php - 在mysqli中把这行代码改成什么

Question

我正在尝试将我的代码从 mysql 更改为 mysqli，但是我在更改这段代码时遇到了问题:

while($sqlrow = mysql_fetch_array($sqlresult)) 
{ 
     $room = $sqlrow['Room']; 
     $roomHTML .= "<option value='".$room."'>" . $room . "</option>".PHP_EOL;  
} 


echo $roomHTML; 

我在下面的 mysqli 代码中没有变量 $sqlrow，我如何将行 $room = $sqlrow['Room']; 更改为下面:

完整代码:

    $building = isset($_POST['building']) ? $_POST['building'] : ''; 

  $sql = "SELECT Room FROM Room WHERE Building = ?"; 

 $sqlstmt=$mysqli->prepare($sql);

 $sqlstmt->bind_param("s",$building);


     $sqlstmt->execute(); 

     $sqlstmt->bind_result($dbRoom);


    $roomHTML  = "";  

     while($sqlstmt->fetch()) { 
         $room = $sqlrow['Room']; 
         $roomHTML .= "<option value='".$room."'>" . $room . "</option>".PHP_EOL;  
    } 


    echo $roomHTML; 

     $sqlstmt->execute(); 

上面的代码试图在下拉菜单中显示所有房间，这些房间与它们所属的建筑物相匹配。

谢谢

Answer 1

您没有使用绑定(bind)结果 var $dbRoom。

// Your output value of interest is $dbRoom as you have bound here:
$sqlstmt->bind_result($dbRoom);
$roomHTML  = "";  

 while($sqlstmt->fetch()) { 
     // $dbRoom holds the value fetched from the result set, since that is what you bound above.
     // Note: added htmlentities() to encode quotes if there are any embedded in the room value.
     $roomHTML .= "<option value='". htmlentities($dbRoom, ENT_QUOTES) ."'>" . htmlspecialchars($dbRoom) . "</option>".PHP_EOL;  
}