php - 使用php删除表的特定行

Question

我设计了一个表格来显示如下数据:

ID  name    Delete
1   abc     Delete
2   def     Delete

上面显示用到的代码是

<?php
$con=mysqli_connect("abc","abc","abc","abc");
// Check connection
if (mysqli_connect_errno()) 
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM student");
echo "<table class='table table-striped table-bordered table-hover'>
<thead>
<tr>
<th>ID</th>
<th>name</th>
<th>delete</th>   
</tr>
</thead>";
while($row = mysqli_fetch_array($result)) 
{
echo "<tbody data-link='row' class='rowlink'>";
echo "<tr>";
echo "<td>" . $row['id'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td><a href='delete.php'>Delete</a></td>";
echo "</tr>";
echo "</tbody>";    
}
echo "</table>";
mysqli_close($con);
?>

delete.php 代码

<?php
$con=mysqli_connect("abc","abc","abc","abc");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"DELETE FROM student WHERE id='$id'");
mysqli_close($con);
header("Location: index.php");
?> 

数据库 View 是

Id  name
1   abc
2   cdf

问题是它没有删除数据，也没有显示任何错误

Answer 1

注意根据HTTP standard , GET 方法不应用于修改数据库中的数据。因此，您应该使用带有 method="POST" 的表单，例如

echo '<td><form method="POST" action="delete.php"><input type="hidden" value="'.$row['id'].'"><input type="submit" value="Delete"></form></td>';

然后对于 delete.php，为了安全起见，您应该考虑使用 mysqli with prepared statements , 或 PDO with prepared statements ，它们更安全。我在下面包含了一个示例。

---

这是一个准备好的语句示例:

<?php 
$DB_HOST = "xxx";
$DB_NAME = "xxx";
$DB_USER = "xxx";
$DB_PASS = "xxx";

mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$con = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);

$id = (int)$_POST['id'];

$update = $con->prepare("DELETE FROM student WHERE id = ?");
$update->bind_param('i', $id);
$update->execute();