Presentation on theme: "Tuning in Action Dennis Shasha. Application Context Each record, when inserted, undergoes several tests to see where and how it should be inserted. Programmer."— Presentation transcript:
Tuning in Action Dennis Shasha
Application Context Each record, when inserted, undergoes several tests to see where and how it should be inserted. Programmer temptation is to treat each record separately and in a loop.
Starting Point Input relation consisting of records to be inserted, alloc. Stored procedure selects rows of alloc into local scalar variables in a loop and then processes each one based on conditions. The record itself may be involved in interior loops.
Example Suppose that extra_price is a scalar variable in the stored procedure derived from alloc.extra_price We have a condition of the form: if (extra_price > 0) then select x, y from R, S where R.A = S.B and R.C = extra_price
What Can We Do? If statement can be translated to where clause, possibly linked to an insert. select x, y, alloc.extra_price from alloc, R, S where R.A = S.B and alloc.extra_price = R.extra_price and R.extra_price > 0
Example 2 An input row translates into many output rows, one for each date between start_date and end_date. How does one generate several rows from a single one? One possibility: do it outside the database, but then interfaces change.
What Can One Do? Set up a table that holds only dates alldates. Then can insert as follows: insert V select alloc.id, alloc.etc, alldates.date from alloc, alldates where alloc.startdate <= alldates.date and alloc.enddate >= alldates.date
Example 3 Each date further gets different price values depending on the day of the week. If weekday(this_date) = 0 then price = Sunday_price
What Can We Do? If there is a record for each date, then again the if can be translated to a where update R set price = Sunday_price where weekday(this_date) = 0
Several updates Would need one update per day of week or even per date Better to have a table of prices S(mydate, price) with seven items and then do a join: update R set price = S.price from R, S where R.this_date = S.mydate
General Structure Issues For each row of alloc and each date, we call a stored procedure that does several joins and inserts. What should we attack first?