使用php刪除表格的特定行使用php

Question

這個問題非常基本，我在stackoverflow中發現了很多類似的問題，但是他們沒有一個適合我。

我已經設計了將顯示數據是這樣的表：

ID name Delete 
1 abc  Delete 
2 def  Delete 

用於上述顯示的代碼是

<?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); 
?> 

爲delete.php代碼

<?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"); 
?> 

查看數據庫是

Id name 
1 abc 
2 cdf 

的問題是，它不刪除數據，也沒有表現出任何差錯

我是新來這個領域，將不勝感激，如果有人可以幫助我

Answer 1

改變這一行：

echo "<td><a href='delete.php'>Delete</a></td>"; 

到

echo "<td><a href=\"delete.php?id=".$row['id']."\">Delete</a></td>"; 

那麼對於delete.php（和原先的評論稱，$id沒有定義）。

<?php 
$con=mysqli_connect("abc","abc","abc","abc"); 
// Check connection 
if (mysqli_connect_errno()) { 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 

$id = $_GET['id']; // $id is now defined 

// or assuming your column is indeed an int 
// $id = (int)$_GET['id']; 

mysqli_query($con,"DELETE FROM student WHERE id='".$id."'"); 
mysqli_close($con); 
header("Location: index.php"); 
?> 

，它會工作。

然而，爲了安全起見，你應該考慮使用mysqli with prepared statements，或PDO with prepared statements，他們更安全。我在下面列出了一個例子。

---

這裏是一個準備好的語句例如：

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

$con = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME); 
if($con->connect_errno > 0) { 
    die('Connection failed [' . $con->connect_error . ']'); 
} 

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

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

Answer 2

一個GET參數添加到您的鏈接：

echo "<td><a href='delete.php?id='".$row['id']."'>Delete</a></td>"; 

然後抓住它在你的delete.php查詢：

$id = $_GET['id']; 
mysqli_query($con,"DELETE FROM student WHERE id='".$id."'"); 

Answer 3

你可能不希望每一個行被刪除，你可以做到這一點使用一個Ajax請求到服務器時刷新頁面快多了。您可以從以下鏈接獲取完整代碼：delete individual row from databse using ajax