Question

我正尽力做好相关链接。所以我有一个链接，该链接转到一个名为post.php的页面，并显示数据库中的正确帖子。现在我想在类别等于该帖子的另一个div中回显所有标题（及其链接）吗？

例如，我要显示的帖子是表中的第一条记录。它应该显示类别1的所有标题。但是，如果我显示的是类别2的页面，那么还将显示类别2的其他标题。

代码我已将页面回显到post.php：

<?php
require_once("db/db.php");
$sql = $db->prepare("
SELECT *
FROM mytable
WHERE slug=?

");
    $sql->bind_param("s",$_GET["slug"]);
    $sql->execute();
    $result = $sql->get_result();
    if ($result->num_rows > 0) {        
        $row = $result->fetch_assoc();
    }

?>

    <h2><?php echo $row['mytitle'];?></h2>
    <?php echo $row['mypost'];?>

+----+-----------------------------+-------------------------------+----------+------------------------+
| id | mytitle                     | mypost                        | category | slug                   |
+----+-----------------------------+-------------------------------+----------+------------------------+
| 1  | title of apost cat1         | it amet, sollicitudin euismod | 1        | my-first-post          |
+----+-----------------------------+-------------------------------+----------+------------------------+
| 2  | post of title cat1          | quam tempor mauris elem       | 1        | post-of-title          |
+----+-----------------------------+-------------------------------+----------+------------------------+
| 3  | Fun title Cat2              | verra mollis justo eget ti    | 2        | fun-title              |
+----+-----------------------------+-------------------------------+----------+------------------------+
| 4  | What about another one Cat2 | et consequat tortor et vive   | 2        | what-about-another-one |
+----+-----------------------------+-------------------------------+----------+------------------------+

Answer 1

您可以尝试

SELECT t1.mytitle,
   t1.mypost,
   t2.titles
FROM mytable AS t1
INNER JOIN
  (SELECT group_concat(mytitle) AS titles,
          category
   FROM mytable
   GROUP BY category) AS t2 ON t1.category=t2.category
WHERE t1.id=1

标题是与该特定帖子类别相关的所有标题。

Answer 2

您可以通过$row['category']获取当前的帖子类别并运行相关查询：

$sql = $db->prepare("
    SELECT *
    FROM mytable
    WHERE category=? AND id NOT IN (?)
    LIMIT 5
");
$sql->bind_param("i",$row['category']);
$sql->bind_param("i",$row['id']);

然后将结果循环到单独的div中。我认为最好通过ajax调用。

PHP mysqli回声相关链接按类别

2 个答案: