小编FTW*_*ngs的帖子

所以我是新的ajax而不是那么"亲"在php,我需要帮助.

我创建了一个ajax调用,在document.ready上显示来自DB的帖子

$.ajax({
        type:"POST",
        url:"testphp.php",
        datatype: 'json',
        success:function myFunction(response) {
            var arr = JSON.parse(response);
            var i;
            var out = " ";

            for(i = 0; i < arr.length; i++) {
                out += 
                    "<div id='"+arr[i].ID+"' class='grid-item "+arr[i].shape+"'>
                        <div class='grid_content'>"+arr[i].img_holder+"</div><div class='grid_title red'>"+arr[i].naslovhj+"</div>
                        <div class='content_grid'></div>
                    </div>"
            }
            out += " ";
            document.getElementById("grids").innerHTML = out;

            console.log("uspjelo"); 
        },
        error:function(){
            $("#ea").html('There was an error updating the settings');
        }   
}); 

testphp.php中的内容是:

$conn = new mysqli($servername, $username, $password, $dbname);
$myArray = array();
if ($result = $conn->query("SELECT * FROM postovi")) …