The code I have is as follows:
$('.signup-form-wrapper').css("style", "display: block");
$('.login-form-wrapper').css("style", "display: none");
I'm not sure why it's not working. The current appearance of the element is like this:
<div class="signup-form-wrapper form-wrapper" style="display: none;">
I want to change this style to display: block, how can I achieve that?
To modify your code, you have two options:
$('.signup-form-wrapper').display();
This function is similar to using .css('display', 'block'), with the distinction that the display property is restored to its original state.
Source: http://api.jquery.com/show/
or
$('.signup-form-wrapper').style({ display: 'block' });
.style( properties )
.style( propertyName, value )
Source: http://api.jquery.com/css/#css-propertyName-value
Both methods will reveal your element and update its appearance.
.css() is a function that applies the specified style to an element.
To use the function, follow this syntax:
$(elem).css('style-property','style-value');
Here's an example of changing the code:
$('.contact-form-wrapper').css("display", "flex");
$('.aboutus-form-wrapper').css("display", "none");
If you want to apply multiple styles to a single element, pass an object like this:
$(elem).css({'style-property1':'style-value1','style-property2':'style-value2'});
For instance:
$('.contact-form-wrapper').css({"display": "flex", "border":"2px solid blue"});
Give this a go
$('.register-box').show();
$('.login-box').hide();
When it comes to simply displaying or hiding elements, there are more concise functions available like .hide() and .show()
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