$post = $wpdb->get_results("SELECT `p`.`ID`, MAX(p.post_date) as `datetime`, `p`.`post_date`, `p`.`post_content`, `p`.`post_title`, `p`.`post_status`, `p`.`post_name`, `p`.`comment_count`, `tax`.`term_taxonomy_id`, `tax`.`term_id`, `tax`.`taxonomy`, `tax`.`parent`, `rel`.`object_id`, `rel`.`term_taxonomy_id`, `t`.`term_id`, `t`.`name`, `t`.`slug`
FROM (`$wpdb->posts` AS p)
INNER JOIN `$wpdb->term_relationships` AS rel ON `rel`.`object_id` = `p`.`ID`
INNER JOIN `$wpdb->term_taxonomy` AS tax ON `tax`.`term_taxonomy_id` = `rel`.`term_taxonomy_id`
INNER JOIN `$wpdb->terms` AS t ON `t`.`term_id` = `tax`.`term_id`
WHERE `tax`.`taxonomy` = 'category'
AND `p`.`post_status` = 'publish'
AND `p`.`post_type` = 'post'
GROUP BY tax.parent
ORDER BY datetime DESC
LIMIT 4");
我需要找到每个类别的最新帖子，然后将结果分组，以便每个类别只有一个最新帖子。
我使用; GROUP BY tax.parent不起作用; ORDER BY datetime DESC
您的 ID 是增量的。 用那个。
select ...
from
.....
where id post_in
(select max(post_id) from table group by category)
如果您知道有多少个类别，则可以使用
where post id in
(select post_id from table where category=1 order by time desc limit 1
union
select post_id from table where category=2 order by time desc limit 1
union
select post_id from table where category=3 order by time desc limit 1
union
select post_id from table where category=4 order by time desc limit 1)
或者你可以使用参数，这将给你一个完美的结果，但查询速度非常慢
select * from
(select
@rn:=if(@prv=category_id, @rn+1, 1) as rId,
@prv:=category_id as category_id,
timestamp,
other columns
from (select category_id, timestamp, other columns from ... )a
join
(select @prv:=0, @rn:=0)tmp
order by
category_id , timestamp desc) a
where rid<=1
尝试GROUP BY t.name而不是 tax.parent，MAX(p.ID) 而不是 MAX(p.post_date) 。 我认为你不能在日期时间使用 MAX（可能是错误的，但它对我不起作用），我也认为按 t.name 分组是你想要的（或t.term_id，或 slug）。
它似乎给了我每个类别的最新帖子，但就我而言，这可能是巧合。