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Home/ Questions/Q 8847711
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T12:15:19+00:00 2026-06-14T12:15:19+00:00

In had a rather large error in my previous question select earliest date from

  • 0

In had a rather large error in my previous question

select earliest date from multiple rows

The answer by horse_with_no_name returns a perfect result, and I am hugely appreciative, however I got my own initial question wrong so I really apologise; if you look at the table below;

circuit_uid  |customer_name     |rack_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.r2    | Customer 1       | 12.01.a1      | 2012-01-02   | 00:01:01     | 4.51 | 229.32 | 1.03 |  21 | 1.03 | 0.85 |    15
cu1.cb1.r2    | Customer 1       | 12.01.a1      | 2012-01-02   | 01:01:01     | 4.18 | 230.3  | 0.96 |  23 | 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.s3    | Customer 2       | 10.01.a1      | 2012-01-02   | 00:01:01     | 7.34 | 228.14 | 1.67 | 121 | 1.67 | 0.88 | 24009
cu1.cb1.s3    | Customer 2       | 10.01.a1      | 2012-01-02   | 01:01:01     | 9.07 |  228.4 | 2.07 | 124 | 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
cu1.cb1.r4    | Customer 3       | 01.01.a1      | 2012-01-02   | 00:01:01     | 7.32 | 229.01 | 1.68 | 215 | 1.68 | 0.89 | 48003 
cu1.cb1.r4    | Customer 3       | 01.01.a1      | 2012-01-02   | 01:01:01     | 6.61 | 228.29 | 1.51 | 217 | 1.51 | 0.88 | 48004

As you can see each customer now has multiple circuits. So the result would now be the sum of each of the earliest kwh readings for each circuit per customer, so the result in this table would be;

customer_name | kwh(sum)
--------------+-----------
customer 1    | 108      (the result of 87 + 21)  
customer 2    | 300      (the result of 179 + 121)  
customer 3    | 438      (the result of 223 + 215)   

There will be more than 2 circuits per customer and the readings can happen at varying times, hence the need for the ‘earliest’ reading.

Would anybody have any suggestions for the revised question?

PostgreSQL 8.4 on CentOs/Redhat.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T12:15:20+00:00Added an answer on June 14, 2026 at 12:15 pm
    SELECT customer_name, sum(kwh) AS kwh_total
    FROM  (
        SELECT DISTINCT ON (customer_name, circuit_uid)
               customer_name, circuit_uid, kwh
        FROM   readings
        WHERE  reading_date = '2012-01-02'::date
        ORDER  BY customer_name, circuit_uid, reading_time
        ) x
    GROUP  BY 1
    

    Same as before, just pick the earliest per (customer_name, circuit_uid).
    Then sum per customer_name.

    Index

    A multi-column index like the following will make this very fast:

    CREATE INDEX readings_multi_idx
    ON readings(reading_date, customer_name, circuit_uid, reading_time);
    
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