如何在变量中存储php查询？

Question

我想从我的DB中获取一个对象，并将该对象存储在一个可验证的$teacher中，稍后在$data中使用它

$teacher = "SELECT CONCAT(teacher.firstname, ' ', teacher.lastname) AS Name 
            FROM teacher 
            LEFT JOIN class ON teacher.id = teacher_fk 
            WHERE class.id = 1";


$data = "<h3 class='kartenUntertitel2'>teacher: '.$teacher.'</h3>";

我在我的网站上得到了以下结果：

<h3 class="kartenUntertitel2">teacher: SELECT CONCAT(teacher.firstname, ' ', teacher.lastname) AS Name FROM teacher LEFT JOIN class ON teacher.id = teacher_fk WHERE class.id = 1</h3>"

但我想要的是

<h3 class="kartenUntertitel2">teacher: Peter Muster</h3>

Answer 1

MySQL数据库的示例：

首先，您必须使用mysqli_connect()建立到数据库的连接。然后，您必须使用mysqli_query()执行查询。如果查询返回结果，则可以将其添加到$data变量中。

最后，您应该使用mysqli_free_result()释放(结果消耗的)内存，并关闭与mysqli_close()的连接。

$data = "";
$con = mysqli_connect("localhost", "my_user", "my_password", "my_db");
$sql = "SELECT CONCAT(teacher.firstname, ' ', teacher.lastname) AS Name 
            FROM teacher 
            LEFT JOIN class ON teacher.id = teacher_fk 
            WHERE class.id = 1";

// Perform query
if ($result = mysqli_query($con, $sql)) {
  $data = "<h3 class='kartenUntertitel2'>teacher: " . $result . "</h3>";
  mysqli_free_result($result);
}

mysqli_close($con);