-2

I am trying to display some information into a table (HTML Table) with PHP MySQLi. I am currently getting an error on the webpage I created. The image shows the image of my page and error.

enter image description here

    <?PHP
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "ip_site_database_project";

    $conn = new mysqli($servername, $username, $password, $dbname);

    if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
    } 
    $sql = "SELECT * FROM site";
    $result = mysqli_query($conn, $sql);

    echo "<table border='1' align ='center'>";
    echo "<tr>";
        echo "<td>Site ID</td>";
        echo "<td>Site Name</td>";
        echo "<td>Site Type</td>";
        echo "<td>Description</td>";
        echo "<td>Site Dial Code</td>";
        echo "<td>Site Number Range (Start)</td>";
        echo "<td>Site Number Range (End)</td>";
        echo "<td>Site Active Since</td>";
        echo "<td>Site Ceased Date</td>";
        echo "</tr>";
    while($rowitem = mysqli_fetch_array($results))
    {
        echo"<tr>";
        echo "<td>" . $rowitem[Site_ID] . "</td>";
        echo "<td>" . $rowitem[Site_Name] . "</td>";
        echo "<td>" . $rowitem[Site_Type] . "/td>";
        echo "<td>" . $rowitem[Description] . "</td>";
        echo "<td>" . $rowitem[Site_Dial_Code] . "/td>";
        echo "<td>" . $rowitem[Site_Number_Range_Start] . "</td>";  
        echo "<td>" . $rowitem[Site_Number_Range_End] . "/td>";
        echo "<td>" . $rowitem[Site_Active_Since] . "</td>";
        echo "<td>" . $rowitem[Site_Ceased_Date] . "/td>";
        echo "</tr>";
    }
    echo"</table>";

    $conn->close();

    ?>
Masivuye Cokile
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1 Answers1

-2

Change this: while($rowitem = mysqli_fetch_array($results)) to this: while($rowitem = mysqli_fetch_array($result))

Sebastian
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