連接兩個表以匹配整數以獲得用戶名

Question

我在弄清楚如何連接兩個表並從不同的數據庫表中SELECT不同的行時遇到了麻煩。

我想添加我的users表以將id與post_creator匹配，然后從users表中獲取ID和post_creator編號均匹配的users username 。

因此，如果我的用戶表的ID為10，而那是吉姆的用戶名。 如果post_creator值為10，我希望找到Jim並將其回顯。

現在，這正在殺死代碼，我不知道自己在做什么錯。

while ($row = $stmt->fetch()) {
        //added in topic title variable
        $topic_title;
//  foreach($stmt as $row) {
        //Prepared SELECT stmt to get forum posts
        $stmt2 = $con->prepare("SELECT forum_posts.id, forum_posts.category_id, forum_posts.topic_id, forum_posts.post_creator, forum_posts.post_content, forum_posts.post_date users.id
        FROM forum_posts LEFT JOIN users WHERE forum_posts.category_id=? AND forum_posts.topic_id=? AND forum_posts.post_creator=? AND users.id=?");
        if($stmt2===false) {
            die();
        } else {
        //var_dump($stmt2);

            $stmt2->bind_param("ii", $cid, $tid);
            $stmt2->execute();
            $stmt2->store_result();
            $stmt2->bind_result($post_id, $post_category_id, $post_topic_id, $post_creator, $post_content, $post_date);
            if (!$stmt2) {
            throw new Exception($con->error);
            }
        }
    $num_rows2 = $stmt2->num_rows;
    if($num_rows2) {
        $count2=0;
        while($stmt2->fetch()) {
        $count2++;
        $post_id;
        $post_category_id;
        $post_topic_id;
        $post_creator;
        $post_content;
         $post_date = fixDate($post_date);
        //$post_date;
            echo "<tr><td valign='top' style='border: 1px solid #000000;'>
            <div style='min-height: 125px;'>".$topic_title. "Post ".$count2."<br />
            by ".$post_creator." - " .$post_date. "<hr />" . $post_content ."</div></td>
            <td width='200' valign='top' align='center' style='border: 1px solid #000000;'>User Info Here!</td></tr>
            <tr><td colspan='2'><hr /></td></tr>";
        }

Answer 1

您可以聯接表...諸如此類（聯接用戶，類別和主題）

SELECT forum_posts.id post_id, forum_posts.category_id post_category_id, forum_posts.topic_id post_topic_id, forum_posts.post_creator, forum_posts.post_content, forum_posts.post_date, forum_post_categories.category_name, forum_post_topics.topic_name, users.id user_id, users.username user_username FROM forum_posts LEFT JOIN users ON users.id = forum_posts.post_creator RIGHT JOIN forum_post_categories ON forum_post_categories.id = forum_posts.category_id RIGHT JOIN forum_post_topics ON forum_post_topics.id = forum_posts.topic_id WHERE forum_posts.category_id=? AND forum_posts.topic_id=? AND forum_posts.post_creator=? AND users.id=?

注意：這未經測試。

Answer 2

select u.*,fp.*  from users u inner join forum_posts fp u.id=fp.post_creator
where  fp.category_id=? AND fp.topic_id=? AND fp.post_creator=? AND  u.id=?

更新的查詢：

select u.user_name,fp.post_creator from 
users u inner join forum_posts fp u.id=fp.post_creator

選中此更新的版本，它只會為您提供user_name和post_creator

Answer 3

嘗試刪除此代碼部分

        if($stmt2===false) {
            die();
        } else {  }

只需：

 $stmt2 = $con->prepare("SELECT forum_posts.id, forum_posts.category_id, forum_posts.topic_id, forum_posts.post_creator, forum_posts.post_content, forum_posts.post_date users.id
        FROM forum_posts LEFT JOIN users WHERE forum_posts.category_id=? AND forum_posts.topic_id=? AND forum_posts.post_creator=? AND users.id=?");
$stmt2->bind_param("ii", $cid, $tid);
            $stmt2->execute();
            $stmt2->store_result();
            $stmt2->bind_result($post_id, $post_category_id, $post_topic_id, $post_creator, $post_content, $post_date);
            if (!$stmt2) {
            throw new Exception($con->error);
            }