MySQL Join 詳解
发表于:2007-07-02来源:作者:点击数:
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MySQL Join 詳解 (C) by Dennis Dll 2004.1.29 還是先 Create table 吧 create table emp( id int not null primary key, name varchar(10) ); create table emp_dept( dept_id varchar(4) not null, emp_id int not null, emp_name varchar(10), primary ke
MySQL Join 詳解
(C) by Dennis Dll 2004.1.29
還是先 Create table 吧
create table emp(
id int not null primary key,
name varchar(10)
);
create table emp_dept(
dept_id varchar(4) not null,
emp_id int not null,
emp_name varchar(10),
primary key (dept_id,emp_id));
insert into emp() values
(1,"Dennis-1"),
(2,"Dennis-2"),
(3,"Dennis-3"),
(4,"Dennis-4"),
(5,"Dennis-5"),
(6,"Dennis-6"),
(7,"Dennis-7"),
(8,"Dennis-8"),
(9,"Dennis-9"),
(10,"Dennis-10");
insert into emp_dept() values
("R&D",1,"Dennis-1"),
("DEv",2,"Dennis-2"),
("R&D",3,"Dennis-3"),
("Test",4,"Dennis-4"),
("Test",5,"Dennis-5");
>> left join
-------------
select a.id,a.name,b.dept_id
from emp a left join emp_dept b on (a.id=b.emp_id);
# 挑出左邊的 table emp 中的所有資料,即使 emp_dept 中沒有的資料也挑出來,沒有的就用 NULL 來顯示,
# 也即顯示資料是以左邊的 table emp 中的資料為基礎
mysql> select a.id,a.name,b.dept_id
-> from emp a left join emp_dept b on (a.id=b.emp_id);
+----+-----------+---------+
| id | name | dept_id |
+----+-----------+---------+
| 1 | Dennis-1 | R&D |
| 2 | Dennis-2 | DEv |
| 3 | Dennis-3 | R&D |
| 4 | Dennis-4 | Test |
| 5 | Dennis-5 | Test |
| 6 | Dennis-6 | NULL |
| 7 | Dennis-7 | NULL |
| 8 | Dennis-8 | NULL |
| 9 | Dennis-9 | NULL |
| 10 | Dennis-10 | NULL |
+----+-----------+---------+
# 挑出 table emp 中有而 table emp_dept 中沒有的人員資料
select a.id,a.name,b.dept_id
from emp a left join emp_dept b on (a.id=b.emp_id)
where b.dept_id IS NULL;
mysql> select a.id,a.name,b.dept_id
-> from emp a left join emp_dept b on (a.id=b.emp_id)
-> where b.dept_id IS NULL;
+----+-----------+---------+
| id | name | dept_id |
+----+-----------+---------+
| 6 | Dennis-6 | NULL |
| 7 | Dennis-7 | NULL |
| 8 | Dennis-8 | NULL |
| 9 | Dennis-9 | NULL |
| 10 | Dennis-10 | NULL |
+----+-----------+---------+
# 把 table emp_dept 放在左邊的情形(當然以 emp_dept 中的數據為基礎來顯示資料,emp 中比emp_dept 中多的資料也就不會顯示出來了):
select a.id,a.name,b.dept_id
from emp_dept b left join emp a on (a.id=b.emp_id);
mysql> select a.id,a.name,b.dept_id
-> from emp_dept b left join emp a on (a.id=b.emp_id);
+------+----------+---------+
| id | name | dept_id |
+------+----------+---------+
| 2 | Dennis-2 | DEv |
| 1 | Dennis-1 | R&D |
| 3 | Dennis-3 | R&D |
| 4 | Dennis-4 | Test |
| 5 | Dennis-5 | Test |
+------+----------+---------+
>> right join
---------------
select a.id,a.name,b.dept_id
from emp a right join emp_dept b on (a.id=b.emp_id);
# 挑資料時以右邊 table emp_dept 中的資料為基礎來顯示資料
mysql> select a.id,a.name,b.dept_id
-> from emp a right join emp_dept b on (a.id=b.emp_id);
+------+----------+---------+
| id | name | dept_id |
+------+----------+---------+
| 2 | Dennis-2 | DEv |
| 1 | Dennis-1 | R&D |
| 3 | Dennis-3 | R&D |
| 4 | Dennis-4 | Test |
| 5 | Dennis-5 | Test |
+------+----------+---------+
5 rows in set (0.00 sec)
# 我們再把 table 的位置交換一下,再用 right join 試試
select a.id,a.name,b.dept_id
from emp_dept b right join emp a on (a.id=b.emp_id);
mysql> select a.id,a.name,b.dept_id
-> from emp_dept b right join emp a on (a.id=b.emp_id);
+----+-----------+---------+
| id | name | dept_id |
+----+-----------+---------+
| 1 | Dennis-1 | R&D |
| 2 | Dennis-2 | DEv |
| 3 | Dennis-3 | R&D |
| 4 | Dennis-4 | Test |
| 5 | Dennis-5 | Test |
| 6 | Dennis-6 | NULL |
| 7 | Dennis-7 | NULL |
| 8 | Dennis-8 | NULL |
| 9 | Dennis-9 | NULL |
| 10 | Dennis-10 | NULL |
+----+-----------+---------+
# 是不是和 left join 一樣了?
>> direct join
--------------
# 如果用right join 同不用 Join 直接挑資料是相同的,它等介於以下的指令
select a.id,a.name,b.dept_id
from emp a ,emp_dept b
where a.id=b.emp_id;
mysql> select a.id,a.name,b.dept_id
-> from emp a ,emp_dept b
-> where a.id=b.emp_id;
+----+----------+---------+
| id | name | dept_id |
+----+----------+---------+
| 2 | Dennis-2 | DEv |
| 1 | Dennis-1 | R&D |
| 3 | Dennis-3 | R&D |
| 4 | Dennis-4 | Test |
| 5 | Dennis-5 | Test |
+----+----------+---------+
怎樣,弄明白了嗎?
Enjoy it!
原文转自:http://www.ltesting.net