Created
March 25, 2015 08:17
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SQL madness daily, by L Thwaits
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Aga works on a horse ranch in Malibu. She grooms eight horses, following this process for each: | |
Wire brush to get off the big dirt | |
Brush with softer brush to get the rest ofthe dust out | |
Brush the tail and mane | |
Hoof picking | |
She has an application that tracks which parts of the grooming process have been completed for each horse. Here are the tables: | |
horses | |
id | name | |
----+-------- | |
1 | Angel | |
2 | Django | |
3 | Blaize | |
4 | Sky | |
5 | Bear | |
6 | Penny | |
7 | Julia | |
8 | Romeo | |
groomings | |
id | description | |
----+--------------------- | |
1 | Big brush | |
2 | Small brush | |
3 | Groom mane and tail | |
4 | Hoof picking | |
horse_groomings | |
horse_id | grooming_id | |
----------+------------- | |
1 | 1 | |
1 | 3 | |
2 | 1 | |
2 | 2 | |
3 | 1 | |
3 | 2 | |
3 | 3 | |
6 | 1 | |
6 | 4 | |
8 | 3 | |
8 | 4 | |
7 | 1 | |
7 | 2 | |
7 | 4 | |
She wants to see what kinds of grooming have been done for each horse. Which SQL statement would she use? |
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A:
SELECT horses.name, groomings.description
FROM horse_groomings
ORDER BY 1;
B:
SELECT horses.name, groomings.description
FROM horses
INNER JOIN horse_groomings ON horses.id = horse_groomings.horse_id
INNER JOIN groomings ON groomings.id = horse_groomings.grooming_id
ORDER BY horses.name;
C:
FROM horses
SELECT horses.name, groomings.description
INNER JOIN horse_groomings ON horses.id = horse_groomings.horse_id
INNER JOIN groomings ON groomings.id = horse_groomings.grooming_id
ORDER BY 1;
😧
SELECT horses.name, groomings.description
FROM horses
INNER JOIN groomings ON groomings.id = horse_groomings.grooming_id
ORDER BY 1;