jQuery - Call one function by hover, then call another function

Question

So I have a submenu that loads content when a main nav button is hovered over. And while that content is loading I want to display a loading graphic.

The problem is I can't get both functions working together.

1st function - load submenu content (page.php)

2nd function - when page.php is loaded, hide loading graphic

<script>
function functionOne() {
jQuery.get('page.php', function(data) { jQuery('.class').html(data); })
}

function functionTwo() {
  jQuery("#loading").hide();    
};

jQuery(".dropdownhover").one('mouseenter', functionOne.done( functionTwo ););
</script>

<a class="dropdownhover">Button</a>
<div id="loading"></div>

With the above code, neither function works. If I call just the first function like this: jQuery(".dropdownhover").one('mouseenter', functionOne ); that works fine. But I of course need to call both, one after the other. Am I just writing the last line of the script wrong?

Rajaprabhu Aravindasamy · Accepted Answer · 2016-12-24 08:07:43Z

1

As per your requirement, the syntax that you have used is completely wrong, there is no such function done available in Function's prototype.

You can achieve that by just writing,

function functionOne() {
 jQuery.get('page.php', function(data) { 
   jQuery('.class').html(data); 
   functionTwo(); 
 });
}

function functionTwo() {
  jQuery("#loading").hide();    
};

jQuery(".dropdownhover").one('mouseenter', functionOne);

answered Dec 24, 2016 at 8:07

Rajaprabhu Aravindasamy

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3 Comments

Rajaprabhu Aravindasamy Over a year ago

@SLePort It was OP's code and OP didn't complain anything like, event is triggering for a single time. So I guess that the code was intentional.

user1610904 Over a year ago

Well that was a lot easier than I thought. This works great thank you. And yes the .one was intentional, just need it once.

Rajaprabhu Aravindasamy Over a year ago

@user1610904 Glad to help! :)

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