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Home/ Questions/Q 6999539
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T20:33:16+00:00 2026-05-27T20:33:16+00:00

I have a little code for dynamically loading part of my web: $(‘a#ajax-call’).click(function (e)

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I have a little code for dynamically loading part of my web:

$('a#ajax-call').click(function (e) {
    $("#container").load('/process.php',{id:myID},function(data){
        return;
    });
    e.preventDefault();
}); 

which returns this html scheme:

<div>my ajax loaded content</div>

and appends to this scheme:

<div id=container>
     <div>php loaded content 1</div>
     <div>php loaded content 2</div>
     <div id="static">php loaded content static</div>
     <div>php loaded content 3</div>
     ... until 12 divs...
</div>

the problem is, load is replacing whole content of #container and I need show this way:

<div id=container>
     <div>ajax loaded content </div>
     <div>php loaded content 1</div>
     <div id="static">php loaded content static</div> <--- remains static
     <div>php loaded content 2</div>
     ... until 12 divs...
 </div>

Any ideas?

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  1. Editorial Team
    Editorial Team
    2026-05-27T20:33:16+00:00Added an answer on May 27, 2026 at 8:33 pm
    $('a#ajax-call').click(function (e) {
        e.preventDefault();
        $.ajax({
        url: '/process.php',
        type: "POST",
        data: "id="+myID,
        cache: true,
        success: function(data){
        $("#container").append(data);           
        }
        });
    });
    

    and modify your anchor tag as

    <a href="javascript:;" id="ajax-call"> ... </a>
    

    if you are not using e.preventDefault(); function in ajax call

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