【发布时间】:2019-01-12 01:34:27
【问题描述】:
我在溢出中的第一个问题 :)
我试图创建一个表来显示来自 SQL 的所有帖子。 我已经完成了下一个代码:
function show_all_posts_in_table(){
global $connection;
$query = "SELECT * FROM posts";
$all_posts_content_query = mysqli_query($connection,$query);
while ($row = mysqli_fetch_assoc($all_posts_content_query)) {
$post_id_display = $row['post_id'];
echo "<tr>";
echo "<td>" . $post_id_display = $row['post_id'] . "</td>";
echo "<td>" . $post_category_display = $row['post_category_id'] . "</td>";
echo "<td>" . $post_title_display = $row['post_title'] . "</td>";
echo "<td>" . $post_author_display = $row['post_author'] . "</td>";
echo "<td>" . $post_date_display = $row['post_date'] . "</td>";
$post_image_display = $row['post_image'];
echo "<td>" . "<img src= '../images/{$post_image_display}' width='200' height='auto'>" . "</td>";
echo "<td>" . $post_content_display = $row['post_content'] . "</td>";
echo "<td>" . $post_tags_display = $row['post_tags'] . "</td>";
echo "<td>" . $post_status_display = $row['post_status'] . "</td>";
echo "<td>{$post_id_display}</td>";
echo "<td> <a href='posts.php?delete={$post_id_display}'>DELETE</a> </td>";
echo "</tr>";
}
}
我看到了我创建的表格中的所有帖子,并且一切正常。
问题是当我尝试删除其中一篇文章时(使用“get”方法删除href)。
由于某种我无法理解的原因,在 chrome 的地址行上我得到了一个额外的</td>。
我已被定向到:posts.php?delete=6</td>。
谁能理解为什么会这样?
谢谢!
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