What could be causing jQuery.css() to not function properly?

Question

What could be causing jQuery.css() to not function properly?

I've been encountering some challenges with a piece of code and haven't been able to make it function as intended. I'm relatively new to JavaScript/jQuery, so bear with me as I navigate through this learning process.

As part of the Free Code Camp course, I am working on creating a Weather App using simpleWeather.js. I'm currently attempting to add some animation to the page to enhance interactivity. Here is the code snippet I'm using:

$('#drop').on('click', function() {
  $('.fa-chevron-circle-down').css('transform', 'rotate(180deg)');
});

My understanding is that this code should rotate the icon 180 degrees upon a click event, but despite multiple attempts, I have not been successful. After spending hours troubleshooting yesterday, I decided it was time to seek assistance! (Once this functionality is achieved, my plan is to animate the elements below so they are only displayed when the button is clicked).

For reference, my codepen can be found here.

Thank you in advance for any guidance or support provided.

javascript jquery css

Answer 1

Answer №1

After the DOM has loaded, your #drop element was dynamically created. In this case, you will need to use event delegation. To do this, attach your click event to the document and then reference your element - #drop.

$(document).on('click','#drop', function() {
  $('.fa-chevron-circle-down').css('transform', 'rotate(225deg)');
});

`Check out the updated PEN`

To simplify things further, if the #weather element already existed during the DOM load, you can attach the event to the #weather element instead of the document. Here's an example:

$('#weather').on('click','#drop', function() {
  $('.fa-chevron-circle-down').css('transform', 'rotate(225deg)');
});

Additionally, I believe the rotation value should be 180deg for a proper turn-around.

Check out the updated PEN 2 with 180deg and #weather

Answer 2

After the DOM has loaded, your #drop element was dynamically created. In this case, you will need to use event delegation. To do this, attach your click event to the document and then reference your element - #drop.

$(document).on('click','#drop', function() {
  $('.fa-chevron-circle-down').css('transform', 'rotate(225deg)');
});

`Check out the updated PEN`

To simplify things further, if the #weather element already existed during the DOM load, you can attach the event to the #weather element instead of the document. Here's an example:

$('#weather').on('click','#drop', function() {
  $('.fa-chevron-circle-down').css('transform', 'rotate(225deg)');
});

Additionally, I believe the rotation value should be 180deg for a proper turn-around.

Check out the updated PEN 2 with 180deg and #weather

Answer 3

Answer №2

I encountered an issue with the jquery .css function while working with code that resembled the original poster's.

Challenge

When I clicked the button with the #correct-email id, the .css jquery function failed to update the css for my .dot-one and .dot-two classes.

Here are the relevant code snippets:

$(document).on('click', '#correct-email', function () {
        $('.dot-one').css('background-color', '$white');
        $('.dot-two').css('background-color', '$gray-darker');
        ...
      });

.dot-one {
  background-color: $gray-darker;
}

.dot-two {
  background-color: $white;
}

Resolution

$(document).on('click', '#correct-email', function () {
        $('.dot-one').css('background-color', '#fff');
        $('.dot-two').css('background-color', '#222');
        ...
      });

It turns out that jquery does not support passing SASS variables into the .css function. This information was not explicitly stated in the documentation. I assumed that jquery would be able to process the SASS variable to compute the new value, but it appears that SASS pre-processes all such variables and does not expose them to css.

Referencing the SASS documentation, it seems this is the likely explanation for the issue.

Answer 4

I encountered an issue with the jquery .css function while working with code that resembled the original poster's.

Challenge

When I clicked the button with the #correct-email id, the .css jquery function failed to update the css for my .dot-one and .dot-two classes.

Here are the relevant code snippets:

$(document).on('click', '#correct-email', function () {
        $('.dot-one').css('background-color', '$white');
        $('.dot-two').css('background-color', '$gray-darker');
        ...
      });

.dot-one {
  background-color: $gray-darker;
}

.dot-two {
  background-color: $white;
}

Resolution

$(document).on('click', '#correct-email', function () {
        $('.dot-one').css('background-color', '#fff');
        $('.dot-two').css('background-color', '#222');
        ...
      });

It turns out that jquery does not support passing SASS variables into the .css function. This information was not explicitly stated in the documentation. I assumed that jquery would be able to process the SASS variable to compute the new value, but it appears that SASS pre-processes all such variables and does not expose them to css.

Referencing the SASS documentation, it seems this is the likely explanation for the issue.

What could be causing jQuery.css() to not function properly?

Answer №1

`Check out the updated PEN`

`Check out the updated PEN 2 with 180deg and #weather`

Answer №2

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