how hide loading image when load jquery done

Question

i have this jquery code

else if ($(this).val() === "5") {
        var codefaqs = $(this).val();
        $('#imgloadfaqs').show();
        $("#faqsmain").attr("disabled", "disabled");
        $('#faqsanswer').load('http://royta.org/somefolder/get.php',{codedgive:codefaqs},function(applycode){
            $('#faqsanswer').html(applycode);
            $('#imgloadfaqs').hide(2000);
            $("#faqsmain").removeAttr("disabled");
        });
    }

and show loading image and disable select box
i want when get.php do its work and done to send resault
loading image gone
but with this code its gone very soon before php show resault
can anyone help me with this?

ρяσѕρєя K · Accepted Answer · 2013-01-12 18:23:25Z

2

move hide method inside function where you are getting result from php file as :

...,codedgive:codefaqs},function(applycode){
            $('#faqsanswer').html(applycode);
            $("#faqsmain").removeAttr("disabled");
            $('#imgloadfaqs').hide(500);
        });

answered Jan 12, 2013 at 18:23

ρяσѕρєя K

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3 Comments

masih arastooyi Over a year ago

not work ....! actutty i dont need $('#faqsanswer').html(applycode); its load in $('#faqsanswer').load('htt') i remove $('#faqsanswer').html(applycode); this line

ρяσѕρєя K Over a year ago

@masiharastooyi : you will need to move hide image logic after code where u are getting result back from server

masih arastooyi Over a year ago

$('#faqsanswer').load('....ler/get.php',{codedgive:codefaqs},function(applycode){ $('#faqsanswer').html(applycode); $('#imgloadfaqs').hide(2000); $("#faqsmain").removeAttr("disabled"); }); but not work its gone very soon yet

Dharman · Accepted Answer · 2013-01-12 18:23:37Z

2

try putting the code that hides the image inside the done function. I believe this would be the right approach:

$('#faqsanswer').load('http://royta.org/module/faqs/get.php',{codedgive:codefaqs},function(applycode){
        $('#faqsanswer').html(applycode);
        $("#faqsmain").removeAttr("disabled");
        $('#imgloadfaqs').hide(500);
    });

answered Jan 12, 2013 at 18:23

Dharman♦

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Comments

Haresh Ambaliya · Accepted Answer · 2013-01-13 06:08:31Z

Use following code snippet to show and hide.

    $(document).ready(function() {



// Setup the ajax indicator

$('body').append('<div id="ajaxBusy"><p><img src="images/loading.gif"></p></div>');



$('#ajaxBusy').css({

display:"none",

margin:"0px",

paddingLeft:"0px",

paddingRight:"0px",

paddingTop:"0px",

paddingBottom:"0px",

position:"absolute",

right:"3px",

top:"3px",

width:"auto"

});

});



// Ajax activity indicator bound to ajax start/stop document events

$(document).ajaxStart(function(){

$('#ajaxBusy').show();

}).ajaxStop(function(){

$('#ajaxBusy').hide();

});

To read more about the same follow http://skfox.com/2008/04/28/jquery-example-ajax-activity-indicator/

if you use this approach on all ajax request you will get spinning image, so you do not need to code for all request and it will lead to save extra efforts.

For working example follow jsfiddle link http://jsfiddle.net/re57j/

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