如何用jQuery显示“忙”指标？

Question

How do I display a spinning "busy" indicator at a specific point in a web page?

I want to start/stop the indicator when an Ajax request commences/completes.

Is it really just a matter of showing/hiding an animated gif, or is there a more elegant solution?

Answer 1

You can just show / hide a gif, but you can also embed that to ajaxSetup, so it's called on every ajax request.

$.ajaxSetup({
    beforeSend:function(){
        // show gif here, eg:
        $("#loading").show();
    },
    complete:function(){
        // hide gif here, eg:
        $("#loading").hide();
    }
});

One note is that if you want to do an specific ajax request without having the loading spinner, you can do it like this:

$.ajax({
   global: false,
   // stuff
});

That way the previous $.ajaxSetup we did will not affect the request with global: false .

More details available at: http://api.jquery.com/jQuery.ajaxSetup

Answer 2

The jQuery documentation recommends doing something like the following:

$( document ).ajaxStart(function() {
  $( "#loading" ).show();
}).ajaxStop(function() {
  $( "#loading" ).hide();
});

Where #loading is the element with your busy indicator in it.

References:

• http://api.jquery.com/ajaxStart/

• http://api.jquery.com/ajaxStop/

• In addition, jQuery.ajaxSetup API explicitly recommends avoiding jQuery.ajaxSetup for these:

  Note: Global callback functions should be set with their respective global Ajax event handler methods— .ajaxStart() , .ajaxStop() , .ajaxComplete() , .ajaxError() , .ajaxSuccess() , .ajaxSend() —rather than within the options object for $.ajaxSetup() .

Answer 3

I tend to just show/hide a IMG as other have stated. I found a good website which generates "loading gifs"

Link I just put it inside a div and hide by default display: none; (css) then when you call the function show the image, once its complete hide it again.

Answer 4

是的，这只是显示/隐藏GIF动画的问题。

Answer 5

To extend Rodrigo's solution a little - for requests that are executed frequently, you may only want to display the loading image if the request takes more than a minimum time interval, otherwise the image will be continually popping up and quickly disappearing

var loading = false;

$.ajaxSetup({
    beforeSend: function () {
        // Display loading icon if AJAX call takes >1 second
        loading = true;
        setTimeout(function () {
            if (loading) {
                // show loading image
            }
        }, 1000);            
    },
    complete: function () {
        loading = false;
        // hide loading image
    }
});

Answer 6

I did it in my project ,

make a div with back ground url as gif , which is nothing but animation gif

<div class="busyindicatorClass"> </div>

.busyindicatorClass
{
background-url///give animation here
}

in your ajax call , add this class to the div and in ajax success remove the class.

it will do the trick thatsit.

let me know if you need antthing else , i can give you more details

in the ajax success remove the class

success: function(data) {
    remove class using jquery
  }

Answer 7

I did it in my project:

Global Events in application.js:

$(document).bind("ajaxSend", function(){
   $("#loading").show();
 }).bind("ajaxComplete", function(){
   $("#loading").hide();
 });

"loading" is the element to show and hide!

References: http://api.jquery.com/Ajax_Events/

Answer 8

I had to use

HTML:
   <img id="loading" src="~/Images/spinner.gif" alt="Updating ..." style="display: none;" />

In script file:
  // invoked when sending ajax request
  $(document).ajaxSend(function () {
      $("#loading").show();
  });

  // invoked when sending ajax completed
  $(document).ajaxComplete(function () {
      $("#loading").hide();
  });

Answer 9

Old thread, but i wanted to update since i worked on this problem today, i didnt have jquery in my project so i did it the plain old javascript way, i also needed to block the content on the screen so in my xhtml

    <img id="loading" src="#{request.contextPath}/images/spinner.gif" style="display: none;"/>

in my javascript

    document.getElementsByClassName('myclass').style.opacity = '0.7'
    document.getElementById('loading').style.display = "block";