Question

我有下表：

+------+------+-----+---------+------------+
| no   | code |eot  |group_id | compulsary |
+------+------+-----+---------+------------+
| 1005 | 101  |  51 |     1   |          1 | 
| 1005 | 102  |  67 |     1   |          1 | 
| 1005 | 121  |  65 |     1   |          1 |
| 1005 | 231  |  82 |     2   |          0 |
| 1005 | 232  |  56 |     2   |          0 |
| 1005 | 233  |  45 |     2   |          1 |
| 1005 | 313  |  80 |     3   |          0 |
| 1005 | 443  |  50 |     4   |          0 |     

|------+------+-----+---------+------------+

现在我想要的是：

1。）返回group_id=1，

所有记录

2。）从group_id = 2返回最好的两个（但是如果compulsary=1然后包含该行，那么剩下的group_id=2中最好的那一行），

3.）从group_id=3和group_id=4返回1行，如果compulsary=1则返回该行

最终结果应该只有七行：

+------+------+-----+---------+------------+
| no   | code |eot  |group_id | compulsary |
+------+------+-----+---------+------------+
| 1005 | 101  |  51 |     1   |          1 | 
| 1005 | 102  |  67 |     1   |          1 | 
| 1005 | 121  |  65 |     1   |          1 |
| 1005 | 231  |  82 |     2   |          0 |
| 1005 | 233  |  45 |     2   |          1 |
| 1005 | 313  |  80 |     3   |          0 |
| 1005 | 443  |  50 |     4   |          0 |     

|------+------+-----+---------+------------+

上面包含compulsary=1行;

到目前为止，我有这个问题，但我不知道如何检查compulsary以获得我想要的内容：

select rg.* 
from 
(
  select * 
  from 
  (
    select 
      rgrade.*,
      @rn := if(@gr=group_id,if(@gr:=group_id,@rn+1,@rn+1),if(@gr:=group_id,1,1)) as rn 
    from rgrade 
    cross join (select @rn:=0,@gr:=0) as vars 
    where admission_no=1005
  ) v 
  where (group_id=1) 
     or (group_id=2 and if(compulsary=1,rn<=1,rn<=2)) 
     or (group_id in (3,4) and rn=1)
) rg  
order by group_id;

查询按预期返回七行，但不检查compulsary中的group_id=2。

任何帮助非常感谢

Answer 1

您在此处描述了几个查询，您想要合并的结果：所有group_id = 1，两个最佳group_id = 2，最好的group_id = 3，以及{{1 }}。所以编写这些查询并将它们与group_id = 4结合起来。用＃34; best＆＃34;定义为UNION ALL首选，然后是最高的，你得到：

compulsary = 1

Answer 2

您正在尝试模仿标准SQL

row_number() over (partition by group_id 
                   order by case when compulsary = 1 then 1 else 0 end desc, eot desc)

用MySQL意味着。我在您的查询中看到了group_id的分区，但我没有看到任何ORDER BY来获得最佳记录。

这是我的尝试。可能有错误;我不是MySQL人。

select
  no, code, eot, group_id, compulsary
from
(
  select 
    no, 
    code,
    eot,
    compulsary,
    @row_number := 
      case when group_id = @last_group_id then @row_number + 1 else 1 end as row_number,
    @last_group_id := group_id as group_id
  from rgrade
  cross join 
  (
    select 
      @row_number := 0, 
      @last_group_id := -1
  ) as vars
  where admission_no = 1005
  order by group_id, (compulsary = 1) desc, eot desc
) ranked
where (group_id = 1)
   or (group_id = 2 and row_number <= 2)
   or (group_id = 3 and row_number = 1)
   or (group_id = 4 and row_number = 1)
order by group_id, code;

mysql有条件地选择前N个记录

2 个答案: