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php - 如何根据mysql中的计数获得排名

转载 作者:行者123 更新时间:2023-11-29 20:02:25 25 4
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我在根据计数获取排名时遇到问题。

我有两张 table :

1)images
2)images_like

现在,如果 images_like 没有条目,那么它在排名中显示“NULL”。请检查查询和输出:

SELECT im.id, 
rank
FROM (SELECT image_id,
@rownum := @rownum + 1 AS rank,
Count(*) AS cnt
FROM images_like,
(SELECT @rownum := 0) r
GROUP BY image_id
ORDER BY `cnt` DESC) AS d
RIGHT JOIN images AS im
ON d.image_id = im.id

enter image description here

我需要显示 3 not null ....

提前致谢!!!!

最佳答案

尝试这样:

SELECT im.id, d.rank  
FROM
(
SELECT t.image_id, @rownum := @rownum + 1 AS rank
FROM
(
SELECT image_id,
Count(*) AS cnt
FROM images_like
GROUP BY image_id
ORDER BY `cnt` DESC) AS d
) t,
(SELECT @rownum := 0) r
) d
RIGHT JOIN images AS im ON d.image_id = im.id

关于php - 如何根据mysql中的计数获得排名,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40483269/

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