SELECT无法获得所需的结果

Question

我是PHP的新手。我正在尝试根据用户查询显示搜索结果。我的问题是我没有得到所有其他类似的搜索结果（只显示了确切的结果）。从安全的角度来看，这是我正在实施的正确方法吗？提前谢谢。

define('HOST','localhost');
define('USER','root');
define('PASSWORD_HOST','');
define('DATABASE','test');

if(defined('HOST') && defined('USER') && defined('PASSWORD_HOST') && defined('DATABASE')){
    $conn = mysqli_connect(HOST, USER, PASSWORD_HOST, DATABASE);


}else{
    die(connection_failed.mysqli_connection_error());
}

这是HTML

<div class="container">
    <div class="row">
        <div class="col-sm-12">
            <form action="" method="POST">
            <h4>Search By</h4>
            <input type="text" name="delName"/>

            <button type="submit" name="submit">search</button>
            </form>
            </div>
        </div>
    </div>

这是PHP

if(isset($_POST['submit'])){

        $delName = "%{$_POST['delName']}%";

    $stmt =$conn->prepare("SELECT id, delName, medName, contact1, contact2, address, pin, creditLimitDealer FROM dealerentrytable WHERE delName LIKE ?");

   $stmt->bind_param("s", $delName); 
   $stmt->execute(); 
   $stmt->bind_result($id, $delName, $medName, $contact1, $contact2,$address,$pin,$creditLimitDealer);

   while ($stmt->fetch()) {

     echo "<table>";
     echo "<tr><td>ID: $id</td>";
     echo "<td>delName: $delName</td>";
      echo "<td>medName: $medName</td>";
       echo "<td>contact1: $contact1</td>";
        echo "<td>contact2: $contact2</td>";
         echo "<td>address: $address</td>";
          echo "<td>pin: $pin</td>";
           echo "<td>creditLimitDealer: $creditLimitDealer</td></tr>";
     echo "</table>";




    }

     $stmt->close();

    }
    ?>