Question

我创建了一个html表单。在那里我有2个下拉列表。下拉列表1从sql数据库表中检索值并在那里显示。现在我想显示另一个下拉列表，其中包含来自下拉列表1的输入。

这是我的代码：

<label for="bname">Select a Building</label>
  <?php
session_start();
include 'db_connection.php';

$conn = OpenCon();
$sql = "SELECT bname FROM building_details";
    $result = mysqli_query($conn, $sql);

echo "<select name='bname'>";
    while ($row = mysqli_fetch_array($result)){
echo "<option value='". $row['bname'] ."'>".$row['bname'] ."</option>";
  }
 echo "</select>";

echo "<label for='rtype'>Select a rtype</label>";
$sql2 = "SELECT rtype FROM room_details WHERE bname='bname'";
$result2 = mysqli_query($conn,$sql2);

echo "<select name='rtype'>";
    while ($row2 = mysqli_fetch_array($result2)){
echo "<option value='". $row2['rtype'] ."'>".$row2['rtype'] ."</option>";
 }
 echo "</select>";

这里一旦bname值从building_details被选中然后根据该输入我需要显示另一个下拉列表，它是来自room_details表的rtype列。有人可以帮我这个???

Answer 1

mysqli_connect_errno（） - 返回上次连接调用的错误代码

mysqli_connect_error（） - 返回上次连接错误的字符串描述

mysqli_errno（） - 返回最近函数调用的错误代码

mysqli_sqlstate（） - 返回先前MySQL操作的SQLSTATE错误

如需更多帮助，请使用以下链接：http://php.net/manual/en/mysqli.error.php

Answer 2

这是我发现的唯一方法。我希望这就是你要找的东西！

<label for="bname">Select a Building</label>
  <?php
session_start();
include 'db_connection.php';

$conn = OpenCon();
$sql = "SELECT bname FROM building_details";
    $result = mysqli_query($conn, $sql);


$bnames = "";
echo "<select name='bname'>";
    while ($row = mysqli_fetch_array($result)){
echo "<option value='". $row['bname'] ."'>".$row['bname'] ."</option>";
$bnames .= $row['bname'] . ", ";
  }

$bnames = substr($bnames, 0, -2);
 echo "</select>";

echo "<label for='rtype'>Select a rtype</label>";

$sql2 = "SELECT rtype FROM room_details WHERE bname IN($bnames)";

$result2 = mysqli_query($conn,$sql2);

echo "<select name='rtype'>";
    while ($row2 = mysqli_fetch_array($result2)){
echo "<option value='". $row2['rtype'] ."'>".$row2['rtype'] ."</option>";
 }
 echo "</select>";

如果需要避免sql注入，请不要忘记使用htmlspecialchars函数转义$bnames变量。

基于另一个下拉列表的输入显示下拉列表时出错

2 个答案: