Php 需要在引导中使用modal编辑表的行
我有一个网站,管理员登录后可以编辑一个“便利设施”列表,显示在他的个人资料上。我已经用下面的代码创建了一个简单的表格来显示便利设施Php 需要在引导中使用modal编辑表的行,php,mysql,sql,mysqli,phpmyadmin,Php,Mysql,Sql,Mysqli,Phpmyadmin,我有一个网站,管理员登录后可以编辑一个“便利设施”列表,显示在他的个人资料上。我已经用下面的代码创建了一个简单的表格来显示便利设施 <?php $con=mysqli_connect("abc.com","abc","abc","abc");// Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } $re
<?php
$con=mysqli_connect("abc.com","abc","abc","abc");// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM amenities");
echo "<table class='table table-striped table-bordered table-hover'>
<thead>
<tr>
<th>amenities</th>
<th>edit</th>
</tr>
</thead>";
while($row = mysqli_fetch_array($result))
{
echo "<tbody data-link='row' class='rowlink'>";
echo "<tr>";
echo "<td>" . $row['amenities'] . "</td>";
echo "<td> <a href='#edit' data-toggle='modal'> edit </a> </td>";
echo "</tr>";
echo "</tbody>";
}
echo "</table>";
mysqli_close($con);
?>
当管理员点击特定的编辑按钮(例如:放置在设施前面的编辑按钮1)时,将显示一个模式,允许管理员仅编辑该设施(在我们的情况下,它是设施1)
用于模态分析的代码为
<div class = "modal fade" id="edit" role="dialog">
<div class = "modal-dialog">
<div class = "modal-content">
<div class = "modal-header">
<h4> Edit Page </h4>
</div>
<div class="modal-body">
<form role="form" action="edit_amenities.php" method="post">
<div class="form-group">
<label for="exampleInputEmail1">Name of Amenities</label>
<input type="email" class="form-control" id="exampleInputEmail1" placeholder="Email" name="amenities">
</div>
<input name="submit" type="submit" value=" Save ">
</form>
</div>
<div class="modal-footer">
<a class="btn btn-primary" data-dismiss="modal"> Close </a>
</div>
</div>
</div>
</div>
我希望的是,当打开模式时,管理员可以输入一个新名称,替换旧名称并保存在数据库中。然而,我无法做到这一点。如果有人能帮助我,我将不胜感激。你为什么不使用这样的东西,举个例子:
<div class="modal-body">
<form action="edit_amenities.php" method="post">
<div class="form-group">
<label for="exampleInputEmail1">Name of Amenities</label>
<input type="text" class="form-control" name="update[amenities]">
</div>
<input name="submit" type="submit" value=" Save ">
</form>
</div>
编辑:在您的示例中:
$sql = "UPDATE amenities SET amenities='".$amenities."'";
使用WHERE,并在输入字段中,提供要更新的设施的ID。当我使用它时,我得到一个错误致命错误:当不在对象上下文中时使用$this。当然,因为您可能没有使用任何MVC。这只是Zend Framework中的一个示例。我已经尝试过使用id。我使用了以下代码:$sql=“更新便利设施集便利设施=”“$professionals.”其中id=”“$id.”;我试着使用它,但它没有显示任何错误,但也没有工作。得到了预期的结果。您是否回应了您的查询?结果是什么?
id amenities
1 q
2 a
<div class="modal-body">
<form action="edit_amenities.php" method="post">
<div class="form-group">
<label for="exampleInputEmail1">Name of Amenities</label>
<input type="text" class="form-control" name="update[amenities]">
</div>
<input name="submit" type="submit" value=" Save ">
</form>
</div>
if ($this->getRequest()->isPost()) {
$post = $this->getRequest()->getPost();
}
$amenitie = $this->amenitie->fetchRow(
$this->amenitie->select()->where('id = ?', $this->params['id'])
);
if ($amenitie) {
foreach ($post['update'] as $key => $val) {
$amenitie->{$key} = $val;
}
$amenitie->save();
}
$sql = "UPDATE amenities SET amenities='".$amenities."'";