如何根据从动态下拉列表中选择的选项在html中显示相应的mysql列 [英] How to display corresponding mysql columns in html based on option selected from dynamic dropdown list
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问题描述
在用户从动态下拉框中选择一个选项后，我想要求帮助显示相应的mysql表列。我真的不知道我在哪里出了问题:(请帮忙：（$ / b>
我有3个mysql表：建筑物，delivery_transaction和位置，全部相互连接。主表是交货交易：
[delivery_transaction表，其中building_ID和location_ID是剩余2个表中的FK] [1]
其中如果用户点击下拉列表中显示的任何建筑物名称，它将只显示我从主表格中查询的列，如下所示。
< form name =bldg_formmethod =postaction => ;
<？php
//建立与db的连接
$ con = new mysqli（localhost， ！，，user_databases）;
if（！$ con）
{
echo无法连接！;
}
//从delivery_transaction，建筑物和位置表中选择列
$ query_ mysqli_query（$ con，SELECT delivery_status，starting_time，
arrival_time，duration，buildings.building_name，
location.location_name from delivery_transaction，buildings，
location where delivery_transaction.building_ID =
buildings.building_ID and delivery_transaction.location_ID =
location.location_ID）;
？>
<！ - 创建下拉框 - >
< select name ='bldg'>
< option value =>选择建筑< / option>;
<？php
while（$ row = mysqli_fetch_assoc（$ query））
{
if（$ row ['building_name'] == $ selectedbldg）
{
echo'< option value = \\'''。$ row ['building_ID']。'
selected>'。$ row ['building_name']。'< / option>' ;
}
else
{
echo'< option value
= \''。$ row ['building_ID']。'>'。$行[ 'building_name'] '< /选项>'。
}
}
？>
< / select>
< input type =submitname =view/>
< / form>
< section class =row text-center placeholders>
< div class =table-responsive>
< p>
< table class =table table-striped>
< thead>
< tr>
< th>传送状态< / th>
< th>开始时间< / th>
< th>到达时间< th>
< th>持续时间< / th>
< th>位置< / th>
< / tr>
< / thead>
< tbody>
< tr>
<？php
if（isset（$ _ POST ['bldg']））
{
while（$ row = mysqli_fetch_row（$ query））
{
echo< tr> ;.< td>。$ row ['delivery_status']。< / td>。
< td>。 $ row ['arrival_time']。< / td>。
< td>。
< td>。$ row ['location_name']。< / td>。< / td> / TR>中;
}
}
else
{
echo没有结果显示;
}
？>
< / tr>
< / tbody>
< / table>
< / p>
< / div>
< / section>
< / main>
我想要做的是如果用户点击一个选项，它会显示
对应表，就像我查询的一样。但是，没有显示:(
[此链接显示用户选择一个选项] [2]
[1]：https：//i.stack.imgur.com/H78Gp.png
[2]：https://i.stack.imgur.com/pA6gI.png
解决方案
如果我理解正确，您可以使用一个下拉菜单来显示所有建筑物。
提交后，这是2个查询（你只有一个）。
这是我如何做到这一点：
//开发开发PHP错误报告绝不是一个坏主意
error_reporting（E_ALL） ;
ini_set（'display_errors'，1）;
$ con = new mysqli（localhost，root，，user_databases）;
//总是查询建筑物bc我们需要它作为下拉菜单
$ bquery = mysqli_query（$ con，SELECT building_ID，building_name FROM buildings）;
$ selectedbldg = null;
//如果表单提交了
if（！emp ty（$ _ POST ['bldg']））{
// store selected building_ID
$ selectedbldg = $ _POST ['bldg'];
//基于建筑物查询交付;
//注意附加条件（我假设building_ID是一个整数）
$ dquery = mysqli_query（$ con，
SELECT delivery_status，starting_time，arrival_time，duration，buildings.building_name，
location.location_name
FROM delivery_transaction，buildings，location
WHERE delivery_transaction.building_ID = buildings.building_ID
AND delivery_transaction.location_ID = location.location_ID
AND buildings.building_ID = {$ selectedbldg}
）;
//尽管它不是您问题范围的一部分，但您应该使用
之前的准备语句
？>
< form name =bldg_formmethod =postaction =>
< select name =bldg>
< option value =>选择建筑< / option>;
<？php while（$ row = mysqli_fetch_assoc（$ bquery））：？>
< option value =<？= $ row ['building_ID']？> <？= $ row ['building_name'] == $ selectedbldg？ 'selected'：''？>><？= $ row ['building_name']？>< / option>
<？php endwhile？>
< / select>
< input type =submitname =view/>
< / form>
< section class =row text-center placeholders>
< div class =table-responsive>
< table class =table table-striped>
< thead>
< tr>
< th>传送状态< / th>
< th>开始时间< / th>
< th>到达时间< th>
< th>持续时间< / th>
< th>位置< / th>
< / tr>
< / thead>
< tbody>
<？php if（isset（$ dquery）&& mysqli_num_rows（$ dquery））：？>
<？php while（$ row = mysqli_fetch_assoc（$ dquery））：？>
< tr>
< td><？= $ row ['delivery_status']？>< / td>
< td><？= $ row ['starting_time']？>< / td>
< td><？= $ row ['arrival_time']？>< / td>
< td><？= $ row ['duration']？>< / td>
< td><？= $ row ['location_name']？>< / td>
< / tr>
<？php endwhile？>
<？php else：？>
< tr>
< td>无结果显示< / td>
< / tr>
<？php endif？>
< / tbody>
< / table>
< / div>
< / section>
好读：
I would like to ask for help in displaying the corresponding mysql table columns after the user selects an option from the dynamic dropdown box. I really don't know where I went wrong :( please help :(
I have 3 mysql tables: buildings, delivery_transaction and location, all connected to each other. The main table is the delivery transaction:
[delivery_transaction table, where building_ID and location_ID are the FKs from the remaining 2 tables][1]
Wherein if user will click on whatever building name is present in dropdown list, it will display only the columns I queried from the main table as follows.
