Question

我似乎无法从使用JQuery链接到我的数据库的下拉框中获取值到SQL语句中？错误在哪里？

由于

这是我的PHP：

Route::post('admin/attributevalue','AttributeController@add_attribute_values');

Answer 1

1）从name="codedrop"赞

移除<option>

echo '<option class="option">'.$code.'</option>';

N.B。：只有<select>具有name属性，而不是<option>。

2）删除第一个<form>，它没用。甚至，不允许嵌套<form>。

3）更改

$codedropOption= $_POST['codedrop'];

要

$codedropOption= $_POST['combobox'];

没有得到

未定义的索引：codedrop

4）更改

$resql = "SELECT * FROM 'ROOMS' WHERE 'roomCode' LIKE '$codedrop%'";

要

$resql = "SELECT * FROM ROOMS WHERE roomCode LIKE '$codedrop%'";

使用backtick代替single quotes '。

5）更改

$resql = "SELECT * FROM 'ROOMS' WHERE 'roomCode' LIKE '$codedrop%'";

要

$resql = "SELECT * FROM ROOMS WHERE roomCode LIKE '$codedropOption%'";

更新代码

<form action="" method="post">
    <p class="timetable-p">Room code:
    <select id="combobox" name="combobox">
        <?php
      echo '<option class="option">Type/Select a room</option>';
      while ($row = $res->fetchRow()) {
          $code = $row['roomcode'];
          $titles[] = $row['park'];
          echo '<option class="option">'.$code.'</option>';
      }?>
    </select>
  <input type="submit" value="subm" name="subm">
</form>

<?php
if( isset( $_POST['subm'] ) )
{
    $codedropOption= $_POST['combobox'];

    $resql = "SELECT * FROM ROOMS WHERE roomCode LIKE '$codedropOption%'";
    $res1 = mysql_query($resql);

    echo "<table>";             
    while($row = mysql_fetch_array($res1)){                         
        echo "<tr><td>" . $row['roomCode'] . "</td>";
        echo "<td>" . $row['Style'] . "</td><td>" . $row['dataProjector'] . "</td>";
        echo "<td>" . $row['Whiteboard'] . "</td><td>" . $row['OHP'] . "</td>";
        echo "<td>" . $row['wheelchairAccess'] . "</td>";
        echo "<td>" . $row['lectureCapture'] . "</td>";
        echo "<td><input type='radio' name='radioSelect' value= '". $row['roomCode']."'></td>";
    }
    echo "</table>"; 
}
?>

使用jQuery从下拉框中定义未定义的索引

1 个答案: