如何在不知道表名的情况下显示表格？

Question

这是我的代码

<?php
$servername = "localhost:3307";
$username = "root";
$password = "";
$dbname = "female"; //database 

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname); //open ng connection 
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 


$username = $_POST['username']; 
$sql = "SELECT * FROM $username";
$result = $conn->query($sql); 
echo "<div class='con'>";
echo "Men";
if ($result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {


        echo "<table id='contestant'>";
        echo "<tr>";
        echo "<td>".$row[0]."</td>"; 
        echo "<td>".$row[6]."</td>";
        echo "</tr>";
        echo "</table>";
        echo "</div>";
                 }

} else 
{
    echo "<div class='con'>";
     echo "<ul id='contestant'>0 Contestant</ul>";
     echo "</div>";

}

?>

$ _POST [＆＃39;用户名＆＃39;]是用户输入用户名的输入名称，因为它用作表名。还要确定应显示哪个表。我的问题是我的代码无法正常工作，这就是为什么我可以在不知道表名的情况下显示表格。