   18:35

# Running totals page 1

The correlated subquery may be used for calculation accumulation running totals — the usual practical problem.

In assumption of some rows` order the running total for each row is a sum of values for some numeric column in this row and for all rows above.

In the other words, the running total for first row in the ordered set would be equal to the numeric value in this row. For any other rows the running total would be equal to sum of numeric value in this row and the running total in the previous rows.

For point 2 in the Outcome_o table for each day get the summary outlay for this day and for all previous days.

Here is the query outputs information about outlays in point 2 in order of ascending of the date:  Console
`SELECT point, date, outFROM Outcome_o oWHERE point = 2ORDER BY date;`

point    date    out
2    2001-03-22 00:00:00.000    1440.00
2    2001-03-29 00:00:00.000    7848.00
2    2001-04-02 00:00:00.000    2040.00

Actually, to solve this problem we need to add one column for accumulating running total (run_tot). Correspondingly the theme, this column would be represent the correlated sub-query. In this subquery the total sum of the values in the out column would be calculated for this date and for all preceding dates.  Console
`SELECT point, date, out,   (SELECT SUM(out)    FROM Outcome_o    WHERE point = o.point AND date <= o.date) run_tot FROM Outcome_o oWHERE point = 2ORDER BY point, date;`

point    date    out    run_tot
2    2001-03-22 00:00:00.000    1440.00    1440.00
2    2001-03-29 00:00:00.000    7848.00    9288.00
2    2001-04-02 00:00:00.000    2040.00    11328.00

In fact, the usage of point 2 is dictated by wish to reduce the resulting set. If we want to get running totals for all points in the Outcome_o table, we need to close in comment this string

`WHERE point = 2`

And if we want to get running total for whole table we need to put off the condition for point`s equality:

`point= o.point`

But in this case we would get one total for different points which working in the same day. Here`s the such fragment from the resulting set:

point    date    out    run_tot
1    2001-03-29 00:00:00.000    2004.00    33599.00
2    2001-03-29 00:00:00.000    7848.00    33599.00

This is not an obstacle if we understand what we want to get. If we want to get the accumulation of the outlay by days we need to exclude the point and summarize the outlay by days.  Console
`SELECT date, SUM(out) out,   (SELECT SUM(out)    FROM Outcome_o    WHERE date <= o.date) run_tot FROM Outcome_o oGROUP BY dateORDER BY date;`

In the other case, we need to assign the order of the addition of the points` outlays if the dates is coincided.
For instance, let`s order it by ascending of its numbers:  Console
`SELECT point, date, out,   (SELECT SUM(out)    FROM Outcome_o    WHERE date < o.date OR ( date = o.date AND point <= o.point)) run_tot FROM Outcome_o oORDER BY date, point;`

Suggested exercises: 69, 101, 134

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