我从我的MYSQL数据库中提取产品列表,并针对每个产品使用删除按钮,以防操作员想要删除产品。
问题是,每当我点击列表中任何产品的删除按钮时,第一个元素都会被删除。
我的代码在下面出了什么问题?
Products
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($rw);
?>
delete_prod.php
:
<?php
if(isset($_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($_POST['delete']);
}
else
{
echo "Delete operation Failed";
}
header('location:show_db.php');
}
?>
我认为我非常遗漏一些简单的观点,但我无法得到它是什么。
答案 0 :(得分:3)
很可能是因为您设置了id="delete"
。通常id属性值不重复。
echo "<td><form action='delete_prod.php' id='delete' method='get'>";
echo "<button type='submit' form = 'delete' class='btn btn-default' name='delete'>Delete</button>";
提交按钮获取第一个ID,从而获得第一个隐藏的输入。
或者,你可以像这样设计你的按钮并作为你的标记:
无需打印每张表格!只需用表格包裹它:
echo "<form action='delete_prod.php' id='delete' method='get'>";
echo '<table>';
while($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>';
然后在PHP处理中:
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!';
}
}
答案 1 :(得分:0)
检查prod_id
是否在表格中正确自动递增。另一件事是当你的表单在循环中时,所有表单的id都将被复制。因此,每次提交第一个表单时,这就是为什么只有第一个产品从您的记录中删除。
答案 2 :(得分:0)
$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;
}
删除delete_prod.php中的操作代码
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');
}
试试这个......