我使用下面的代码
update
contracts a,
contracts_history b
set
a.name_surname=b.name_surname 我的表有64列,我正在寻找一种解决方案来复制所有数据,而不必指定列名-沿着以下行:
$sql = "INSERT INTO `contracts_history`
SELECT * FROM `contracts` WHERE id='$contract_id'";发布于 2012-09-10 12:33:42
只要表的字段数与数据类型匹配的字段数完全相同,就可以使用insert into select...语法-否则,必须使用列名来指定要复制的内容。
我只是运行以下示例来向您展示语法:
mysql> use test
Database changed
mysql> show tables;
Empty set (0.00 sec)
mysql> create table test1 (id int(2), varry varchar(3));
Query OK, 0 rows affected (0.08 sec)
mysql> create table test2 (id int(2), barry varchar(3));
Query OK, 0 rows affected (0.05 sec)
mysql> insert into test2 values(1,'aaa');
Query OK, 1 row affected (0.00 sec)
mysql> select * from test1;
Empty set (0.00 sec)
mysql> insert into test1 (select * from test2);
Query OK, 1 row affected (0.06 sec)
Records: 1 Duplicates: 0 Warnings: 0
mysql> select * from test1;
+------+-------+
| id | varry |
+------+-------+
| 1 | aaa |
+------+-------+
1 row in set (0.00 sec)
mysql> alter table test2 add column third int(1);
Query OK, 1 row affected (0.06 sec)
Records: 1 Duplicates: 0 Warnings: 0
mysql> update test2 set barry='ccc';
Query OK, 1 row affected (0.00 sec)
Rows matched: 1 Changed: 1 Warnings: 0
mysql> insert into test1 (select * from test2);
ERROR 1136 (21S01): Column count doesn't match value count at row 1
mysql>https://stackoverflow.com/questions/12345299
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