select data where date is max(date) less than x

Question

I have 2 tables The first one has exchange rates:

  | date  | ratio | currency |
------------------------------
 1|  9/09 |   1.0 | EUR      |
 2|  9/09 |   1.1 | USD      | -- no weekend
 3| 12/09 |   1.0 | EUR      | -- goes from 9 to 12
 4| 12/09 | 120.0 | JPY      |

The second one has transactions

  | date  | amount | currency |
------------------------------
 1|  9/09 |  20.0   | EUR      |
 2|  9/09 | 101.0   | USD      | -- weekend
 3| 10/09 |   1.0   | USD      | -- has 10/09 which is a saturday
 4| 10/09 |  10.0   | USD      |

Both contain the date and the currency. As it stands my exchange rates are not updated during the weekend, and that won't change.

I'm looking for a performant way to select the last available data to be put into the exchange_rate table. In other words, the last day before the missing day.(10/09 in the example)

I'm using the transaction table to get a list of days that need the exchange-rate information, so that I can convert everything to EUR.

the full result wanted should be something like

  | date  | amount | currency | ratio |
----------------------------------------
 1|  9/09 |  20.0   | EUR      |  1.0  |
 2|  9/09 | 101.0   | USD      |  1.1  |   -- already exists in exchange_rate
 3| 10/09 |   1.0   | USD      |  1.1  |   -- selected because 9/09 is last available line
 4| 10/09 |  10.0   | USD      |  1.1  |

Alternatively I am fine with a query that updates the exchange_rate table with the needed data as well, because the final query would be cleaner and easier to maintain later on

Answer 1

You can do this using a correlated subquery:

select t.*,
       (select er.ratio
        from exchangerates er
        where er.date <= e.date and er.currency = t.currency
        order by er.date desc
        limit 1
       ) as ratio
from transactions t;

For performance, you want an index on exchangerates(currency, date, ratio) .

I would start with this and see if it meets your needs.