Question

我有一个名为result的表，其中包含列name, rollno, sub1, sub2, sub3, max1, max2等。 sub1, sub2, sub3将存储已获得的商标。我想找到sub1, sub2, sub3的最大值并将其存储在max1中，并找到sub1, sub2, sub3的第二个最大值，并将其存储在max2中。

示例

sub1  sub2  sub3      max1  max2
10     15    20        20    15
40     10    25        40    25
33     64    51        64    51

有人可以告诉我这个sql代码吗？

更新

我希望将最大数量和第二个最大数量除以2，而不是将最大数量存储到max1，将第二个最大数量存储到max2，而不是将其存储在average列中将其存储在max1和max2中。

表示我不想要两个额外的列max1和max2来存储最大值和第二个最大值，然后添加它并将其除以2然后将其存储在average中。我想做平均直接。

请更新代码。

实施例

sub1  sub2  sub3   average

10     15    20     17.5   ( (Maximum + Second Maximum)/2 )
40     10    25     32.5   
33     64    51     57.5

Answer 1

考虑以下

mysql> create table test (sub1 int, sub2 int , sub3 int);
Query OK, 0 rows affected (0.11 sec)

mysql> insert into test values (20,30,40),(10,40,50),(30,10,20);
Query OK, 3 rows affected (0.08 sec)
Records: 3  Duplicates: 0  Warnings: 0

mysql> select * from test ;
+------+------+------+
| sub1 | sub2 | sub3 |
+------+------+------+
|   20 |   30 |   40 |
|   10 |   40 |   50 |
|   30 |   10 |   20 |
+------+------+------+

因此，要从列中获取max1和max2，您可以使用greatest函数。

select * ,
greatest(sub1,sub2,sub3) as max1 , 
greatest(
 case 
  when greatest(sub1,sub2,sub3) = sub1 then 0 else sub1 
 end,
 case 
  when greatest(sub1,sub2,sub3) = sub2 then  0 else sub2 
 end, 
 case 
  when greatest(sub1,sub2,sub3) = sub3 then 0 else sub3 
 end
) as max2 from test ;

这会给你一些东西

+------+------+------+------+------+
| sub1 | sub2 | sub3 | max1 | max2 |
+------+------+------+------+------+
|   20 |   30 |   40 |   40 |   30 |
|   10 |   40 |   50 |   50 |   40 |
|   30 |   10 |   20 |   30 |   20 |
+------+------+------+------+------+

您可以将此命令用作更新命令

update table_name
set 
max1 = greatest(sub1,sub2,sub3),
max2 = greatest(
 case 
  when greatest(sub1,sub2,sub3) = sub1 then 0 else sub1 
 end,
 case 
  when greatest(sub1,sub2,sub3) = sub2 then  0 else sub2 
 end, 
 case 
  when greatest(sub1,sub2,sub3) = sub3 then 0 else sub3 
 end
)

获取max1和max2的平均值并更新为

update table_name
set 
`average`
= ( 
   greatest(sub1,sub2,sub3)+
   greatest(
    case 
      when greatest(sub1,sub2,sub3) = sub1 then 0 else sub1 
    end,
    case 
      when greatest(sub1,sub2,sub3) = sub2 then  0 else sub2 
    end, 
    case 
      when greatest(sub1,sub2,sub3) = sub3 then 0 else sub3 
    end
     )
)/2 ;

SQL查询：查找最多2个3号并将其存储在另一列中

1 个答案: