Question

我想创建一个单选按钮，它将具有一个独特的roomCode作为基于每一行的选择器。因此，如果roomCode = 1，该行中的唯一标识符将为1.这样我就可以提交一个单选按钮，根据所选的roomCode显示时间表。

这是我的PHP代码：

$numeroOption = $_POST['numero'];
$roomtype = $_POST['roomtype'];
$selectOption = $_POST['parkname'];
$query = "SELECT * FROM `ROOMS` WHERE `Capacity` < '$numeroOption' AND `Park` LIKE '$selectOption%' AND `dataProjector` LIKE '$proj_check%' AND `Whiteboard` LIKE '$white_check%' AND `OHP` LIKE '$ohp_check%' AND `WheelchairAccess` LIKE '$wheel_check%' AND `lectureCapture` LIKE '$cap_check%' AND `Style` LIKE '$roomtype%'"; 
$result = mysql_query($query);

if ($result == FALSE) 
    die ("could not execute statement $query<br />");

echo "<table>"; 
while($row = mysql_fetch_array($result))
{   
    echo "<tr><td>" . $row['roomCode'] . "</td><td>" . $row['Park'] . "</td><td>" . $row['Capacity'] . "</td><td>" . $row['Style'] . "</td><td>" . $row['dataProjector'] . "</td><td>" . $row['Whiteboard'] . "</td><td>" . $row['OHP'] . "</td><td>" . $row['wheelchairAccess'] . "</td><td>" . $row['lectureCapture'] . "</td></tr>";
}
echo "</table>"; 
mysql_close(); 
}

Answer 1

对于初学者，你应该使用mysqli或PDO。如果你将使用多个数据库（mysql和sql等），PDO是最好的，但如果没有，我更喜欢mysqli，因为我觉得它更有效。

解决问题...

    <form>
      while($row = mysql_fetch_array($result))
      { 
          echo "<tr><td><input type="radio" name="radioSelect" value= "" checked="" /></td><td>" . $row['roomCode'] . "</td><td>" . $row['Park'] . "</td><td>" . $row['Capacity'] . "</td><td>" . $row['Style'] . "</td><td>" . $row['dataProjector'] . "</td><td>" . $row['Whiteboard'] . "</td><td>" . $row['OHP'] . "</td><td>" . $row['wheelchairAccess'] . "</td><td>" . $row['lectureCapture'] . "</td></tr>";
    }
     }
    <input type="submit"/>
    </form>

为每个SQL结果行创建一个唯一的单选按钮

1 个答案: