Python Day45多表連接查詢


一、多表連接查詢

1 交叉連接:不適用任何匹配條件。生成笛卡爾積

mysql> select * from employee,department;
+----+------------+--------+------+--------+------+--------------+
| id | name       | sex    | age  | dep_id | id   | name         |
+----+------------+--------+------+--------+------+--------------+
|  1 | egon       | male   |   18 |    200 |  200 | 技術         |
|  1 | egon       | male   |   18 |    200 |  201 | 人力資源     |
|  1 | egon       | male   |   18 |    200 |  202 | 銷售         |
|  1 | egon       | male   |   18 |    200 |  203 | 運營         |
|  2 | alex       | female |   48 |    201 |  200 | 技術         |
|  2 | alex       | female |   48 |    201 |  201 | 人力資源     |
|  2 | alex       | female |   48 |    201 |  202 | 銷售         |
|  2 | alex       | female |   48 |    201 |  203 | 運營         |
|  3 | wupeiqi    | male   |   38 |    201 |  200 | 技術         |
|  3 | wupeiqi    | male   |   38 |    201 |  201 | 人力資源     |
|  3 | wupeiqi    | male   |   38 |    201 |  202 | 銷售         |
|  3 | wupeiqi    | male   |   38 |    201 |  203 | 運營         |
|  4 | yuanhao    | female |   28 |    202 |  200 | 技術         |
|  4 | yuanhao    | female |   28 |    202 |  201 | 人力資源     |
|  4 | yuanhao    | female |   28 |    202 |  202 | 銷售         |
|  4 | yuanhao    | female |   28 |    202 |  203 | 運營         |
|  5 | liwenzhou  | male   |   18 |    200 |  200 | 技術         |
|  5 | liwenzhou  | male   |   18 |    200 |  201 | 人力資源     |
|  5 | liwenzhou  | male   |   18 |    200 |  202 | 銷售         |
|  5 | liwenzhou  | male   |   18 |    200 |  203 | 運營         |
|  6 | jingliyang | female |   18 |    204 |  200 | 技術         |
|  6 | jingliyang | female |   18 |    204 |  201 | 人力資源     |
|  6 | jingliyang | female |   18 |    204 |  202 | 銷售         |
|  6 | jingliyang | female |   18 |    204 |  203 | 運營         |
+----+------------+--------+------+--------+------+--------------+

2 內連接:只連接匹配的行

#找兩張表共有的部分,相當於利用條件從笛卡爾積結果中篩選出了正確的結果
#department沒有204這個部門,因而employee表中關於204這條員工信息沒有匹配出來
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee inner join department on employee.dep_id=department.id; 
+----+-----------+------+--------+--------------+
| id | name      | age  | sex    | name         |
+----+-----------+------+--------+--------------+
|  1 | egon      |   18 | male   | 技術         |
|  2 | alex      |   48 | female | 人力資源     |
|  3 | wupeiqi   |   38 | male   | 人力資源     |
|  4 | yuanhao   |   28 | female | 銷售         |
|  5 | liwenzhou |   18 | male   | 技術         |
+----+-----------+------+--------+--------------+

#上述sql等同於
mysql> select employee.id,employee.name,employee.age,employee.sex,department.name from employee,department where employee.dep_id=department.id;

3 外鏈接之左連接:優先顯示左表全部記錄

#以左表為准,即找出所有員工信息,當然包括沒有部門的員工
#本質就是:在內連接的基礎上增加左邊有右邊沒有的結果
mysql> select employee.id,employee.name,department.name as depart_name from employee left join department on employee.dep_id=department.id;
+----+------------+--------------+
| id | name       | depart_name  |
+----+------------+--------------+
|  1 | egon       | 技術         |
|  5 | liwenzhou  | 技術         |
|  2 | alex       | 人力資源     |
|  3 | wupeiqi    | 人力資源     |
|  4 | yuanhao    | 銷售         |
|  6 | jingliyang | NULL         |
+----+------------+--------------+

4 外鏈接之右連接:優先顯示右表全部記錄

#以右表為准,即找出所有部門信息,包括沒有員工的部門
#本質就是:在內連接的基礎上增加右邊有左邊沒有的結果
mysql> select employee.id,employee.name,department.name as depart_name from employee right join department on employee.dep_id=department.id;
+------+-----------+--------------+
| id   | name      | depart_name  |
+------+-----------+--------------+
|    1 | egon      | 技術         |
|    2 | alex      | 人力資源     |
|    3 | wupeiqi   | 人力資源     |
|    4 | yuanhao   | 銷售         |
|    5 | liwenzhou | 技術         |
| NULL | NULL      | 運營         |
+------+-----------+--------------+

5 全外連接:顯示左右兩個表全部記錄

全外連接:在內連接的基礎上增加左邊有右邊沒有的和右邊有左邊沒有的結果
#注意:mysql不支持全外連接 full JOIN
#強調:mysql可以使用此種方式間接實現全外連接
select * from employee left join department on employee.dep_id = department.id
union
select * from employee right join department on employee.dep_id = department.id
;
#查看結果
+------+------------+--------+------+--------+------+--------------+
| id   | name       | sex    | age  | dep_id | id   | name         |
+------+------------+--------+------+--------+------+--------------+
|    1 | egon       | male   |   18 |    200 |  200 | 技術         |
|    5 | liwenzhou  | male   |   18 |    200 |  200 | 技術         |
|    2 | alex       | female |   48 |    201 |  201 | 人力資源     |
|    3 | wupeiqi    | male   |   38 |    201 |  201 | 人力資源     |
|    4 | yuanhao    | female |   28 |    202 |  202 | 銷售         |
|    6 | jingliyang | female |   18 |    204 | NULL | NULL         |
| NULL | NULL       | NULL   | NULL |   NULL |  203 | 運營         |
+------+------------+--------+------+--------+------+--------------+

#注意 union與union all的區別:union會去掉相同的紀錄

二、子查詢

    

  1:子查詢是將一個查詢語句嵌套在另一個查詢語句中。
  2:內層查詢語句的查詢結果,可以為外層查詢語句提供查詢條件。
  3:子查詢中可以包含:IN、NOT IN、ANY、ALL、EXISTS 和 NOT EXISTS等關鍵字
  4:還可以包含比較運算符:= 、 !=、> 、<等



1 帶IN關鍵字的子查詢
查詢employee表,但dep_id必須在department表中出現過
select * from employee
    where dep_id in
        (select id from department);

2 帶比較運算符的子查詢

#比較運算符:=、!=、>、>=、<、<=、<>
#查詢平均年齡在25歲以上的部門名
select id,name from department
    where id in 
        (select dep_id from employee group by dep_id having avg(age) > 25);

#查看技術部員工姓名
select name from employee
    where dep_id in 
        (select id from department where name='技術');

#查看不足1人的部門名
select name from department
    where id in 
        (select dep_id from employee group by dep_id having count(id) <=1);

3 帶EXISTS關鍵字的子查詢

EXISTS關字鍵字表示存在。在使用EXISTS關鍵字時,內層查詢語句不返回查詢的記錄。
而是返回一個真假值。True或False
當返回True時,外層查詢語句將進行查詢;當返回值為False時,外層查詢語句不進行查詢

#department表中存在dept_id=203,Ture
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=200);
+----+------------+--------+------+--------+
| id | name       | sex    | age  | dep_id |
+----+------------+--------+------+--------+
|  1 | egon       | male   |   18 |    200 |
|  2 | alex       | female |   48 |    201 |
|  3 | wupeiqi    | male   |   38 |    201 |
|  4 | yuanhao    | female |   28 |    202 |
|  5 | liwenzhou  | male   |   18 |    200 |
|  6 | jingliyang | female |   18 |    204 |
+----+------------+--------+------+--------+

#department表中存在dept_id=205,False
mysql> select * from employee
    ->     where exists
    ->         (select id from department where id=204);
Empty set (0.00 sec)

 




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