Question

我有两个表，我正在尝试完成一个完整的外连接。

availability
availability_id   availability_desc
1                 Monday Morning
2                 Monday Afternoon
3                 Monday Evening
etc

user_availability
user_id          availability_id
1                1
1                3
etc

SELECT * from availability a left outer join user_availability u 
on a.availability_id=u.availability_id;

给出了结果：

availability_id; availability_desc;        id;     user_id;       availability_id
1;               Monday Morning;           1;      1;             1
1;               Monday Morning;           15;     2;             1
1;               Monday Morning;           40;     4;             1
1;               Monday Morning;           57;     5;             1
2;               Monday Afternoon;         2;      1;             2
3;               Monday Evening;           16;     2;             3
3;               Monday Evening;           42;     4;             3

相同表格上的右外连接导致：

availability_id     availability_desc   id  user_id availability_id
1                   Monday Morning      1   1        1
2                   Monday Afternoon    2   1        2
4                   Tuesday Morning     3   1        4

右外连接和左外连接的并集给出了以下结果：

availability_id availability_desc   id  user_id availability_id
1               Monday Morning      1   1       1
1               Monday Morning      15  2       1
1               Monday Morning      40  4       1
1               Monday Morning      57  5       1
2               Monday Afternoon    2   1       2
2               Monday Afternoon    30  3       2
2               Monday Afternoon    41  4       2
3               Monday Evening      16  2       3
3               Monday Evening      42  4       3

当其他表中没有匹配项时，我没有返回任何空值。我希望生成如下所示的结果：

availability_desc     user_id
Monday Morning        1
Monday Afternoon      1
Monday Evening        (null)

不确定我做错了什么。

Answer 1

如果要为每个用户显示一条记录，对于每个可用性，是否有匹配的记录，这都可以解决问题。这假设有一个名为user的表，其中包含系统中的所有用户。使用CROSS JOIN可以获取每个可用性记录的每个用户记录。然后将这两个表连接到user_availability表将查看用户是否可用。

SELECT * 

FROM   availability a 

       cross join user u

       LEFT JOIN user_availability ua
       on u.user_id = ua.user_id
       AND a.availability_id = ua.availability_id
;

这是一个更易读的版本，加上一个sqlfiddle。

SELECT availability_desc, 
       user_name, 
       CASE WHEN ua.user_id IS NULL THEN 'Not Available' ELSE 'Available' END AS availability

FROM   availability a 

       cross join user u

       LEFT JOIN user_availability ua
       on u.user_id = ua.user_id
       AND a.availability_id = ua.availability_id
;

sqlfiddle：http://www.sqlfiddle.com/#!2/626489/4

mysql外连接不显示null结果

1 个答案: