jQuery multiple events to trigger the same function

Question

Is there a way to have keyup, keypress, blur, and change events call the same function in one line or do I have to do the

Answer 1

You could define the function that you would like to reuse as below:

var foo = function() {...}

And later you can set however many event listeners you want on your object to trigger that function using on('event') leaving a space in between as shown below:

$('#selector').on('keyup keypress blur change paste cut', foo);

Answer 2

Is there a way to have keyup, keypress, blur, and change events call the same function in one line?

It's possible using .on(), which accepts the following structure: .on( events [, selector ] [, data ], handler ), so you can pass multiple events to this method. In your case it should look like this:

$('#target').on('keyup keypress blur change', function(e) {
    // "e" is an event, you can detect the type of event using "e.type"
});

And here is the live example:

$('#target').on('keyup keypress blur change', function(e) {
  console.log(`"${e.type.toUpperCase()}" event happened`)
});

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

Answer 3

$("element").on("event1 event2 event..n", function() {
   //execution
});

This tutorial is about handling multiple events.

Answer 4

It's simple to implement this with the built-in DOM methods without a big library like jQuery, if you want, it just takes a bit more code - iterate over an array of event names, and add a listener for each:

function validate() {
  // ...
}

const element = document.querySelector('#element');
['keyup', 'keypress', 'blur', 'change'].forEach((eventName) => {
  element.addEventListener(eventName, validate);
});

Answer 5

As of jQuery 1.7, the .on() method is the preferred method for attaching event handlers to a document. For earlier versions, the .bind() method is used for attaching an event handler directly to elements.

$(document).on('mouseover mouseout',".brand", function () {
  $(".star").toggleClass("hovered");
})