Question

SELECT htpid AS parentid 
    FROM likehot WHERE htpid 
    IN (SELECT id FROM bultin WHERE  DATE  >= '1317108749') 
    GROUP BY htpid ORDER BY COUNT( htpid ) 
    DESC

给出结果

parentid
16060
16059
16058
16057

和其他查询

SELECT app_id,bultin.id,photo_album_id,entcmmnt_id,link_url,youtubeLink,
       link_image,id, mem_id, subj, body, bultin.date,parentid, 
       from_id, visible_to, image_link,post_via 
FROM bultin 
WHERE id IN ('16062','16059','16058','16057') 
ORDER BY FIELD('16062','16059','16058','16057') 
LIMIT 5

实际上我想要与IN中相同的序列，所以我使用了FIELD运算符

请建议感谢

Answer 1

我无法理解你的问题..如果你可以请解释一下更好的方式..我可以给你解决方案..我想说你不能在第一个方面使用超过一个值查询“SELECT id FROM bultin WHERE DATE＆gt; ='1317108749'）”。还有一件事Y你不使用ist查询中的值，就像这样

SELECT htpid AS parentid 来自likehot WHERE htpid IN（16062'，'16059'，'16058'，'16057'）

Answer 2

如果我没有错过任何内容，这应该可以胜任：

SELECT b.app_id, b.id, b.photo_album_id, b.entcmmnt_id, b.link_url, b.youtubeLink,
    b.link_image, b.mem_id, b.subj, b.body, b.date, b.parentid,
    b.from_id, b.visible_to, b.image_link, b.post_via
FROM bultin AS b
JOIN (
    SELECT htpid, COUNT( htpid ) AS htpid_count
    FROM likehot
    GROUP BY htpid ORDER BY COUNT( htpid )
    DESC
) AS l ON l.htpid = b.id
WHERE DATE  >= '1317108749'
ORDER BY l.htpid_count
LIMIT 5

如果您需要有关此查询的更多详细信息，请与我们联系。

如何结合这两个查询？

2 个答案: