Server Side datatables using PHP and Ajax

Question

This is my DataTable code:

$('#open').DataTable( {
        select: true,
        "processing": true,
        "sAjaxSource": "booked1.php",
        "serverside": true,
        "columns" :[ {
            "data" : "name"
        }, {
            "data" : "date1"
        }, {
            "data" : "bookingtoken"
        }, {
            "data" : "insurance"
        }]
    } );

This is my ajax call :

$("#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');

            }


        });
    }

    });

This is my PHP Script:

 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);
}

This is the first time I am using server side data tables. I am getting error like "DataTables warning: table id=open - Invalid JSON response. For more information about this error, please see http://datatables.net/tn/1" Can anyone guide me how to use data tables server side for my code.

Shouldn't the echo json response be after the while loop closing brace? Also you should array push — Rotimi
– Rotimi, Commented Aug 22, 2017 at 6:29
@RahulMeshram could you please help in code where i have to change — karthik
– karthik, Commented Aug 22, 2017 at 6:37
It clearly indicate that Invalid JSON response. In your php you also echo <td> tags which obviously not josn format echo '<tr>'; echo "<td>$name </td>"; echo "<td> $date1 </td>"; echo "<td>$bookingtoken </td>"; echo "<td>$insurance </td>"; echo '</tr>'; — B. Desai
– B. Desai, Commented Aug 22, 2017 at 7:10

Rahul · Accepted Answer · 2017-08-22 06:40:12Z

0

Try to change this code,

$result = [];
$result[]=array(
        "name" => $name,
        "date1" => $date,
        "bookingtoken" => $bookingtoken,
        "insurance" => $insurance
    );
echo json_encode($result);
die;

Once check in network->xhr whether you are getting any response or not.

answered Aug 22, 2017 at 6:40

Rahul

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1 Comment

karthik Over a year ago

still same dude

davidkonrad · Accepted Answer · 2017-08-22 07:18:26Z

0

You really overcomplicate this. You should never echo out the actual <tr><td>.. markup, this is why you get the warning. And use mysqli_fetch_assoc instead :

$data = array();
while( $row1 = mysqli_fetch_assoc($query) ){
  $row1['name'] = $row1['fname'].' '.$row1['lname'];
  $row1['date1'] = $row1['date']; //??
  $data[] = $row1;
}
echo json_encode($data);

Update. You will probably need to use

echo json_encode( array('data' => $data) );

If you not set dataSrc to ''.

answered Aug 22, 2017 at 7:18

davidkonrad

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