[英]insert delete button of specific row from a database displayed in table
您好,我有以下代碼顯示數據庫結果,我想在表的第3個TD中插入刪除按鈕,該按鈕將刪除該按鈕旁邊的數據的表記錄,第3個TD中的代碼應該是什么刪除按鈕的標簽?
<?php
$con=mysqli_connect("localhost","table","password","database");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM recetas_galletas");
echo "<table border='1'>
<tr>
<th>Title</th>
<th>Description</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['title'] . "</td>";
echo "<td>" . $row['description'] . "</td>";
echo "<td>" . DELETE BUTTON "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
您可以建立一個鏈接,將ID傳遞到另一個執行刪除操作的php頁面。 因此,鏈接將類似於:
echo "<td><a href='http://www.site.com/delete.php?id=" . $row['id'] . "'></td>";
然后在您的delete.php中:
if (is_int($_GET["id"]) {
$query = "DELETE FROM recetas_galletas WHERE id = " . $_GET["id"];
$result = mysqli_query($con, $query);
// Check the result and post confirm message
}
這是一種非常簡單的方法,但是對您而言可能不是最佳選擇。如果您不驗證此頁面的用戶,則任何人都可以輕松地操作此代碼並刪除表的任何行。
其他方法可能是導航到同一頁面並在同一頁面中執行相同的過程,或者您可以使用ajax,但這是另一個問題。
希望這對我有幫助,請原諒我的英語。
存檔的方法有多種,首先也是最重要的一點是,您需要在數據庫表上有一個字段,您可以在其中標識要刪除的記錄,例如ID形式的主鍵或唯一鍵。
您可以通過創建帶有指向delete.php
頁面的文本的鏈接來完成此操作,或者可以使用JQuery和AJAX,也可以使用內部表單。
您還希望只允許授權用戶使用這些頁面,因此您還需要一個帶有會話的登錄頁面。
最簡單的一個是到刪除頁面的鏈接,請參見此處的示例:
<?php
$con = mysqli_connect("localhost","table","password","database");
// Check connection
if (mysqli_connect_errno())
{
die("Failed to connect to MySQL: " . mysqli_connect_error());
}
if (!$result = mysqli_query($con,"SELECT * FROM recetas_galletas"))
{
die("Error: " . mysqli_error($con));
}
?>
<table border='1'>
<tr>
<th>Title</th>
<th>Description</th>
</tr>
<?php
while($row = mysqli_fetch_array($result))
{
?>
<tr>
<td><?php echo $row['title']; ?></td>
<td><?php echo $row['description']; ?></td>
<td><a href="delete.php?id=<?php echo $row['id']; ?>">Delete</a></td>
</tr>
<?php
}
mysqli_close($con);
?>
</table>
然后,在您的刪除頁面上,您將看到以下內容:
<?php
// Your database info
$db_host = '';
$db_user = '';
$db_pass = '';
$db_name = '';
if (!isset($_GET['id']))
{
echo 'No ID was given...';
exit;
}
$con = new mysqli($db_host, $db_user, $db_pass, $db_name);
if ($con->connect_error)
{
die('Connect Error (' . $con->connect_errno . ') ' . $con->connect_error);
}
$sql = "DELETE FROM recetas_galletas WHERE id = ?";
if (!$result = $con->prepare($sql))
{
die('Query failed: (' . $con->errno . ') ' . $con->error);
}
if (!$result->bind_param('i', $_GET['id']))
{
die('Binding parameters failed: (' . $result->errno . ') ' . $result->error);
}
if (!$result->execute())
{
die('Execute failed: (' . $result->errno . ') ' . $result->error);
}
if ($result->affected_rows > 0)
{
echo "The ID was deleted with success.";
}
else
{
echo "Couldn't delete the ID.";
}
$result->close();
$con->close();
echo "
<form action='user.php' method='post'>
<table cellpadding='2' cellspacing='2' border='2' >
<tr>
<td>Id</td>
<td>Name</td>
<td>Gender</td>
<td>Action</td>
</tr>
";
$select = "select * from user";
$result = mysqli_query($con,$select);
while($r = mysqli_fetch_row($result))
{
echo "
<tr>
<td>$r[0]</td>
<td>$r[1]</td>
<td>$r[2]</td>
<td><input type='submit' value='Delete $r[0]' style='width:53px;' name='submit' ></td>
</tr>
";
}
echo "
</table>
</form>
";
if ($_POST['submit'])
{
$id = $_POST['submit'];
$id = end(explode(" ",$id));
$delete = "delete from user where id=$id";
mysqli_query($con,$delete);
header("Location:user.php");
}
?>
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