Question

我有这个查询

SELECT concours_photo_like.id , COUNT(concours_photo_like.id) AS nb_like , (@rank := @rank + 1) AS rank 
FROM ( SELECT @rank := 0 ) as r , concours_photo_like
JOIN concours_photo ON concours_photo.id = concours_photo_like.id_concours_photo
WHERE concours_photo.id_concours = 67
GROUP BY concours_photo_like.id_concours_photo

结果

| id     | nb_like    | rank |
|--------|------------|------|
|    110 |          2 |    2 |
|    104 |          3 |    1 |
|    134 |          1 |    4 |
|    176 |          1 |    5 |
|    113 |          2 |    3 |

如果我想要ID 134的等级

，我如何获得特定ID的等级

我试过像

这样的东西

SELECT concours_photo_like.id , COUNT(concours_photo_like.id) AS nb_like , (@rank := @rank + 1) AS rank 
FROM ( SELECT @rank := 0 ) as r , concours_photo_like
JOIN concours_photo ON concours_photo.id = concours_photo_like.id_concours_photo
WHERE concours_photo.id_concours = 67
AND concours_photo_like.id = 134
GROUP BY concours_photo_like.id_concours_photo

但结果是

| id     | nb_like    | rank |
|--------|------------|------|
|    134 |          1 |    1 |

，结果应为4而不是1

Answer 1

如何用另一个SELECT语句包装原始查询，然后按id进行过滤，如下所示：

SELECT * FROM 
(SELECT concours_photo_like.id , COUNT(concours_photo_like.id) AS nb_like , (@rank := @rank + 1) AS rank 
FROM ( SELECT @rank := 0 ) as r , concours_photo_like
JOIN concours_photo ON concours_photo.id = concours_photo_like.id_concours_photo
WHERE concours_photo.id_concours = 67
GROUP BY concours_photo_like.id_concours_photo) AS nb
WHERE nb.id = 134

执行此查询可能有更优雅的方式，但这就是现在想到的。

Answer 2

MySQL在为选择列表中的用户变量赋值之前执行where子句。

您必须过滤having子句或将查询包装到子查询中并在外部查询中进行过滤。

有：

SELECT concours_photo_like.id , COUNT(concours_photo_like.id) AS nb_like , (@rank := @rank + 1) AS rank 
FROM ( SELECT @rank := 0 ) as r , concours_photo_like
JOIN concours_photo ON concours_photo.id = concours_photo_like.id_concours_photo
WHERE concours_photo.id_concours = 67
GROUP BY concours_photo_like.id_concours_photo
HAVING concours_photo_like.id = 134

子查询：

SELECT *
FROM
    (SELECT concours_photo_like.id , COUNT(concours_photo_like.id) AS nb_like , (@rank := @rank + 1) AS rank 
     FROM ( SELECT @rank := 0 ) as r , concours_photo_like
     JOIN concours_photo ON concours_photo.id = concours_photo_like.id_concours_photo
     WHERE concours_photo.id_concours = 67
     GROUP BY concours_photo_like.id_concours_photo
    ) t1
WHERE t1.id = 134

使用特定ID排名mysql

2 个答案: