# 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 12:24, 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