作为一名php程序员

对sql的使用仅局限一些简单的增删改查操作，和一些关联查询

可别忘了sql本身就是一门编程语言，有时候一条sql就可以搞定了，没必要使用脚本来实现

今天说的是sql的关联更新操作

mysql> create table test1(id int primary key auto_increment,name varchar(50),age int);
Query OK, 0 rows affected (0.02 sec)

mysql>
mysql>
mysql> create table test2(id int primary key auto_increment,name varchar(50),age int);
Query OK, 0 rows affected (0.03 sec)

mysql> insert into test1(id) values(1);
Query OK, 1 row affected (0.01 sec)

mysql> insert into test1(id) values(2);
Query OK, 1 row affected (0.00 sec)

mysql> select *from test1;
+----+------+------+
| id | name | age  |
+----+------+------+
|  1 | NULL | NULL |
|  2 | NULL | NULL |
+----+------+------+
2 rows in set (0.00 sec)

mysql> insert into test2(id,name,age) values(1,'xiaoming',10);
Query OK, 1 row affected (0.01 sec)

mysql> insert into test2(id,name,age) values(2,'xiaohong',8);
Query OK, 1 row affected (0.00 sec)

mysql>
mysql> select *from test1 left join test2 on test1.id=test2.id;
+----+------+------+------+----------+------+
| id | name | age  | id   | name     | age  |
+----+------+------+------+----------+------+
|  1 | NULL | NULL |    1 | xiaoming |   10 |
|  2 | NULL | NULL |    2 | xiaohong |    8 |
+----+------+------+------+----------+------+
2 rows in set (0.00 sec)

mysql> update test1,test2 set test1.name=test2.name,test1.age=test2.age where test1.id=test2.id;
Query OK, 2 rows affected (0.00 sec)
Rows matched: 2  Changed: 2  Warnings: 0

mysql>
mysql>
mysql> select *from test1 left join test2 on test1.id=test2.id;
+----+----------+------+------+----------+------+
| id | name     | age  | id   | name     | age  |
+----+----------+------+------+----------+------+
|  1 | xiaoming |   10 |    1 | xiaoming |   10 |
|  2 | xiaohong |    8 |    2 | xiaohong |    8 |
+----+----------+------+------+----------+------+
2 rows in set (0.00 sec)

是不是非常方便与简单

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