【发布时间】：2018-09-14 10:31:24
【问题描述】：
我有两张桌子，posts 和 comments。 posts 表列是：id, body, user_id, likes, dislikes, time。 comments 表列是：id, body, post_id, user_id, likes, dislikes, time。
让我们以两个帖子（A 和 B）为例。 Post-A 有 1 条评论，有 10 条喜欢和 2 条不喜欢，Post-B 有 1 条评论，有 5 条喜欢和 12 条不喜欢。
通过 API 端点 category 订购帖子时，如果请求是针对 most disliked 评论的，我如何实现以 Post-B 开头的 ORDER BY comment.likes/dislikes DESC 语句。或者如果请求是针对most liked 评论，则从 Post-A 开始。
这就是我当前的查询，它按任何评论请求的发布 cmets 数量排序，看起来像。请注意，我没有从 comments 表中选择，因为在获取帖子后会为每个帖子 ID 加载 cmets。
<?
if (isset($_GET['ordertags'])) {
//post tags e.g general/work/school etc
$orderTags = $_GET['ordertags'];
}else {
$orderTags = "alltags";
}
if (isset($_GET['orderreactions'])) {
//post reactions, e.g date time/ most post-likes/dislikes/comments & comment-likes/dislikes etc
$orderReactions = $_GET['orderreactions'];
}else {
$orderReactions = "pdt";
}
//declare vars
$orderBy = "";
//get start offset to load the first 10 results
$start = (int)$_GET['start'];
switch ($orderReactions) {
case "pdt":
$orderBy = "ORDER BY posts.posted_at";
break;
case "mlp":
$orderBy = "AND posts.likes != 0 ORDER BY posts.likes";
break;
case "mdp":
$orderBy = "AND posts.dislikes != 0 ORDER BY posts.dislikes";
break;
case "mcp":
$orderBy = "AND posts.comments != 0 ORDER BY posts.comments";
break;
case "mlc":
$orderBy = "AND posts.comments != 0 ORDER BY posts.comments";
break;
case "mdc":
$orderBy = "AND posts.comments != 0 ORDER BY posts.comments";
break;
default:
$orderBy = "ORDER BY posts.posted_at";
break;
}
//posts from users
if ($orderTags == "alltags") {
$followingposts = $db->query('SELECT posts.id, posts.body, posts.posted_at, posts.likes, posts.dislikes, posts.tags, users.`username`, users.`profileimg` FROM users, posts
WHERE users.id = posts.user_id
'.$orderBy.' DESC LIMIT 10 OFFSET '.$start.';');
}else {
$followingposts = $db->query('SELECT posts.id, posts.body, posts.posted_at, posts.likes, posts.dislikes, posts.tags, users.`username`, users.`profileimg` FROM users, posts
WHERE users.id = posts.user_id
AND tags = :tag
'.$orderBy.' DESC LIMIT 10 OFFSET '.$start.';', array(':tag'=>$orderTags));
}
?>
这就是我的comments 端点的样子，它可以毫无问题地检索最喜欢/最不喜欢的 cmets。
<?
if (isset($_GET['action'])) {
//comment action request i.e most liked/disliked/regular i.e postDate
$action = $_GET['action'];
}else {
$action = "reg";
}
//declare vars
$orderBy = "";
switch ($action) {
case "mdc":
$orderBy = "ORDER BY comments.dislikes DESC";
break;
case "mlc":
$orderBy = "ORDER BY comments.likes DESC";
break;
case "reg":
$orderBy = "ORDER BY comments.posted_at ASC";
break;
default:
$orderBy = "ORDER BY comments.posted_at ASC";
break;
}
//fetch comments from db
$comments = $db->query('SELECT comments.id, comments.comment, comments.post_id, comments.posted_at, comments.likes, comments.dislikes, users.username, users.profileimg FROM comments, users WHERE comments.post_id = :postid AND comments.user_id = users.id '.$orderBy.';', array(':postid'=>$_GET['postid']));
?>
所以问题是我如何在ORDER BY DESC 语句中选择具有最高喜欢/不喜欢的 cmets 的帖子，同时仍然排除带有 0 个喜欢/不喜欢的 cmets 的帖子。谢谢。
【问题讨论】：
-
鉴于您构建数据的方式，任何喜欢或不喜欢都将是永久性的，因为它们与帖子和 cmets 相关联，而不是与喜欢和不喜欢的用户相关联。您如何检查某人是否已经喜欢或不喜欢帖子或评论？至于您的问题：与您的
$orderBy类似，您可以在 switch 案例中添加$where条件。在某些情况下，它将是comments.likes != 0，或comments.dislikes != 0，而在其他情况下，它只是一个空字符串。