PHP Mysql删除查询无法正常工作
我正在从MYSQL数据库中提取产品列表,并对每个产品使用删除按钮,以防操作员想要删除该产品 问题是每次我点击列表中任何产品的delete按钮时,第一个元素都会被删除 我下面的代码有什么问题PHP Mysql删除查询无法正常工作,php,mysql,mysqli,Php,Mysql,Mysqli,我正在从MYSQL数据库中提取产品列表,并对每个产品使用删除按钮,以防操作员想要删除该产品 问题是每次我点击列表中任何产品的delete按钮时,第一个元素都会被删除 我下面的代码有什么问题 产品页面: <?php $link=mysqli_connect("localhost","root","","smartcart"); $prod="select * from products"; $rw=mysqli_query($link,$prod) or die(mysqli_errno()
产品
页面:
<?php
$link=mysqli_connect("localhost","root","","smartcart");
$prod="select * from products";
$rw=mysqli_query($link,$prod) or die(mysqli_errno()."in query $prod");
$count=1;
while($row=mysqli_fetch_assoc($rw))
{
echo "<tr>";
echo "<td>".$count."</td>";
echo "<td>".$row['prod_id']."</td>";
echo "<td>".$row['prod_name']."</td>";
echo "<td>".$row['prod_price']."</td>";
echo "<td><form action='delete_prod.php' id='delete' method='get'>";
echo "<input type='hidden' name='prod_id' value='".$row['prod_id']."' />";
echo "<button type='submit' form = 'delete' class='btn btn-default' name='delete'>Delete</button>";
echo "</form></td>";
$count=$count+1;
}
mysqli_free_result($rw);
?>
我想我非常遗漏了一些简单的要点,但我无法理解它是什么。很可能是因为您设置了
id=“delete”
。通常id属性值不会重复
echo "<td><form action='delete_prod.php' id='delete' method='get'>";
echo "<button type='submit' form = 'delete' class='btn btn-default' name='delete'>Delete</button>";
检查
prod\u id
是否正确自动递增或不在表中。另一件事是,当表单处于循环中时,所有表单的id都将被复制。所以每次它提交第一个表单时,这就是为什么只从您的记录中删除第一个产品
$link=mysqli_connect("localhost","root","","smartcart");
$prod="select * from products";
$rw=mysqli_query($link,$prod) or die(mysqli_errno()."in query $prod");
$count=1;
while($row=mysqli_fetch_assoc($rw))
{
echo "<tr>";
echo "<td>".$count."</td>";
echo "<td>".$row['prod_id']."</td>";
echo "<td>".$row['prod_name']."</td>";
echo "<td>".$row['prod_price']."</td>";
echo "<td><form action='delete_prod.php' method='get'>";
echo "<input type='hidden' name='prod_id' value='".$row['prod_id']."' />";
echo "<input type='submit' value='Delete' class='btn btn-default' name='delete'/>";
echo "</form></td>";
$count=$count+1;
}
试试这个…我真的希望这不是互联网上的代码-这是一个即将发生的安全事件!切勿直接使用来自
$\u GET
、$\u POST
或$\u REQUEST
的输入!先把它清理干净,即使你确定它没问题!您是否在delete\u prod.php上收到$\u请求['prod\u id']?你有什么错误吗?应该在本地机器上使用。从不上网。是的,我收到$\u请求['prod\u id']
,但它是第一个产品的id,而不是我要删除的产品的id。否,无错误。页面上列出了多个产品,每个产品都有一个删除按钮,带有一个隐藏的输入字段,其中包含产品的prod\u id
。当我点击delete按钮时,这个特定的prod\u id
应该被带到delete\u prod.php
页面。但是,第一个列出的产品的prod\u id
被携带了出来。我还没有尝试将表格包装到表单中。将form='id'
从按钮中删除。另外,我还不熟悉您建议的面向对象的形式,但我一定会尝试一下。你能解释一下为什么form='id'
会导致错误吗?而且$\u GET['prod'u id']
现在应该是$\u GET['delete']
,因为你现在已经给按钮分配了值,它的名字是delete
?@user3736335是的,很抱歉,我修改了错误的代码indexing@user3736335对这就是问题所在,问题在于该id,所发生的是您的按钮被指定用于id为id=“delete”
的表单。当您循环时,您的标记将明显重复。然后,当您单击delete按钮时,它会得到第一个id为delete的表单,然后它会得到第一个表单,因此只指向第一个隐藏的表单input@user3736335我还建议适应mysqli的面向对象接口,这样您就不需要将$link
提供给那些需要它的函数。永远不要在查询字符串中直接注入用户输入变量。同时使用prepared语句。它将提交其按钮被单击的表单,而不是第一个表单。使用“查看页面源代码”,您将发现每个表单都有相同的id。虽然您单击的是最后一个表单,但它将调用id首先匹配的表单。您可以尝试此代码-echo“”;回声“;回应“删除”;回声“;
echo "<form action='delete_prod.php' id='delete' method='get'>";
echo '<table>';
while($row = mysqli_fetch_assoc($result)) {
$prod_id = $row['prod_id'];
echo "<tr>";
echo "<td>".$count."</td>";
echo "<td>".$row['prod_id']."</td>";
echo "<td>".$row['prod_name']."</td>";
echo "<td>".$row['prod_price']."</td>";
echo "<td>";
// each id is assigned to each button, so that when its submitted you get the designated id, the one that you clicked
echo "<button type='submit' value='$prod_id' class='btn btn-default' name='delete'>Delete</button>";
echo "</td>";
echo '</tr>';
}
echo '</table>';
echo '</form>';
if(isset($_GET['delete'])) // as usual
{
include "connection.php";
$prod_id = $_GET['delete']; // get the id
// USE PREPARED STATEMENTS!!!
$del="DELETE FROM products WHERE prod_id = ?";
$delete = $link->prepare($del);
$delete->bind_param('i', $prod_id);
$delete->execute();
// don't echo anything else, because you're going to use header
if($delete->affected_rows > 0) {
header('location:show_db.php');
} else {
echo 'Sorry delete did not push thru!';
}
}
$link=mysqli_connect("localhost","root","","smartcart");
$prod="select * from products";
$rw=mysqli_query($link,$prod) or die(mysqli_errno()."in query $prod");
$count=1;
while($row=mysqli_fetch_assoc($rw))
{
echo "<tr>";
echo "<td>".$count."</td>";
echo "<td>".$row['prod_id']."</td>";
echo "<td>".$row['prod_name']."</td>";
echo "<td>".$row['prod_price']."</td>";
echo "<td><form action='delete_prod.php' method='get'>";
echo "<input type='hidden' name='prod_id' value='".$row['prod_id']."' />";
echo "<input type='submit' value='Delete' class='btn btn-default' name='delete'/>";
echo "</form></td>";
$count=$count+1;
}
if(isset($_GET['delete']))
{
include "connection.php";
$prod_id=$_REQUEST['prod_id'];
$del="delete from products where prod_id=$prod_id";
if (mysqli_query($link,$del))
{
echo "Successfully deleted";
unset($_GET['delete']);
}
else
{
echo "Delete operation Failed";
}
header('location:show_db.php');
}