基于其他列的MySQL更新列

我有这张桌子：

table_a
+----+--------+
| id | status |
+----+--------+
| 10 | ERROR  |
+----+--------+

table_b
+------------+----------------+
| trading_id | trading_status |
+------------+----------------+
|         10 | CLOSED         |
+------------+----------------+

我还有一张桌子：

table_a
+----+--------+
| id | status |
+----+--------+
| 10 | ERROR  |
+----+--------+

table_b
+------------+----------------+
| trading_id | trading_status |
+------------+----------------+
|         10 | CLOSED         |
+------------+----------------+

如何基于

表b.交易状态

更新表a.状态，使结果如下：

table_a
+----+--------+
| id | status |
+----+--------+
| 10 | CLOSED |
+----+--------+

试试这个：

UPDATE table_a
SET status = COALESCE((SELECT trading_status
                       FROM table_b 
                       WHERE trading_id = table_a.id), table_a.status)

上述查询假设

表b

中最多有一条记录与

表a

中的每条记录相匹配

---

您可以使用更新连接语法：

UPDATE table_a
JOIN   table_b ON table_a.id = table_b.id
SET    table_a.status = table_b.trading_status