sql實戰 多表
阿新 • • 發佈：2018-12-14
實戰
有三張表
| trips | |
|---|---|
| id | INTEGER |
| date | TEXT |
| pickup | TEXT |
| dropoff | TEXT |
| rider_id | INTEGER |
| car_id | INTEGER |
| type | TEXT |
| cost | INTEGER |
| riders | |
|---|---|
| id | INTEGER |
| first | TEXT |
| last | TEXT |
| username | TEXT |
| rating | INTEGER |
| total_trips | INTEGER |
| referred | INTEGER |
| cars | |
|---|---|
| id | INTEGER |
| model | TEXT |
| OS | TEXT |
| status | TEXT |
| trips_completed | INTEGER |
1.將司機名字與汽車型號做一個簡單的cross Join
SELECT riders.first,riders.last,cars.model
FROM riders,cars;| first | last | model |
|---|---|---|
| Sonny | Li | Ada |
| Sonny | Li | Ada |
| Sonny | Li | Turing XL |
| Sonny | Li | Akira |
| Laura | Breiman | Ada |
| Laura | Breiman | Ada |
| Laura | Breiman | Turing XL |
| Laura | Breiman | Akira |
2.將罰單與人對應 使用left join
SELECT *
FROM trips
LEFT JOIN riders
ON trips.rider_id=
riders.id;
！這樣寫也是沒問題的
SELECT trips.date,
trips.pickup,
trips.dropoff,
trips.type,
trips.cost,
riders.first,
riders.last,
riders.username
FROM trips
LEFT JOIN riders
ON trips.rider_id = riders.id;
| id | date | pickup | dropoff | rider_id | car_id | type | cost | id | first | last | username | rating | total_trips | referred |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 1001 | 2017-12-05 | 06:45 | 07:10 | 102 | 1 | X | 28.66 | 102 | Laura | Breiman | @lauracle | 4.99 | 687 | 101 |
| 1002 | 2017-12-05 | 08:00 | 08:15 | 101 | 3 | POOL | 9.11 | 101 | Sonny | Li | @sonnynomnom | 4.66 | 352 | |
| 1003 | 2017-12-05 | 09:30 | 09:50 | 104 | 4 | X | 24.98 | 104 | Yakov | Kagan | @yakovkagan | 4.52 | 1910 | 103 |
| 1004 | 2017-12-05 | 13:40 | 14:05 | 105 | 1 | X | 31.27 | |||||||
| 1005 | 2017-12-05 | 15:15 | 16:00 | 103 | 2 | POOL | 18.95 | 103 | Kassa | Korley | @kassablanca | 4.63 | 42 | |
| 1006 | 2017-12-05 | 18:20 | 18:55 | 101 | 3 | XL | 78.52 | 101 | Sonny | Li | @sonnynomnom | 4.66 | 352 |
3. 3.將罰單與車對應 使用inner join
select *
from trips
JOIN cars
ON trips.car_id=cars.id;