When my website loads, the popup appears - I need to make it automatically close after a specific time.
$(document).ready(function () {
//select the POPUP FRAME and show it
$("#popup").hide().fadeIn(1000);
//close the POPUP if the button with id="close" is clicked
$("#close").on("click", function (e) {
e.preventDefault();
$("#popup").fadeOut(1000);
});
});
There is already a button but i need to remove it.
You can use the delay()
function for that:
$(document).ready(function() { $("#popup").hide().fadeIn(1000).delay(5000).fadeOut(1000); $("#close").on("click", function(e) { e.preventDefault(); $("#popup").fadeOut(1000); }); });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div id="popup">popup</div>
Please mind the note that is given on the documentation:
The .delay() method is best for delaying between queued jQuery effects. Because it is limited—it doesn't, for example, offer a way to cancel the delay—.delay() is not a replacement for JavaScript's native setTimeout function, which may be more appropriate for certain use cases.
Javascript have option setTimeout .setTimeout is a native JavaScript function (although it can be used with a library such as jQuery, as we'll see later on), which calls a function or executes a code snippet after a specified delay (in milliseconds).
setTimeout(function() {
$("#popup").fadeOut(1000);
}, 1000);
or in jquery use .delay()
. Set a timer to delay execution of subsequent items in the queue.
$("#popup").delay(1000).fadeOut(1000);
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