jquery ajax process success then continue load html code

Question

Is it possible to make. jquery ajax success, then continue load page's html?

For explain: I need when open the page. fist loading css, js code. then loading near the jquery ajax code, make a stop. stay the jquery ajax data poccess finished, then continue loading the <body> code. show <div id="data"></div> and other html code.

<script type="text/javascript">
$.ajax({
    url: "pageb.php", 
    dataType: "html",
    type: 'POST', 
    data: "value="+ value, 
    success: function(data){ 
        //stop open page, after data back success, then continue loading the html code.
        $("#data").html(data);        
        }
});
</script>
</head>
<body>
<div id="data"></div>
...
</body>

Michael Mior · Accepted Answer · 2011-07-04 22:30:03Z

1

If you want that way , you should go for async:false as others suggested.

But async:false will block the page , but if you want that way its fine.

But if you have time try this plugin , nice one

http://jquery.malsup.com/block/

edited Jul 4, 2011 at 22:30

Michael Mior

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answered Jul 4, 2011 at 22:02

kobe

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Comments

Claudio · Accepted Answer · 2011-07-04 22:20:08Z

0

Unless the whole page is loaded via AJAX, there is no way to tell the browser "Ehi, wait rendeering the page until I say so".

answered Jul 4, 2011 at 22:20

Claudio

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1 Comment

Michael Mior Over a year ago

Sure there is. <body style="display:none"> $(document).ready(function() { $('body').show(); } (with the JS placed within a script tag of course). I wouldn't really recommend this approach however.

SeanCannon · Accepted Answer · 2011-07-04 21:21:43Z

0

Well the idea of AJAX is that it's asynchronous so it does not care about what everything else on the page is doing/what is being loaded, etc. Try setting async:false in your $.ajax() params and see what happens.

answered Jul 4, 2011 at 21:21

SeanCannon

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