[英]How can I combine these looped MySQL queries?
我最近开始对某个应用程序发表评论/回复部分。
我最初的方法是查询数据库,以获取链接到特定项目的评论,如下所示:
$commentquery = "SELECT projects_comments.*, users.user_url, users.display_name
FROM ".$wpdb->prefix."projects_comments projects_comments
LEFT JOIN ".$wpdb->prefix."users users on users.ID=projects_comments.userid
WHERE projectid = '$projectid'
ORDER BY projects_comments.commentid DESC";
然后,我使用以下查询开始了一个foreach
循环,以获取答复:
if($comments) {
foreach ( $comments as $c ) {
$replyquery = "SELECT project_replies.*, users.user_url, users.display_name
FROM ".$wpdb->prefix."project_replies project_replies
LEFT JOIN ".$wpdb->prefix."users users on users.ID=project_replies.uid
WHERE project_replies.cid = '$c->commentid'
ORDER BY project_replies.id DESC";
}
}
在实践中,这一切都很好,但是令我惊讶的是,这效率极低,并且必须有一种方法可以在一个查询中检索所有数据。 我担心的是,在页面上有100条评论时,我将仅对评论执行100条查询。
解决此问题的最佳方法是什么,我将如何组合这些查询以生成一个大对象/数组以从中提取数据。
或者,如何至少防止“环内查询”。
我倾向于避免这种设置的方法是:
SELECT ...
WHERE `keyfield` IN (".implode(",",$keyvalues).")
SELECT c.*, cu.user_url AS cu_url, cu.display_name AS cu_name,
r.*, ru.user_url AS ru_url, ru.display_name AS ru_name
FROM projects_comments AS c
LEFT JOIN users AS cu ON cu.ID = c.userid
JOIN project_replies AS r ON r.cid = c.commentid
LEFT JOIN users AS ru ON ru.ID = r.uid
WHERE c.projectid = $projectid
ORDER BY c.commentid DESC, r.id DESC
连接Project_Replies和Project_Comments时,唯一需要担心的细节是JOIN而不是可能使用LEFT JOIN。 这只会挑选至少有一个回复的评论; 使用LEFT JOIN,您还会收到没有回复的评论。
您可以像执行users
表一样JOIN
他们
$commentquery = "SELECT projects_comments.*, project_replies.*, users.user_url, users.display_name, users2.user_url as comment_user_url, users2.display_name as comment_display_name
from ".$wpdb->prefix."projects_comments projects_comments
LEFT JOIN ".$wpdb->prefix."project_replies project_replies ON project_replies.cid = projects_comments.comment_id
LEFT JOIN ".$wpdb->prefix."users users on users.ID=project_replies.uid
left join ".$wpdb->prefix."users users2 on users2.ID=projects_comments.userid
WHERE projectid = '$projectid'
ORDER BY projects_comments.commentid desc, project_replies.id DESC";
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