Forum Discussion
MouradSahbeni
Jun 11, 2022Copper Contributor
problem displaying data in php form
Hello everyone,
I am creating a page where I want to display a table containing the data of a sql server 2017 table. The problem is I cannot display them.. Can anyone help me solve this problem?
here is my php code
<!doctype html>
<html lang="en">
<head>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1, shrink-to-fit=no">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/css/bootstrap.min.css">
<link rel="stylesheet" href="https://cdn.datatables.net/1.10.25/css/dataTables.bootstrap4.min.css">
<title>Home</title>
</head>
<body>
<h1>Liste Collisage</h1>
<nav class="navbar navbar-light bg-light">
<a class="navbar-brand">
<pre>Delivery note Nr. C20220019 Crings Lieferung <?php echo "" . date("Y/m/d") . "<br>";?></pre>
<a>
</nav>
<div class="container-fluid">
<div class="row">
<div class="col-md-12">
<table id="dest_data" class="table table-striped table-bordered" style="width:100%">
<thead>
<tr>
<th>Code à Barres</th>
<th>Article Name</th>
<th>Désignation</th>
<th>Quantity par Size</th>
<th>Quantité Totale</th>
<th>Article Number</th>
<th>CRINGS Number</th>
<th>Order Number</th>
<th>Order</th>
<th>Box NR</th>
</tr>
</thead>
<tbody>
<?php
error_reporting(E_ALL);
ini_set('display_errors', '1');
$serverName = "SRV"; //serverName\instanceName
$connectionInfo = array( "Database"=>"phils", "UID"=>"sa", "PWD"=>"Passw0rd");
$conn = sqlsrv_connect( $serverName, $connectionInfo);
if( $conn ) {
echo "Connexion établie.<br />";
}else{
echo "La connexion n'a pu être établie.<br />";
die( print_r( sqlsrv_errors(), true));
}
function ReadData()
{
try
{
$conn = OpenConnection();
$tsql = "SELECT Code à Barres,
Article,
Désignation,
Quantité par Unité,
Quantité Totale,
Groupe2,
Propriété3,
OF,
NUMORDRE,
NUMPALETTE
FROM phils.Ligne Livraison Client réf";
$getProducts = sqlsrv_query($conn, $tsql);
if ($getProducts == FALSE)
die(FormatErrors(sqlsrv_errors(),TRUE));
$productCount = 0;
while($row = sqlsrv_fetch_array($getProducts, SQLSRV_FETCH_ASSOC))
{
echo "Code à Barres: ".$row[0]."\n";
echo "Article: ".$row[1]."\n";
echo "Désignation: ".$row[2]."\n";
echo "Quantité par Unité: ".$row[3]."\n";
echo "Quantité Totale: ".$row[4]."\n";
echo "Groupe2: ".$row[5]."\n";
echo "Propriété3: ".$row[6]."\n";
echo "OF: ".$row[7]."\n";
echo "NUMORDRE: ".$row[8]."\n";
echo "NUMPALETTE: ".$row[9]."\n";
echo "-----------------\n";
$productCount++;
}
sqlsrv_free_stmt($getProducts);
sqlsrv_close($conn);
}
catch(Exception $e)
{
echo("Error!");
}
}
?>
</tbody>
</table>
</div>
</div>
</div>
<script src="https://code.jquery.com/jquery-3.5.1.js"></script>
<script src="https://cdnjs.cloudflare.com/ajax/libs/popper.js/1.12.9/umd/popper.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/4.0.0/js/bootstrap.min.js"></script>
<script src="https://cdn.datatables.net/1.10.25/js/jquery.dataTables.min.js"></script>
<script src="https://cdn.datatables.net/1.10.25/js/dataTables.bootstrap4.min.js"></script>
<script>
$('#dest_data').DataTable();
</script>
</body>
</html>
1 Reply
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- olafhelperBronze Contributor
MouradSahbeni , your SQL statement is more the a bit wrong and will cause errors. If an object name contains special characters, even simple spaces, you must set them in brackets, see below.
Better avoid spaces in object name.
$tsql = "SELECT [Code à Barres], Article, Désignation, [Quantité par Unité], [Quantité Totale], Groupe2, Propriété3, OF, NUMORDRE, NUMPALETTE FROM phils.[Ligne Livraison Client réf]";