how to save ajax call request data in a variable and show it?

Question

$.ajax({
    url: "test.html",
    context: document.body
}).done(function() {
    //show data
});

how to save data from ajax request into a variable and then show it? and how to show a message "loading..." while it is proccessing?

possible duplicate of http://stackoverflow.com/questions/905298/jquery-storing-ajax-response-into-global-variable — Cole Tobin, Sep 29 '13 at 18:35

Gildas.Tambo · Accepted Answer · 2013-09-30T04:43:27.950

1

 $.ajax({
    url: "test.html",
    context: document.body
}).done(function(data) { 
    alert(data);
});

UPDATED:

$.ajax({
    url: "test.html",
    context: document.body
}).done(function(data) {
   $('#loading').hide();    
   // alert(data);
});

markup:

<div id='loading'></div>

edited Sep 30 '13 at 04:43

answered Sep 29 '13 at 18:35

Gildas.Tambo

22,173
7
50
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thanks how to show a "loading..." message when a request is send ? – M1X Sep 29 '13 at 18:39

score 1 · Answer 2 · answered Sep 29 '13 at 18:43

1

For loading message, use beforeSend() in ajax.

  beforeSend : function () {
      $('body').html('Loading...');
  }

answered Sep 29 '13 at 18:43

Rajesh

3,743
1
24
31

score 1 · Answer 3 · answered Sep 29 '13 at 18:46

1

You can try something like this. In your function you can show loading div

function your_ajax() {
  $('#loading').show();
  $.ajax({
    url: "test.html",
    context: document.body
    }).done(function(data) {
     $('#loading').hide();      
     alert(data);

  });
}

Please add this part in your html

<div id="loading" style="display:none;">Loading </div>

answered Sep 29 '13 at 18:46

Shafeeque

2,039
2
13
28

how to wrap this in a timeout request for example refresh every 5 sec? – M1X Sep 29 '13 at 19:34

how to save ajax call request data in a variable and show it?

3 Answers3