.hover（...）和on。（“悬停”...）的行为不同

问题描述

使用JQuery我试图在元素具有悬停状态时链接几个函数。

我通常会使用 .on（）只能使用1个函数，因此您必须询问传入的事件来检查事件是什么。试试这个：

  $（element）。on（hover，function（e）{
 if （e.type ==mouseenter）{
 console.log（one）; 
} 
 else {// mouseleave 
 console.log（two） ; 
} 
}）; 
  

或者，您可以拆分构成 hover（） 方法 - mouseenter 和 mouseleave ：

（）（mouseenter，function（）{console.log（one））

;}）。on（'mouseleave'，function（）{console.log（two）;}）;

  #element {background-color：black; height：100px; margin：100px; width：100px;}  

< script src =https ：//ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js>< / script>< div id =element>< / div>

Using JQuery I am trying to chain a couple of functions when an element has a hover state.

I would normally use the .hover event function, but after reading some tutorials I read that using .on is better as you can use one event handler to monitor all bubbling events within a document.

However, I am having problems when I chain two functions together like so:

$("element").on( "hover", function() {
    console.log("one");
}, function() {
    console.log("two");
});

I expected the result to be one two (which was the case when using .hover) but instead I get two two.

Can anyone explain what I am doing wrong or whether this is expected behaviour and why?

Reproduced using .hover(...): http://jsfiddle.net/gXSdG/

Reproduced using .on(hover...): http://jsfiddle.net/gXSdG/1/

解决方案

.on() can only take 1 function, so you have to interrogate the passed event to check what the event was. Try this:

$("element").on("hover", function(e) {
    if (e.type == "mouseenter") {
        console.log("one");
    }
    else { // mouseleave
        console.log("two");
    }
});

Example fiddle

Alternatively you can split the two events which constitute the hover() method - mouseenter and mouseleave:

$("#element").on("mouseenter", function() {
  console.log("one");
}).on('mouseleave', function() {
  console.log("two");
});

#element {
  background-color: black;
  height: 100px;
  margin: 100px;
  width: 100px;
}

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div id="element"></div>

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