Here's my code so far:
<form name="bldg_form" method="post" action="">
<?php
//establish sql connection with db
$con = new mysqli("localhost" ,"root" ,"" ,"user_databases");
if(!$con)
{
echo "Failed to connect!";
}
//select columns from delivery_transaction, buildings and location table
$query = mysqli_query($con, "SELECT delivery_status, starting_time,
arrival_time, duration, buildings.building_name,
location.location_name from delivery_transaction, buildings,
location where delivery_transaction.building_ID =
buildings.building_ID and delivery_transaction.location_ID =
location.location_ID");
?>
<!--Creates dropdown box-->
<select name = 'bldg'>
<option value = "">Choose Building</option>;
<?php
while($row = mysqli_fetch_assoc($query))
{
if($row['building_name'] == $selectedbldg)
{
echo '<option value = \"'.$row['building_ID'].'"
selected>'.$row['building_name'].'</option>';
}
else
{
echo '<option value
=\"'.$row['building_ID'].'">'.$row['building_name'].'</option>';
}
}
?>
</select>
<input type="submit" name="view"/>
</form>
<section class="row text-center placeholders">
<div class="table-responsive">
<p>
<table class="table table-striped">
<thead>
<tr>
<th>Delivery Status</th>
<th>Starting Time</th>
<th>Arrival Time</th>
<th>Duration</th>
<th>Location</th>
</tr>
</thead>
<tbody>
<tr>
<?php
if(isset($_POST['bldg']))
{
while($row = mysqli_fetch_row($query))
{
echo "<tr>"."<td>".$row['delivery_status']."</td>"."
<td>".$row['starting_time']."</td>"."
<td>".$row['arrival_time']."</td>"."
<td>".$row['duration']."</td>"."
<td>".$row['location_name']."</td>"."</tr>";
}
}
else
{
echo "No results to display";
}
?>
</tr>
</tbody>
</table>
</p>
</div>
</section>
</main>
What I want to do is if user clicks on an option, it will display the
corresponding table just like I queried. However, nothing displays :(
[This link shows user choosing an option][2]
[1]: https://i.stack.imgur.com/H78Gp.png
[2]: https://i.stack.imgur.com/pA6gI.png
解决方案
If I understood correctly, you have one dropdown to show all the buildings.
Upon submission, show a table of deliveries based on the selected building.
That is 2 queries (you only have one).
Here's how I would do it:
// never a bad idea to turn on your PHP error reporting in development
error_reporting(E_ALL);
ini_set('display_errors', 1);
$con = new mysqli("localhost" ,"root" ,"" ,"user_databases");
// always query buildings bc we need it for the dropdown
$bquery = mysqli_query($con, "SELECT building_ID, building_name FROM buildings");
$selectedbldg = null;
// if the form was submitted
if (!empty($_POST['bldg'])) {
// store selected building_ID
$selectedbldg = $_POST['bldg'];
// query deliveries based on building;
// note the additional condition (I'm assuming building_ID is an integer)
$dquery = mysqli_query($con, "
SELECT delivery_status, starting_time, arrival_time, duration, buildings.building_name,
location.location_name
FROM delivery_transaction, buildings, location
WHERE delivery_transaction.building_ID = buildings.building_ID
AND delivery_transaction.location_ID = location.location_ID
AND buildings.building_ID = {$selectedbldg}
");
// though it is not part of the scope of your question,
// you should probably use prepared statement above
}
?>
<form name="bldg_form" method="post" action="">
<select name="bldg">
<option value="">Choose Building</option>;
<?php while ($row = mysqli_fetch_assoc($bquery)) : ?>
<option value="<?= $row['building_ID'] ?>" <?= $row['building_name'] == $selectedbldg ? 'selected' : '' ?>><?= $row['building_name'] ?></option>
<?php endwhile ?>
</select>
<input type="submit" name="view" />
</form>
<section class="row text-center placeholders">
<div class="table-responsive">
<table class="table table-striped">
<thead>
<tr>
<th>Delivery Status</th>
<th>Starting Time</th>
<th>Arrival Time</th>
<th>Duration</th>
<th>Location</th>
</tr>
</thead>
<tbody>
<?php if (isset($dquery) && mysqli_num_rows($dquery)) : ?>
<?php while($row = mysqli_fetch_assoc($dquery)) : ?>
<tr>
<td><?= $row['delivery_status'] ?></td>
<td><?= $row['starting_time'] ?></td>
<td><?= $row['arrival_time'] ?></td>
<td><?= $row['duration'] ?></td>
<td><?= $row['location_name'] ?></td>
</tr>
<?php endwhile ?>
<?php else : ?>
<tr>
<td>No results to display</td>
</tr>
<?php endif ?>
</tbody>
</table>
</div>
</section>
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