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-- Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

-- +---------+------------+------------------+

-- | Id(INT) | Date(DATE) | Temperature(INT) |

-- +---------+------------+------------------+

-- | 1 | 2015-01-01 | 10 |

-- | 2 | 2015-01-02 | 25 |

-- | 3 | 2015-01-03 | 20 |

-- | 4 | 2015-01-04 | 30 |

-- +---------+------------+------------------+

-- For example, return the following Ids for the above Weather table:

-- +----+

-- | Id |

-- +----+

-- | 2 |

-- | 4 |

-- +----+

select w1.Id from Weather w1, Weather w2 where w1.Temperature > w2.Temperature and to_days(w1.Date) - to_days(w2.date) = 1;