这是我的数据表代码：
$('#open').DataTable( {
select: true,
"processing": true,
"sAjaxSource": "booked1.php",
"serverside": true,
"columns" :[ {
"data" : "name"
}, {
"data" : "date1"
}, {
"data" : "bookingtoken"
}, {
"data" : "insurance"
}]
} );
这是我的ajax调用：
$("#submit").on('click', function () {
$('#loadarModal').modal({backdrop: 'static', keyboard: false});
var date = $("#date").val();
//alert(date);
if (date == '') {
$("#dateText").show();
$("#dateText").html("Please select date");
$("#loadarModal").modal('hide');
} else {
$("#dateText").hide();
//alert("can processd");
var data = $("#form").serialize();
$.ajax({
type: 'POST',
url: 'booked1.php',
data: {
date: date
},
cache: false,
dataType: "html",
success: function (response) {
alert(response);
if(response==''){
}
$("#booking").html(response);
$("#loadarModal").modal('hide');
}
});
}
});
这是我的PHP脚本：
include 'd_b_con.php';
if(isset($_POST['date'])){
$date=$_POST['date'];
$query=mysqli_query($conn,"select date as date ,tokenno as tokenno ,inusrance as inusrance,bookingtoken as bookingtoken ,
fname as fname,lname as lname , status as status from at_booking where date='$date'");
$data=array();
while($row1=mysqli_fetch_array($query)){
$data[] = $row1;
$date1=$row1['date'];
$tokenno=$row1['tokenno'];
$bookingtoken=$row1['bookingtoken'];
$fname=$row1['fname'];
$lname=$row1['lname'];
$status=$row1['status'];
$insurance=$row1['inusrance'];
$name=$fname.' '.$lname;
echo '<tr>';
echo "<td>$name </td>";
echo "<td> $date1 </td>";
echo "<td>$bookingtoken </td>";
echo "<td>$insurance </td>";
echo '</tr>';
$result=array(
"name" => $name,
"date1" => $date,
"bookingtoken" => $bookingtoken,
"insurance" => $insurance
);
echo json_encode($result);
}
这是我第一次使用服务器端数据表。我得到的错误，如"数据表警告：表id=open-无效的JSON响应。有关此错误的详细信息，请参阅http://datatables.net/tn/1"有人能指导我如何将数据表服务器端用于我的代码吗？
尝试更改此代码，
$result = [];
$result[]=array(
"name" => $name,
"date1" => $date,
"bookingtoken" => $bookingtoken,
"insurance" => $insurance
);
echo json_encode($result);
die;
一旦检查在网络-
你真的把事情复杂化了。您永远不应该呼出实际的
$data = array();
while( $row1 = mysqli_fetch_assoc($query) ){
$row1['name'] = $row1['fname'].' '.$row1['lname'];
$row1['date1'] = $row1['date']; //??
$data[] = $row1;
}
echo json_encode($data);
更新。你可能需要使用
echo json_encode( array('data' => $data) );
如果未将dataSrc设置为'。
