Question

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

我要做的是生成一个结果，该结果在该日期最早（min(reading_time)）为每个客户提供KWH读数，用户将在网络表单中选择日期。

结果将是/应该类似于;

Customer 1   87
Customer 2   179
Customer 3   223

此处显示的每天行数超过，客户数量也会增加，客户数量会定期变化。

我没有太多的SQL经验，我查看过子查询等但是我没有找到如何根据每个客户的最早阅读进行排列，然后输出kwh列。

这是在Redhat / CentOS上的PostgreSQL 8.4中运行。

Answer 1

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

作为替代方案：

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：为什么要将日期和时间存储在两个单独的列中？您是否意识到使用timestamp列可以更好地处理此问题？

Answer 2

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

似乎是另一个应用程序：

Select first row in each GROUP BY group?

Answer 3

   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。

sql为多行选择最早的日期

3 个答案: