我需要将combined_streets中的table1插入combined_streets中的table2。共同密钥为stop_id但请注意stop_id在table1中是唯一的，但在stop_id中有table2的重复。
table1
+---------+-----------+--------------+-----------+------------+-----------------+---------+---------------+----------------------+
| stop_id | on_street | cross_street | boardings | alightings | month_beginning | daytype | location | combined_streets |
+---------+-----------+--------------+-----------+------------+-----------------+---------+---------------+----------------------+
| 137 | JACKSON | KOLMAR | 4.6 | 14.2 | 10/01/2012 | Weekday | "(41.87696338 | JACKSON & KOLMAR |
| 138 | JACKSON | KENTON | 5 | 15.2 | 10/01/2012 | Weekday | "(41.87696546 | JACKSON & KENTON |
| 139 | JACKSON | KILPATRICK | 38.9 | 242.1 | 10/01/2012 | Weekday | "(41.87692666 | JACKSON & KILPATRICK |
| 140 | JACKSON | CICERO | 0.2 | 1.7 | 10/01/2012 | Weekday | "(41.87685893 | JACKSON & CICERO |
| 141 | JACKSON | 4900 WEST | 0 | 0 | 10/01/2012 | Weekday | "(41.87682102 | JACKSON & 4900 WEST |
| 142 | JACKSON | LAVERGNE | 0 | 0 | 10/01/2012 | Weekday | "(41.87680393 | JACKSON & LAVERGNE |
| 143 | JACKSON | 5056 WEST | 0 | 0 | 10/01/2012 | Weekday | "(41.87675729 | JACKSON & 5056 WEST |
| 144 | JACKSON | LEAMINGTON | 0 | 0 | 10/01/2012 | Weekday | "(41.87672858 | JACKSON & LEAMINGTON |
| 145 | JACKSON | LARAMIE | 59.1 | 172.9 | 10/01/2012 | Weekday | "(41.87673038 | JACKSON & LARAMIE |
| 146 | JACKSON | LOCKWOOD | 14.3 | 92 | 10/01/2012 | Weekday | "(41.87667680 | JACKSON & LOCKWOOD |
+---------+-----------+--------------+-----------+------------+-----------------+---------+---------------+----------------------+
table2
+---------+-------+------------------+
| stop_id | route | combined_streets |
+---------+-------+------------------+
| 74 | 126 | NULL |
| 74 | 132 | NULL |
| 74 | 143 | NULL |
| 74 | 147 | NULL |
| 75 | 1 | NULL |
| 75 | 7 | NULL |
| 75 | X28 | NULL |
| 75 | 126 | NULL |
| 75 | 129 | NULL |
| 76 | 3 | NULL |
+---------+-------+------------------+
答案 0 :(得分：0)
UPDATE table2 t2 LEFT JOIN table1 t1 ON t2.stop_id = t1.stop_id SET t2.combined_streets = t1.combined_streets;
答案 1 :(得分：0)
UPDATE table1, table2 SET table2.combined_streets = table1.combined_streets WHERE
table1.stop_id=table2.stop_id;