Question

您好我有以下代码显示数据库结果，在表的第3个TD中我想插入删除按钮，这将删除按钮旁边的数据的表记录，应该是什么代码删除按钮的tabe的第3个TD？

<?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);
?>

Answer 1

您可以创建一个将id传递给另一个预先形成删除的php页面的链接。所以链接将是：

echo "<td><a href='http://www.site.com/delete.php?id=" . $row['id'] . "'></td>";

然后在你的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
}

这是一种非常简单的方法，但它可能不是最适合您的。如果您没有验证此页面的用户，任何人都可以轻松操作此代码并删除该表的任何行。 / p>

其他方式可能导航到同一页面并在同一页面中执行相同的过程，或者您可以使用ajax但这是另一个问题。

希望这有助于我的英语。

Answer 2

有几种方法可以存档，首先，最重要的是您需要在数据库表上使用一个字段来识别要删除的记录，例如ID或唯一形式的主键键。

您可以通过创建带有delete.php页面文本的链接来完成此操作，也可以使用JQuery和AJAX，也可以使用内部表单。

您还希望只有授权用户才能使用这些页面，因此您还需要一个包含会话的登录页面。

You can see here an example of login page with sessions.

最简单的是删除页面的链接，请参见此处的示例：

<?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>

然后在删除页面上你会有这样的东西：

<?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();

Answer 3

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");
}
?>

从表中显示的数据库中插入特定行的删除按钮

3 个答案: