当我单击“删除”按钮时，“下一步”图像将被删除，而不是我在php mysql映像库中单击的图像。

Question

表单以上载图像

 <form method='post' action='' enctype='multipart/form-data'>
 <input type='file' name='file'>
 <input type='submit' value='Upload'>
 </form>

php编码，用于插入和显示数据库会话中的图片库以存储要删除的图像的id。

<?php 
session_start();
$conn = mysqli_connect("localhost","root","","img") or die ("connection failed ");
if ($_FILES)
{
$name = $_FILES['file']['name'];
$type = $_FILES['file']['type'];
$insert = "insert into image (image) values ('".$name."')";
$sql = mysqli_query($conn,$insert);
}
$view = mysqli_query($conn,"select * from image");
while($row = mysqli_fetch_array($view))
{   
extract($row);
echo "<form action=\"\" method=\"post\"><input type=\"submit\" name=\"del\" value=\"submit\"></form><a href=\"$image\"><img src=\"$image\" /></a>";
if(isset($_POST['del']))
{   
$_SESSION['del_id'] = $id;
header("location:del.php");
}
}
?>

del.php会话id将被删除

<?php
session_start();
$conn = mysqli_connect("localhost","root","","img") or die ("connection failed ");
$id = $_SESSION['del_id'];
$sql = mysqli_query($conn,"delete from image where id = '$id' ");
header("location:x.php");
?>

Answer 1

但是，我不知道从哪里获取id值来设置del_id.，您需要这样的东西：

取代：

 echo "<form action=\"\" method=\"post\"><input type=\"submit\" name=\"del\" value=\"submit\"></form><a href=\"$image\"><img src=\"$image\" /></a>";

在这方面：

echo "<form action=\"\" method=\"post\">";
echo "<input type=\"submit\" name=\"del\" value=\"submit\">";
echo "<input type=\"hidden\" name=\"del_id\" value=\"$row['id']\">";
echo "</form>";
echo "<a href=\"$image\"><img src=\"$image\" /></a>";

并将$_SESSION['del_id'] = $id;更改为$_SESSION['del_id'] = $_POST['del_id'];

Answer 2

$_SESSION['del_id'] = $id;

你在哪里找到这个$id的？