5

我有一个如下所示的数据库；

circuit_uid   |  customer_name   | location      | reading_date | reading_time | amps | volts  |  kw  | kwh | kva  |  pf  |  key
--------------------------------------------------------------------------------------------------------------------------------------
cu1.cb1.r1    | Customer 1       | 12.01.a1      | 2012-01-02   | 00:01:01     | 4.51 | 229.32 | 1.03 |  87 | 1.03 | 0.85 |    15
cu1.cb1.r1    | Customer 1       | 12.01.a1      | 2012-01-02   | 01:01:01     | 4.18 | 230.3 | 0.96 |  90 | 0.96 | 0.84 |    16
cu1.cb1.s2    | Customer 2       | 10.01.a1      | 2012-01-02   | 00:01:01     | 7.34 | 228.14 | 1.67 | 179 | 1.67 | 0.88 | 24009
cu1.cb1.s2    | Customer 2       | 10.01.a1      | 2012-01-02   | 01:01:01     | 9.07 |  228.4 | 2.07 | 182 | 2.07 | 0.85 | 24010
cu1.cb1.r1    | Customer 3       | 01.01.a1      | 2012-01-02   | 00:01:01     | 7.32 | 229.01 | 1.68 | 223 | 1.68 | 0.89 | 48003 
cu1.cb1.r1    | Customer 3       | 01.01.a1      | 2012-01-02   | 01:01:01     | 6.61 | 228.29 | 1.51 | 226 | 1.51 | 0.88 | 48004

我要做的是产生一个结果，该结果使每个客户的 KWH 读数从该min(reading_time)日期的最早 ( ) 开始，用户将在 Web 表单中选择该日期。

结果将/应该类似于；

Customer 1   87
Customer 2   179
Customer 3   223

这里每天显示的行数超过了这里显示的行数，并且客户数量更多，并且客户数量会定期变化。

我对 SQL 没有太多经验，我看过子查询等，但我没有能力通过每个客户的最早阅读来弄清楚如何安排它，然后只输出kwh列。

这是在 Redhat/CentOS 上的 PostgreSQL 8.4 中运行的。

4

3 回答 3

3 
select customer_name,
       kwh,
       reading_date, 
       reading_time
from (
   select customer_name,
          kwh,
          reading_time,
          reading_date,
          row_number() over (partition by customer_name order by reading_time) as rn
   from readings
   where reading_date = date '2012-11-17'
) t
where rn = 1

作为备选：

select r1.customer_name,
       r1.kwh, 
       r1.reading_date,
       r1.reading_time
from readings r1
where reading_date = date '2012-11-17'
and reading_time = (select min(r2.reading_time)
                    from readings
                    where r2.customer_name = r1.customer_name
                    and r2.read_date = r1.reading_date);

但我希望第一个更快。

顺便说一句：您为什么将日期和时间存储在两个单独的列中？您是否知道这可以通过timestamp列更好地处理？

于 2012-11-17T18:11:11.707 回答
3

这应该是最快的解决方案之一：

SELECT DISTINCT ON (customer_name)
       customer_name, kwh  -- add more columns as needed.
FROM   readings
WHERE  reading_date = user_date
ORDER  BY customer_name, reading_time

似乎是另一个应用：

于 2012-11-17T18:16:27.123 回答
0 
   SELECT rt.circuit_uid ,  rt.customer_name, rt.kwh
   FROM READING_TABLE rt JOIN  
       (SELECT circuit_uid, reading_time
       FROM READING_TABLE
       WHERE reading_date = '2012-01-02'
       GROUP BY customer_uid
       HAVING MIN(reading_time) = reading_time) min_time
   ON (rt.circuit_uid = min_time.circuit_uid 
      AND rt.reading_time = min_time.reading_time);

参数化上述查询中的 reading_date 值。

于 2012-11-17T18:10:08.923 回答