Question

我想计算客户请求和订购的时间，只显示前10个输出。我使用count和rownum函数，如下所示，但是订单的计数未在代码中实现。请在下面查看我的代码，对于解决该问题的建议，我将不胜感激。谢谢。

SET SERVEROUT ON
create or replace procedure total is
begin
for a in (select customer_no, count(*) as total from orders where rownum <= 
10 group by customer_no)
loop
dbms_output.put_line('customer number '||a.customer_no|| ' total orders 
'||a.total);

end loop;
end;
/
execute total;
drop procedure total;

输出具有行数的示例

Procedure TOTAL compiled

customer number 1062 total orders 1
customer number 1054 total orders 1
customer number 1051 total orders 1
customer number 1060 total orders 1
customer number 1052 total orders 1
customer number 1059 total orders 1
customer number 1061 total orders 1
customer number 1055 total orders 1
customer number 1053 total orders 1
customer number 1058 total orders 1


PL/SQL procedure successfully completed.
Procedure TOTAL dropped.

不包含行号的输出示例：

Procedure TOTAL compiled

customer number 1098 total orders 25
customer number 1041 total orders 11
customer number 1000 total orders 18
customer number 1003 total orders 16



PL/SQL procedure successfully completed.


Procedure TOTAL dropped.

Answer 1

您只需要10行，而不必关心它们处于哪10个或什么顺序？

SELECT *
FROM (SELECT customer_no, count(*) AS total
      FROM orders
      GROUP BY customer_no)
WHERE rownum <= 10;

Example

（据说Oracle 12支持FETCH FIRST语法来返回给定数量的行，但是我没有任何测试方法）

在rownum中使用count（*）

1 个答案: