sandshakimi
asked on
Resolve a "Warning: mysqli_fetch_array()...
Here's my code. How to resolve this Warning. The page actually works on my local environment, but doesn't parse when uploaded to dev.
<?php
define('DB_IP', 'localhost'); // MySQL Server
define('DB_USERNAME', 'xxx'); //database username
define('DB_PASSWORD', 'xxx'); //database password
define('DB_DATABASE', 'navigator_data'); //database
class test_object // proxy for web service object
{
public $contractVehicle = array(
0=>'ITSchedule70',
1=>'ITCommodityProgram',
2=>'Army CHESS',
3=>'NASA SEWP'
);
function get_contract()
{
return $this->contractVehicle;
}
}
$var = new test_object(); // the variable get back from webservice
$connection = mysqli_connect(DB_IP,DB_USERNAME,DB_PASSWORD,DB_DATABASE); // connection to navigator_data
$qry_nav_data = "SELECT * FROM contract_vehicle"; // formulate the SQL query
$result = mysqli_query($connection, $qry_nav_data); // execute the query
// display results in HTML table
echo "<table border='1'>
<tr>
<th>contract_vehicle_id</th>
<th>vehicle_name</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['contract_vehicle_id'] . "</td>";
echo "<td>" . $row['vehicle_name'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($connection);
?>
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SOLUTION
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