Question

我有一个MySQL表 “评级”，

   an ID column
   a column called like_dislike (holds either null, 0, or 1),
   and a column called lesson_id (a foreign key from lessons).

MySQL表，“课程”，带

   an ID column
   a teacher_id column

我需要选择此信息：

   "SELECT r.like_dislike FROM ratings r INNER JOIN lessons l on l.lesson_id = r.lesson_id";

然而，这实际上是一个更大的SQL语句的一部分，我想做的是：

   Foreach lesson_id, if like_dislike == 0, SELECT count(like_dislike) as like

和

   Foreach lesson_id, if like_dislike == 1, SELECT count(like_dislike) as dislike

我不知道如何将这个伪代码转换为SQL。我还需要在SQL中执行此操作，而不是像PHP那样，因为它是更大的SQL语句的一部分，它转换为格式正确的数组非常麻烦。

Answer 1

您应该能够通过分组来完成此任务。例如：

SELECT r.lesson_id, COUNT(*) AS like
FROM ratings r
INNER JOIN lessons l ON l.lesson_id = r.lesson_id
WHERE r.like_dislike = 0
GROUP BY r.lesson_id;

对于不喜欢，只需将WHERE子句更改为

即可

WHERE r.like_dislike = 1

修改

通过添加另一级别的分组，可以将其组合成一个查询请求：

SELECT r.lesson_id, r.like_dislike, COUNT(*) AS count FROM ratings r INNER JOIN lessons l ON l.lesson_id = r.lesson_id GROUP BY r.lesson_id, r.like_dislike;

This will give you output, for example: +-----------+--------------+-------+ | lesson_id | like_dislike | count | +-----------+--------------+-------+ | 1 | 0 | 12 | | 1 | 1 | 7 | | 2 | 0 | 1 | | 2 | 1 | 4 | +-----------+--------------+-------+

因此对于1的lesson_id，有12个喜欢，7个不喜欢等...

编辑2：

要为每个lesson_id获取一行，您可以稍微修改一下该语句：

SELECT r.lesson_id, CASE WHEN r.like_dislike = 0 THEN COUNT(*) END AS like, CASE WHEN r.like_dislike = 1 THEN COUNT(*) END AS dislike FROM ratings r INNER JOIN lessons l ON l.lesson_id = r.lesson_id GROUP BY r.lesson_id, r.like_dislike;

就此而言，您根本不需要加入课程表，除非您以某种方式获得未链接到课程的评分。如果您想要包含没有评分的课程，则必须更改为OUTER加入：

SELECT l.lesson_id, CASE WHEN r.like_dislike = 0 THEN COUNT(*) END AS like, CASE WHEN r.like_dislike = 1 THEN COUNT(*) END AS dislike FROM lessons l LEFT JOIN ratings r ON r.lesson_id = l.lesson_id GROUP BY l.lesson_id, r.like_dislike;

Answer 2

也许您可以使用case语句，如下所示：

SELECT r.lesson_id,
       case when r.like_dislike == 0 then (count(*)) end as like,
       case when r.like_dislike == 1 then (count(*)) end as dislike
FROM ratings r INNER JOIN lessons l ON l.lesson_id = r.lesson_id
GROUP BY r.lesson_id, r.like_dislike

我没有测试过，但你可以看到这个想法。此外，您必须将case设置为1，将另一个设置为0，因为like_dislike可以为null。

Answer 3

试试这个

SELECT if(r.like_dislike =0 ,count(like_dislike) as like , if( r.like_dislike =1 , count(like_dislike) as dislike, 'its null'))
FROM ratings r 
INNER JOIN lessons l on l.lesson_id = r.lesson_id

Answer 4

如果要在yr数据表上添加条件，则按以下方式编写代码：

count_inversions_fast(array, values.size()) << endl;

希望有所帮助

SQL中的条件和计数

4 个答案: