Mysql多表查询，多表插入和多表更新

多表查询：
CREATE TABLE IF NOT EXISTS contact(
contact_id int(11) NOT NULL AUTO_INCREMENT,
user_name varchar(255),
nom varchar(255),
prenom varchar(255),
mail varchar(64),
passcode char(64),
PRIMARY KEY(contact_id)
);
CREATE TABLE IF NOT EXISTS droit(
droit_id int( 11 ) NOT NULL AUTO_INCREMENT ,
droit varchar(255),
PRIMARY KEY(droit_id)
);
CREATE TABLE IF NOT EXISTS contactdroit(
contactdroit_id int(11) NOT NULL AUTO_INCREMENT,
contact_id int( 11 ),
droit_id int( 11 ),
PRIMARY KEY( contactdroit_id )
);
Insert into contact(contact_id, user_name) values(1,'user1');
Insert into contact(contact_id, user_name) values(2,'user2');
Insert into contact(contact_id, user_name) values(3,'user3');
Insert into droit(droit_id, droit) values(1,'admin');
Insert into droit(droit_id, droit) values(2,'superuser');
Insert into contactdroit(contact_id, droit_id) values(1, 1);
Insert into contactdroit(contact_id, droit_id) values(2, 1);
Insert into contactdroit(contact_id, droit_id) values(3, 2);

SELECT c.contact_id, d.droit_id, d.droit FROM contact c, contactdroit cd, droit d
where c.contact_id = cd.contact_id
and cd.droit_id = d.droit_id;
结果：
contact_id     droit_id     droit
1                      1           admin
2                      1           admin
3                      2          superuser


多表联查例子：
两个方法都可以，inner join on 更好点。表结构没贴出来，但比较好懂了。
简单方法：

select c.nom, e.nom
from consultant c, affaire a, besoin b, salarie sa, site s, entreprise e
where c.consultant_id=a.consultant_id and a.besoin_id=b.besoin_id and b.salarie_id=sa.salarie_id and sa.site_id=s.site_id and s.entreprise_id=e.entreprise_id

inner join方法：

select c.nom, e.nom
from consultant c
inner join affaire a on c.consultant_id=a.consultant_id
inner join besoin b on a.besoin_id=b.besoin_id
inner join salarie sa on b.salarie_id=sa.salarie_id
inner join site s on sa.site_id=s.site_id
inner join entreprise e on s.entreprise_id=e.entreprise_id


多表插入：

<?php
$conn = mysql_connect("localhost","charles","charles");
mysql_select_db("test");
$query = "INSERT INTO contact(user_name,nom, prenom, mail, passcode) values('sa','se','sf', '[e-mail]', '123')";
$result = mysql_query($query) or die("insert contact failed:".mysql_error());
$lastid = mysql_insert_id(); //得到上一个 插入的id值
echo "last insert id :".$lastid."<br>";
$query2 = "INSERT INTO contactdroit(contact_id, droit_id) values('$lastid','11')";
echo $query2."<br>";
$result2 = mysql_query($query2) or die("insert contactdroit failed: ".mysql_error());
if(isset($result) && isset($result2)){
echo "Good Insertion<br>";
echo $lastid;
}
?>
需注意的是：

The mysql_insert_id() function returns the AUTO_INCREMENT ID generated from the previous INSERT operation.
mysql_insert_id()函数的作用是：取得上一步 INSERT 操作产生的 ID。

This function returns 0 if the previous operation does not generate an AUTO_INCREMENT ID, or FALSE on MySQL connection failure.
如果先前的操作不产生一个自动增加的ID[AUTO_INCREMENT ID]，那么，函数返回0；如果MySQL连接失败，将返回False。



多表更新：
update contact c, contactdroit cd
set c.user_name = '$username', c.nom = '$lastname', c.prenom = '$firstname', c.passcode = '$password', cd.droit_id = '$droitid'
where c.contact_id = '$id' and c.contact_id = cd.contact_id;
示例：
【以下为引用：http://www.javaeye.com/problems/340 】
mysql> create table one(id int(10), name varchar(20));
Query OK, 0 rows affected (0.03 sec)

mysql> create table two(id int(10), name varchar(20));
Query OK, 0 rows affected (0.05 sec)

mysql> insert one value(1, '1');
Query OK, 1 row affected (0.00 sec)

mysql> insert two value(22, '22');
Query OK, 1 row affected (1.02 sec)

mysql> update one o, two t set o.name='oo', t.name='tt';
Query OK, 2 rows affected (0.00 sec)
Rows matched: 2  Changed: 2  Warnings: 0

mysql> select * from one;
+------+------+
| id   | name |
+------+------+
|    1 | oo   |
+------+------+
1 row in set (0.00 sec)

mysql> select * from two;
+------+------+
| id   | name |
+------+------+
|   22 | tt   |
+------+------+
1 row in set (0.00 sec)