甲骨文按客户和日期分组

Question

我有一个有效的休闲查询：

SELECT count(cand_id) as candidates, SB.client_id as client
            FROM REPLIES r
            JOIN clients sb on sb.main_acc_id = r.acc_id
            where reply_dt >= TO_DATE('2021-02-08 00:00:00', 'YYYY-MM-DD HH24:MI:SS')
            AND reply_dt <= TO_DATE('2021-02-15 23:59:59', 'YYYY-MM-DD HH24:MI:SS')
            AND sb.status = 'A'
            GROUP BY client_id

它产生的结果如下：

 Candidates| Clients
     2     |  client 1
     4     |  cleint 2
    56     |  client whatever..

我将如何着手在给定日期范围内按天和客户细分候选人数量？所以它会显示那天，没有候选人，客户

Answer 1

这是你想要的吗？

SELECT TRUNC(r.reply_dt), sb.client_id, COUNT(cand_id) as candidates
FROM REPLIES r JOIN
     clients sb 
     ON sb.main_acc_id = r.acc_id
WHERE sb.status = 'A'
GROUP BY TRUNC(r.reply_dt), client_id

甲骨文按客户和日期分组

Oracle group by client and date

sql

oracle

oracle9i