<tr>
<td><label class="label label-success">Building Name: </label></td>
<td>
<select class="form-control" name="building_name" required>
<option value="">Select Building</option>
<?php
include '../Database/db.php';
$res=mysql_query("SELECT * FROM building");
while($row = mysql_fetch_array($res)){
?>
<option value="<?php echo $row['building_name']; ?>"><?php echo $row['building_name']; ?></option>
<?php } ?>
</select>
</td>
</tr>
<tr>
<td><label class="label label-success">Floor: </label></td>
<td>
<select class="form-control" name="floor" required>
<option value="">Select Floor</option>
<option>></option>
</select>
</td>
</tr>
谁能帮我 我想从数据库中显示所选建筑物的楼层。 如果我选择建筑物,楼层将显示在选择选项楼层上。
在不知道您的数据库设置或结构的情况下,下面是如何实现此目的的示例:
<?php require_once('../Database/db.php'); ?>
<tr>
<td><label class="label label-success">Building Name: </label></td>
<td>
<select class="form-control" name="building_name" required">
<option value="">Select Building</option>
<?php
$res=mysql_query("SELECT * FROM building");
while($row = mysql_fetch_array($res)){
?>
<option value="<?php echo $row['building_name']; ?>"><?php echo $row['building_name']; ?></option>
<?php } ?>
</select>
</td>
</tr>
<tr>
<td><label class="label label-success">Floor: </label></td>
<td>
<select class="form-control" name="floor" required>
<option value="">Select Floor</option>
<?php
$building_name = mysql_real_escape_string($_POST['building_name']);
$res2=mysql_query("SELECT * FROM floor WHERE building_id = '{$building_name}'");
while($row = mysql_fetch_array($res2)){
?>
<option value="<?php echo $row['floor_name']; ?>"><?php echo $row['floor_name']; ?></option>
<?php } ?>
</select>
</td>
</tr>
您需要将发布的参数传递给 SQL 中的 WHERE 子句。请注意,您应该从发布的参数中转义恶意代码。
我还假设您在 form
标签中具有此 html 代码以及提交按钮
您应该使用 AJAX 获取要在 DROPDOWN 中显示的数据库值
$('#building_name').change(function(){
var building_name= $('#building_name').val();
$.ajax({
url: "index.php", //Change it yours
type: "post",
data: {building_name: building_name},
success: function(data) {
if(data != "false")
{
$('#Floor').html(data);
}
else{
alert("No value");}
}
});
});
而 PHP 函数应该是
$building_name = mysql_real_escape_string($_POST['building_name']);
$res2=mysql_query("SELECT * FROM building WHERE building_name = '{$building_name}'");
while($row = mysql_fetch_array($res2)){
echo '<option value="'.$row['floor_name'].'">'.$row['floor_name'].'</option>';
}