Difference between revisions of "Multiply sets"
| Line 6: | Line 6: | ||
yr INTEGER, | yr INTEGER, | ||
rate INTEGER ); | rate INTEGER ); | ||
| − | INSERT INTO interest VALUES (2002, | + | INSERT INTO interest VALUES (2002,5); |
| − | INSERT INTO interest VALUES (2003, | + | INSERT INTO interest VALUES (2003,4); |
| − | INSERT INTO interest VALUES (2004, | + | INSERT INTO interest VALUES (2004,5); |
| − | INSERT INTO interest VALUES (2005, | + | INSERT INTO interest VALUES (2005,3); |
</source> | </source> | ||
<div> | <div> | ||
| Line 16: | Line 16: | ||
As the added up rate is not accurate 5 + 4 + 5 + 3 = 17% | As the added up rate is not accurate 5 + 4 + 5 + 3 = 17% | ||
We need to instead find the logarithm of the compound interest and then we need to sum that. | We need to instead find the logarithm of the compound interest and then we need to sum that. | ||
| − | SELECT SUM(LN( | + | SELECT SUM(LN((rate/100)+1)) FROM interest |
Then we inverse or take the exponent of the logarithm with | Then we inverse or take the exponent of the logarithm with | ||
| − | SELECT EXP(SUM(LN( | + | SELECT EXP(SUM(LN((rate/100)+1))) FROM interest |
and then finally to get the amount after 4 years we times this amount by 100 ($100). | and then finally to get the amount after 4 years we times this amount by 100 ($100). | ||
</div> | </div> | ||
| − | <source lang='sql' class='def'>SELECT EXP(SUM(LN( | + | <source lang='sql' class='def'>SELECT EXP(SUM(LN((rate/100)+1)))*100 |
FROM interest</source> | FROM interest</source> | ||
<div class="ecomm e-mysql" style="display: none"></div> | <div class="ecomm e-mysql" style="display: none"></div> | ||
</div> | </div> | ||
{{Hacks Ref}} | {{Hacks Ref}} | ||
Revision as of 11:24, 30 July 2012
Multiply across a result set.
DROP TABLE interest
CREATE TABLE interest( yr INTEGER, rate INTEGER ); INSERT INTO interest VALUES (2002,5); INSERT INTO interest VALUES (2003,4); INSERT INTO interest VALUES (2004,5); INSERT INTO interest VALUES (2005,3);
In this example we are attempting to see how much money has been obtained over 4 years due to interest. As the added up rate is not accurate 5 + 4 + 5 + 3 = 17% We need to instead find the logarithm of the compound interest and then we need to sum that. SELECT SUM(LN((rate/100)+1)) FROM interest Then we inverse or take the exponent of the logarithm with SELECT EXP(SUM(LN((rate/100)+1))) FROM interest and then finally to get the amount after 4 years we times this amount by 100 ($100).
SELECT EXP(SUM(LN((rate/100)+1)))*100 FROM interest
Hack 10 Converting subqueries into joins
Hack 11 Converting aggregate subqueries into joins
Hack 16 Search for a String across columns
Hack 24 Multiply Across a Result Set
Hack 25.5 Splitting and combining columns
Hack 26 Include the rows your JOIN forgot
Hack 30 Calculate the maximum/minimum of two fields
Hack 33 Get values and subtotals in one shot
Hack 50 Combine tables containing different data
Hack 51/52 Display rows as columns
Hack 55 Import Someone Else's Data
Hack 62 Issue Queries Without Using a Table
Hack 63 Generate rows without tables
Hack 72 Extract a subset of the results
Hack 78 Break it down by Range
Hack 88 Test two values from a subquery
