Question

您好我正在寻找一种编写SQL语句的方法，该语句将产生以下结果： -

让我们说我们有Dept＆amp; Emp ID我想从第3部分为前两行生成类似的记录，然后是部门2，然后是一行，然后继续部门3，依此类推：

DEPT         EMPID
-----        ------
3              1
3              2
2              3
3              7
3              8
2              9

谢谢。

Answer 1

你可以使用这样的东西

SELECT 
  DEPT,
  EMPID
FROM (
      SELECT 
        *,
        ceil((row_number() OVER (PARTITION BY dept ORDER BY EMPID ))/ 2::numeric(5,2)) AS multiple_row_dept,
        row_number() OVER (PARTITION BY dept ORDER BY EMPID ) AS single_row_dept
      FROM 
        test_data2
  ) sub_query 
ORDER BY 
  CASE 
    WHEN DEPT = 2 THEN single_row_dept
    ELSE multiple_row_dept 
  END,
  DEPT DESC,
  EMPID

single_row_dept个细节，哪些部门只应出现一次，在这种情况下，其DEPT 2后跟多个其他部门

Answer 2

首先在子查询中按empid排序表，
然后计算剩余的rowid除以3，
然后根据计算结果，使用case表达式返回2或3，
像这样

SELECT 
   CASE REMAINDER( rownum, 3 )
     WHEN 0 THEN 2
            ELSE 3
   END As DeptId,
   empid
FROM (
  SELECT empid
  FROM table1
  ORDER BY empid
)

演示：http://sqlfiddle.com/#!4/bd1bb/3

Answer 3

以下代码有一些限制：

1）源表中只有2个DepId

2）具有不同DepId的记录之间的比率恰好是2：1

3）应该更改在查询的一个位置的排序，这取决于DeptId是否有更多记录自然大于另一个

with t as (
select 3 dept_id, 1 emp_id from dual
union all
select 3, 2 from dual
union all
select 3, 7 from dual
union all
select 3, 8 from dual
union all
select 2, 3 from dual
union all
select 2, 9 from dual)

select dept_id, emp_id
  from (select dept_id,
               emp_id,
               dense_rank() over(partition by dept_id order by in_num + mod(in_num, out_num)) ord
          from (select dept_id,
                       emp_id,
                       row_number() over(partition by dept_id order by emp_id) in_num,
/*in the following ORDER BY change to DESC or ASC depending on which dept_id has more records*/
                       dense_rank() over(order by dept_id) out_num
                  from t))
 order by ord, dept_id desc;

   DEPT_ID     EMP_ID
---------- ----------
         3          1
         3          2
         2          3
         3          8
         3          7
         2          9

Oracle SQL生成交叉的SQL结果

3 个答案: