
    setTimeout(() => {
        setShow(true);
    }, 5000)
}, []);

Now, in your button: Toggle visibility based on the state using a ternary operator.

Answer 6

let [show, setShow] = useState(false);

To update visibility after the page has loaded:

useEffect(() => {
    setTimeout(() => {
        setShow(true);
    }, 5000)
}, []);

Now, in your button: Toggle visibility based on the state using a ternary operator.

Guide on creating a delayed button display in a react component

Answer №1

Answer №2

Answer №3

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