Reporting Problem

Reporting Problem

vinkey33vinkey33 Posts: 7Questions: 4Answers: 0

Hello, I have a problem of index not defined and I do not know how
`<?php
$servername = "localhost";
$username = "root";
$password = "Mm101010";
$dbname = "smartphone";

$conn = mysqli_connect($servername, $username, $password, $dbname) or die("Connection failed: " . mysqli_connect_error());

$requestData = $_REQUEST;

$columns = array(
0 => 'USER_ID',
1 => 'Nom',
2 => 'Prenom',
3 => 'Num_SIM',
4 => 'PIN_Terminal',
5 => 'PIN_SIM',
6 => 'Num_IMEI',
7 => 'Date_Debut',
8 => 'Date_Fin',
8 => 'Vitre',
10 => 'Coque',
11 => 'Support_Vehicule',
12 => 'Actif',
13 => 'Statut'

);

$sql = "SELECT Or_Affectation ";
$sql.=" FROM vu_affect_empl";
$query=mysqli_query($conn, $sql) or die("Affectation.php: get employees");
$totalData = mysqli_num_rows($query);
$totalFiltered = $totalData;

$sql = "SELECT Or_Affectation, USER_ID, Nom, Prenom, Num_SIM, PIN_Terminal, PIN_SIM, Num_IMEI, Date_Debut, Date_Fin, Vitre, Coque, Support_Vehicule, Actif, Statut ";
$sql.=" FROM vu_affect_empl WHERE 1=1";

if( !empty($requestData['search']['value']) ) {
$sql.=" AND ( USER_ID LIKE '".$requestData['search']['value']."%' ";
$sql.=" OR Num_SIM LIKE '".$requestData['search']['value']."%' ";

$sql.=" OR Nom LIKE '".$requestData['search']['value']."%' )";

}
$query=mysqli_query($conn, $sql) or die("Affectation.php: get employees");
$totalFiltered = mysqli_num_rows($query);
$query=mysqli_query($conn, $sql) or die("Affectation.php: get employees");

$data = array();
$i=1+$requestData['start'];
while( $row=mysqli_fetch_array($query) ) {
$nestedData=array();

$nestedData[] = "<input type='checkbox'  class='deleteRow' value='".$row['Or_Affectation']."'  /> #".$i ;
$nestedData[] = $row["USER_ID"];
$nestedData[] = $row["Nom"];
$nestedData[] = $row["Prenom"];
$nestedData[] = $row["Num_SIM"];
$nestedData[] = $row["PIN_Terminal"];
$nestedData[] = $row["PIN_SIM"];
$nestedData[] = $row["Num_IMEI"];
$nestedData[] = $row["Date_Debut"];
$nestedData[] = $row["Date_Fin"];
$nestedData[] = $row["Vitre"];
$nestedData[] = $row["Coque"];
$nestedData[] = $row["Support_Vehicule"];
$nestedData[] = $row["Actif"];
$nestedData[] = $row["Statut"];

$data[] = $nestedData;
$i++;

}

$json_data = array(
"draw" => intval( $requestData['draw'] ),
"recordsTotal" => intval( $totalData ),
"recordsFiltered" => intval( $totalFiltered ),
"data" => $data
);

echo json_encode($json_data);

<?php > ` Here is the error image ?>

thx

4.PNG 13.1K

Answers

  • allanallan Posts: 63,893Questions: 1Answers: 10,531 Site admin

    Have you enabled serverSide in your DataTables initialisation?

    Allan

  • vinkey33vinkey33 Posts: 7Questions: 4Answers: 0

    yes, But I do not see what identified the variable with respect to the variable

  • allanallan Posts: 63,893Questions: 1Answers: 10,531 Site admin

    I don't really understand I'm afraid. The serverSide option tells DataTables to enter into server-side processing mode, whereby it will send the parameters described here to the server with each request (defined by ajax).

    Allan

This discussion has been closed.