假设我有三个表项目，并且我想用php从数据库中提取一个具有正确值的html。
我尝试过这样：
select (SELECT group_concat(DISTINCT wine_name) FROM wine) as wine_name UNION SELECT username FROM user;
评估表格
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| ev_id | user_id | wine_id | point
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| 1 | 1 | 1 | 80
| 2 | 2 | 1 | 67
| 3 | 1 | 2 | 87
| 5 | 2 | 2 | 97
| 6 | 2 | 3 | 81
| 7 | 3 | 2 | 99
| 8 | 3 | 3 | 66
| 9 | 3 | 1 | 79
-----------------------------------------------------------------------------
酒桌
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| wine_id | wine_name | wine_type | wine_color
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| 1 | Dominio Campo | sweet | red
| 2 | Guitian Godello | dry | white
| 3 | Cosme Palacio | dry | red
| 4 | Azpilicueta | sweet | red
| 5 | Parotet Vermell | sweet | white
-----------------------------------------------------------------------------
User Table
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| user_id | user name | name | email
-----------------------------------------------------------------------------
| 1 | user1 | dsa | asd@asd.com
| 2 | user2 | dsd | abd@asd.com
| 3 | user3 | dss | acd@asd.com
| 4 | user4 | sdd | add@asd.com
| 5 | user5 | ssd | aed@asd.com
-----------------------------------------------------------------------------
在这里我要从三个表中查询并显示结果。
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| | Dominio Campo | Guitian Godello | Cosme Palacio |
-----------------------------------------------------------------------------
| user1 | 80 | 87 | 77 |
| user2 | 67 | 97 | 81 |
| user3 | 79 | 99 | 66 |
-----------------------------------------------------------------------------
答案 0 :(得分：1)
这就是您需要的所有查询：
SELECT w.*
, e.user_id
, e.point
FROM wine w
JOIN evaluate e
ON e.wine_id = w.wine_id
ORDER
BY e.wine_id
, user_id;
其他所有事情都可以并且应该在您的应用程序代码中完成。