根据条件在sql中创建列？

Question

表：

  Student_Id    Roll_No    Date   _Rank
1    101         10101    201905    3
2    101         10101    201910    6
3    101         10101    201912    9
4    101         10102    201905    15
5    101         10102    201910    9
6    101         10102    201912    3
7    103         10103    201905    3
8    103         10103    201910    9
9    103         10103    201912    9
10   103         10104    201905    3
11   103         10104    201910    9
12   103         10104    201912    9

我需要根据条件创建一个结果列。

1. 检查特定的 Student_Id,Roll_No 如果 rank 是 9 并且 rank> 9 没有 根据日期出现在 9 之前，然后出现 1，否则为 0。

2. 如果某个 Student_Id,Roll_No 有多个 9，则仅标记第一次出现。

结果：

  Student_Id    Roll_No    Date   _Rank   Result
1    101         10101    201905    3       0
2    101         10101    201910    6       0
3    101         10101    201912    9       1 
4    101         10102    201905    15      0 ---
5    101         10102    201910    9       0 --- 15 > 9, came before 9 for that Roll_No
6    101         10102    201912    3       0
7    103         10105    201905    9       1
8    103         10103    201910    9       1
9    103         10103    201912    9       0
10   103         10104    201905    3       0
11   103         10104    201910    9       1 --- first occurrence should 1
12   103         10104    201912    9       0

查询：

SELECT Student_Id, Roll_Number, Date, _Rank,
ROW_NUMBER() OVER( PARTITION BY Roll_Number ORDER BY Date) as rw_number
FROM table t
-- this is giving the row_number which is appeared first.

Answer 1

你可以使用窗口函数：

select 
    t.*, 
    case 
        when _rank = 9 and row_number() over(partition by roll_no, _rank order by date) = 1
        then 1
        else 0
    end result
from mytable t

Answer 2

我想你想要：

select t.*, 
       (case when _rank = 9 and 
                  row_number() over (partition by roll_no, _rank order by date) = 1 and
                  max(_rank) over (partition by roll_no order by date) <= 9
             then 1 else 0
        end) as result
from t
order by roll_no, date;

max(_rank) 条件检查在给定行之前没有出现高于 9 的排名。

Here 是一个 dbfiddle。