Question

我正在尝试一个mysql查询来通过距离对帖子进行排序让我们说在半径250英里范围内，一切正常，但我需要以特色帖子始终位于顶部的方式订购帖子。这是我的mysql查询（我在wordpress上做的）。

    SELECT SQL_CALC_FOUND_ROWS wp_posts.*, ( 3959 * acos( cos( radians(40.140711) ) * cos(      radians( latitude.meta_value ) ) * cos( radians( longitude.meta_value ) - radians(-74.20619299999998) ) + sin( radians(40.140711) ) * sin( radians( latitude.meta_value ) ) ) ) AS distance , 
latitude.meta_value AS latitude , 
longitude.meta_value AS longitude
    FROM wp_posts 
    INNER JOIN wp_postmeta ON (wp_posts.ID = wp_postmeta.post_id) 
    INNER JOIN wp_postmeta AS latitude ON wp_posts.ID = latitude.post_id 
    INNER JOIN wp_postmeta AS longitude ON wp_posts.ID = longitude.post_id 
    WHERE 1=1 
    AND wp_posts.post_type = 'event' 
    AND (wp_posts.post_status = 'publish' OR wp_posts.post_status = 'recurring') 
    AND (wp_postmeta.meta_key = 'st_date' ) 
    AND latitude.meta_key="geo_latitude" 
    AND longitude.meta_key="geo_longitude" 
    AND wp_posts.ID in (SELECT tr.object_id FROM wp_term_relationships tr JOIN wp_term_taxonomy t on t.term_taxonomy_id=tr.term_taxonomy_id WHERE t.term_id IN ('3','4','5','6','7','8','9','10')) 
    AND (wp_posts.ID IN (SELECT wp_postmeta.post_id FROM wp_postmeta WHERE wp_postmeta.meta_key='st_date' 
    AND date_format(wp_postmeta.meta_value,'%Y-%m-%d') >'2013-09-23 14:35:58')) 
    GROUP BY wp_posts.ID HAVING distance <= 250 
    ORDER BY (SELECT wp_postmeta.meta_value FROM wp_postmeta, wp_posts AS p where wp_postmeta.post_id=p.ID AND wp_postmeta.meta_key = 'featured_h') ASC, distance ASC LIMIT 0, 10

问题在于ORDER BY，它给出了错误“Subquery返回超过1行”

如果我删除ORDER BY（从wp_postmeta中选择wp_postmeta.meta_value，wp_posts为p，其中wp_postmeta.post_id = p.ID和wp_postmeta.meta_key ='featured_h'

它工作正常，但它不会显示最好的帖子，无论如何修复它。

Answer 1

在结果集中放置一列显示featured状态。你相信，你必须再次加入wp_postmeta来实现这一目标。

然后使用

 ORDER BY status, distance

或者

 ORDER BY STATUS='featured', distance

这将使您的精选帖子首先按距离排序，然后是其余部分。

（将子查询放在ORDER BY子句中非常奇怪。）

Answer 2

试试这个，我从子查询中选择了custom_order_by列并按此顺序

SELECT SQL_CALC_FOUND_ROWS wp_posts.*, ( 3959 * acos( cos( radians(40.140711) ) * cos(      radians( latitude.meta_value ) ) * cos( radians( longitude.meta_value ) - radians(-74.20619299999998) ) + sin( radians(40.140711) ) * sin( radians( latitude.meta_value ) ) ) ) AS distance , 
latitude.meta_value AS latitude , 
longitude.meta_value AS longitude,
(SELECT wp.meta_value FROM wp_postmeta  wp where wp.post_id=wpm.post_id AND wp.meta_key = 'featured_h') AS custom_order_by
FROM wp_posts 
        INNER JOIN wp_postmeta AS wpm ON (wp_posts.ID = wp_postmeta.post_id) 
        INNER JOIN wp_postmeta AS latitude ON wp_posts.ID = latitude.post_id 
        INNER JOIN wp_postmeta AS longitude ON wp_posts.ID = longitude.post_id 
        WHERE 1=1 
        AND wp_posts.post_type = 'event' 
        AND (wp_posts.post_status = 'publish' OR wp_posts.post_status = 'recurring') 
        AND (wpm.meta_key = 'st_date' ) 
        AND latitude.meta_key="geo_latitude" 
        AND longitude.meta_key="geo_longitude" 
        AND wp_posts.ID in (SELECT tr.object_id FROM wp_term_relationships tr JOIN wp_term_taxonomy t on t.term_taxonomy_id=tr.term_taxonomy_id WHERE t.term_id IN ('3','4','5','6','7','8','9','10')) 
        AND (wp_posts.ID IN (SELECT wp_postmeta.post_id FROM wp_postmeta WHERE wp_postmeta.meta_key='st_date' 
        AND date_format(wp_postmeta.meta_value,'%Y-%m-%d') >'2013-09-23 14:35:58')) 
        GROUP BY wp_posts.ID HAVING distance <= 250 
        ORDER BY custom_order_by ASC, distance ASC LIMIT 0, 10

Wordpress按距离排序和特色帖子查询

2 个答案: