如何显示一个搜索过滤值的查询表？

Question

我的搜索和过滤 PHP 代码没有显示我搜索和过滤的值的查询表。我该如何解决这个问题？

PHP

if(isset($_POST['search']))
{
    $valueToSearch = $_POST['valueToSearch'];
    // search in all table columns
    // using concat mysql function
    $query = "SELECT * FROM `details` WHERE CONCAT('id','date_of_entry','region','province_of_provider') LIKE '%".$valueToSearch."%'";
    $search_result = filterTable($query);
}
 else {
    $query = "SELECT * FROM `details`";
    $search_result = filterTable($query);
}

// function to connect and execute the query
function filterTable($query)
{

$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "dbname";

    $connect = mysqli_connect($servername, $username, $password, $dbname);
    $filter_Result = mysqli_query($connect, $query);
    return $filter_Result;
}

while ($trainee = mysqli_fetch_array($search_result)){

    echo "<tr>";
    echo "<td>".$trainee['id']."</td>"; 
    echo "<td>".$trainee['date_of_entry']."</td>";
    echo "<td>".$trainee['region']."</td>";
    echo "<td>".$trainee['province_of_provider']."</td>";
    echo "</tr>";

}// End while

?>

当我从textbox 搜索和过滤时，表格行不包含任何值，包括其边框，但每列的表格标题仍然存在。此外，当我将搜索留空并单击过滤器按钮时，所有表行都包含在数据库表中找到的所有数据的所有值。

第一次编辑：

我改变了我的：

$query = "SELECT * FROM `details` WHERE CONCAT('id','date_of_entry','region','province_of_provider') LIKE '%".$valueToSearch."%'";

到这里：

$query = "SELECT * FROM `details` WHERE CONCAT_WS(' ','id','date_of_entry','region','province_of_provider') LIKE '%".$valueToSearch."%'";

还是一样，我的搜索结果是一个空表，但只有表列标题。

Answer 1

替换这部分：

$query = "SELECT * FROM `details` WHERE CONCAT('id','date_of_entry','region','province_of_provider') LIKE '%".$valueToSearch."%'";

这样：

$query = "SELECT * FROM `details` WHERE CONCAT(`id`, `date_of_entry`, `region`, `province_of_provider`) LIKE '%".$valueToSearch."%'";