Php 通过生成的带有记录的表从mysql中删除行

我用了2页。在一个页面上，它生成一个带有记录和删除按钮的表。按delete键后，进入第二页，该页应删除记录。但事实并非如此。下面是我正在使用的代码

附：代码改编自我不久前通过谷歌找到的一个教程

删除_overzicht.php

<?php
// Load Joomla! configuration file
require_once('../../../configuration.php');
// Create a JConfig object
$config = new JConfig();
// Get the required codes from the configuration file
$server = $config->host;
$username   = $config->user;
$password   = $config->password;
$database = $config->db;
// Connect to db
$con = mysqli_connect($server,$username,$password,$database);
if (!$con){
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,$database);
// Get results
$result = mysqli_query($con,"SELECT * FROM cypg8_overzicht");

echo "<table border='1' id='example' class='tablesorter'><thead><tr><th>Formulier Id</th><th>Domeinnaam</th><th>Bedrijfsnaam</th><th>Datum</th><th>Periode</th><th>Subtotaal</th><th>Dealernaam</th><th>Verwijderen</th></tr></thead><tbody>";

while($row = mysqli_fetch_array($result))
  {
  echo "<tr>";
  echo "<td>" . $row['formuliernummer'] . "</td>";
  echo "<td>" . $row['domeinnaam'] . "</td>";
  echo "<td>" . $row['bedrijfsnaam'] . "</td>";
  echo "<td>" . $row['datum'] . "</td>";
  echo "<td>" . $row['periode'] . "</td>";
  echo "<td> &euro; " . $row['subtotaal'] . "</td>";        
  echo "<td>" . $row['dealercontactpersoon'] . "</td>";     
  echo "<td><a href='delete.php?id=" . $row['id'] . "'>Verwijderen </a></td>";
  echo "</tr>";
  }
echo "</tbody></table>";

mysqli_close($con);
?>

delete.php

<?php
// Load Joomla! configuration file
require_once('../../../configuration.php');
// Create a JConfig object
$config = new JConfig();
// Get the required codes from the configuration file
$server = $config->host;
$username   = $config->user;
$password   = $config->password;
$database = $config->db;
// Connect to db
$con = mysqli_connect($server,$username,$password,$database);
if (!$con){
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,$database);

// Check whether the value for id is transmitted
if (isset($_GET['id'])) {

// Put the value in a separate variable
$id = $_GET['id'];

// Query the database for the details of the chosen id
$result = mysqli_query($con,"DELETE * FROM cypg8_overzicht WHERE id = $id");

} else {
die("No valid id specified!");
}
?>

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感谢所有愿意帮忙的人

以下答案来自用户Andrewsi。这一答案已张贴在评论中

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DELETE的语法是

DELETE FROM

-您需要删除

*

DELETE的语法是

DELETE FROM

-您需要删除

*

不要使用GET查询执行删除操作：@andrewski感谢您的回答。这是正确的钱。它工作得很好。你介意把它作为答案贴出来吗。这样我才能选择它作为正确答案。@MarcB谢谢你的警告。我很高兴你能帮助我。如果你能告诉我怎么做的话，我会很高兴的。因为我不知道。再次感谢。@MarcB执行删除操作的安全性如何？