SQL查询计数选择

我需要如下所述生成表。我有这些样本，但它不起作用。任何请帮助。SQL查询计数选择

 
SELECT id, room, bed, 
bed -(SELECT count(bed) FROM tb_student WHERE room_id = id) as FREE 
FROM tb_rooms 

tb_rooms 
+----+-------+-----+ 
| ID | ROOM | BED | 
+----+-------+-----+ 
| 1 | A111 | 4 | 
| 2 | A112 | 2 | 
| 3 | A113 | 2 | 
| 4 | A114 | 2 | 
+----+-------+-----+ 

tb_student 
+----+---------+----------+ 
| ID | STUD_ID | ROOM_ID | 
+----+---------+----------+ 
| 1 | 211 | 3  | 
| 2 | 212 | 1  | 
| 3 | 213 | 1  | 
| 4 | 214 | 2  | 
+----+----------+---------+ 

I need something like this... 
+----+-------+------+-----+ 
| ID | ROOM | BED |FREE | 
+----+-------+------+-----+ 
| 1 | A111 | 4 | 2 | 
| 2 | A112 | 2 | 1 | 
| 3 | A113 | 2 | 1 | 
| 4 | A114 | 2 | 2 | 
+----+-------+------+-----+

来源

2016-01-02 John Arzaga

你目前收到了什么结果？ –

我从tb_students全部所有房间中扣除。 –

试试这个

SELECT id as id, room as room, bed as bed, 
bed -(SELECT count(Room_ID) FROM tb_student where room_id = soh.id) as FREE 
FROM tb_rooms soh

好运

来源

2016-01-02 08:23:35

谢谢Soheil Karami ...但它不起作用。 –

tb_student中的theres no（bed）列。它说未知的专栏。 –

哦，对不起，我已经意识到，等我解决这个问题 –

John：您好您的查询权仅创建表的别名在这里看到您的查询：

SELECT id as id, room as room, bed as bed, 
bed -(SELECT count(*) FROM tb_student where room_id = tr.id) as FREE 
FROM tb_rooms tr

来源

2016-01-02 09:01:59

你可以试试这个：

SELECT 
    r.ID, 
    r.ROOM, 
    r.BED, 
    r.BED - COUNT(DISTINCT s.ID) AS FREE 
FROM tb_rooms AS r 
    LEFT JOIN tb_student AS s ON r.ID = s.ROOM_ID 
GROUP BY r.ID, r.ROOM, r.BED

这是SQLFIDDLE。

来源

2016-01-02 09:10:08 Beginner

谢谢你Guyz！ –

SQL查询计数选择

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