Question

我很难尝试查询以某种方式显示结果。我需要的是每个用户在一行中显示所有用户答案，我无法连接所有答案，因为我的程序需要访问每个答案的结果和我理解的连接将返回一个字符串。

这是表格的定义方式：

这是存储用户信息的地方。用户

+------+---------+-------------+
| u_id | email   | name        |
+------+---------+-------------+
|    1 | bob@b   | bob         |
|    2 | r@e.com | sam         |
|    3 | ra@se   | steve       |
|    4 | tv@we   | rob         |
|    5 | tr@sd   | ted         |
|    6 | qw@as   | john        |
+------+---------+-------------+

Questions
+------+---------+-------------+
| q_id | question_name         |
+------+---------+-------------+
|    1 | Age range?            |
|    2 | Do you use Amazon?    |
|    3 | Do you use Facebook?  |
|    4 | Interested in toys?   |
+------+---------+-------------+

这些是用户可以选择的答案。

Option_Choices
+------+---------+-------------+
|oc_id | opt_choice_name       |
+------+---------+-------------+
|    1 | YES                   |
|    2 | NO                    |
|    3 | 18-24 Years           |
|    4 | 25-35 Years           |
|    5 | Very Interested       |
|    6 | Not Interested        |
+------+---------+-------------+

这是一个问题与每个可能的问题相关的地方回答这个具体问题。

Question_Options
+------+---------+-------------+
|qo_id | q_id    | oc_id       |
+------+---------+-------------+
|    1 | 1       | 3           | // Age range?: 18-24
|    2 | 1       | 4           | // Age range?: 25-35
|    3 | 2       | 1           | // Do you use Amz? Yes
|    4 | 2       | 2           | ...
|    5 | 3       | 1           | ...
|    6 | 3       | 2           | ...
|    7 | 4       | 5           | ...
|    8 | 4       | 6           |
+------+---------+-------------+

这是存储每个用户给出的答案的地方。

Answers
+------+---------+-------------+
| a_id | u_id    | qo_id       |
+------+---------+-------------+
|    1 | 1       | 2           |
|    2 | 1       | 4           |
|    3 | 1       | 6           |
|    4 | 1       | 7           |
|    1 | 2       | 1           |
|    2 | 2       | 3           |
|    3 | 2       | 6           |
|    4 | 2       | 8           |
|    1 | 3       | 2           |
|    2 | 3       | 4           |
|    3 | 3       | 5           |
|    4 | 3       | 8           |
+------+---------+-------------+

我现在需要的是构建一个查询来显示信息以类似的方式：

+---------+-------------+-------------+-------------+---------------------+
| email   | Age range?  | Use Amazon? | Use FB?     | Interested in toys? |
+---------+-------------+-------------+-------------+---------------------+
| bob@b   | 18-24 Years |    YES      |    NO       | Very Interested     |
| r@e.com | 25-35 Years |    NO       |    YES      | Not Interested      |
| ra@se   | 25-35 Years |    NO       |    YES      | Not Interested      |
| tv@we   | 18-24 Years |    YES      |    YES      | Not Interested      |
| tr@sd   | 18-24 Years |    YES      |    NO       | Not Interested      |
| qw@as   | 25-35 Years |    YES      |    YES      | Very Interested     |
+------+---------+--------------------+-------------+---------------------+

但是只使用JOIN我会得到类似的东西“

+---------+----------------+-----------------+
| email   | question_name  | opt_choice_name |
+---------+----------------+-----------------+
| bob@b   | Age range?     |  18-24 Years    |
| bob@b   | Use Amazon?    |    YES          |
| bob@b   | Use FB?        |    NO           |
| bob@b   | Inter in toys? |Very Interested  |
| r@e.com | Age range?     |  25-35 Years    |
| r@e.com | Use Amazon?    |    NO       |
| r@e.com | Use FB?        |    YES          |
| r@e.com | Inter in toys? | Not Interested  |

   ...         ...               ...
+---------+----------------+-----------------+

我尝试使用带有这样的查询的枢轴：

SELECT u.email,
(CASE q.question_name WHEN q_id = 1 THEN oc.opt_choice_name ELSE 0) AS 'Age_Range',
(CASE q.question_name WHEN q_id = 2 THEN oc.opt_choice_name ELSE 0) AS 'Amazon',
(CASE q.question_name WHEN q_id = 3 THEN oc.opt_choice_name ELSE 0) AS 'FB',
(CASE q.question_name WHEN q_id = 4 THEN oc.opt_choice_name ELSE 0) AS 'Toys'
FROM Users u
INNER JOIN Ansers a ON u.u_id = a.u_id
INNER JOIN Question_Options qo ON a.qo_id = qo.qo_id
INNER JOIN Questions q ON qo.q_id = q.q_id
INNER JOIN Option_Choices oc ON qo.oc_id = oc.oc_id
GROUP BY u.email

但我的结果好坏参半。

任何建议都将不胜感激。

感谢。

我尝试使用MAX

SELECT u.email,
MAX( CASE q.question_name WHEN q.question_id = 1 THEN oc.option_choice_name ELSE 0 END) AS 'AgeRange',
MAX( CASE q.question_name WHEN q.question_id = 2 THEN oc.option_choice_name ELSE 0 END) AS 'Amazon',
MAX( CASE q.question_name WHEN q.question_id = 3 THEN oc.option_choice_name ELSE 0 END) AS 'FB',
MAX( CASE q.question_name WHEN q.question_id = 4 THEN oc.option_choice_name ELSE 0 END) AS 'toys',
FROM Users u
INNER JOIN Answers a ON u.u_id = a.u_id
INNER JOIN Question_Options qo ON a.qo_id = qo.qo_id
INNER JOIN Questions q ON qo.q_id = q.q_id
INNER JOIN Option_Choices oc ON qo.oc_id = oc.oc_id
GROUP BY u.email

但我得到的结果是：

+---------+-------------+-------------+-------------+---------------------+
| email   | Age range?  | Use Amazon? | Use FB?     | Interested in toys? |
+---------+-------------+-------------+-------------+---------------------+
| bob@b   | YES         |    YES      |    NO       |          NO         |
| r@e.com | YES         |    YES      |    YES      | YES                 |
| ra@se   | YES         |    YES      |    YES      | YES                 |
| tv@we   | YES         |    YES      |    YES      | YES                 |
+------+---------+--------------------+-------------+---------------------+

Answer 1

虽然你可以用MySQL做到这一点，但它效率不高，因为MySQL索引的行不是列。

索引用于快速查找具有特定列值的行。如果没有索引，MySQL必须从第一行开始然后读取通过整个表来查找相关的行。越大了表，这个成本越多。

http://dev.mysql.com/doc/refman/5.0/en/mysql-indexes.html

大多数人会这样做的方法是用JOIN列举所有问题，例如：

+---------+----------------+-----------------+
| email   | question_name  | opt_choice_name |
+---------+----------------+-----------------+
| bob@b   | Age range?     |  18-24 Years    |
| bob@b   | Use Amazon?    |    YES          |
| bob@b   | Use FB?        |    NO           |
| bob@b   | Inter in toys? |Very Interested  |
| r@e.com | Age range?     |  25-35 Years    |
| r@e.com | Use Amazon?    |    NO       |
| r@e.com | Use FB?        |    YES          |
| r@e.com | Inter in toys? | Not Interested  |

   ...         ...               ...
+---------+----------------+-----------------+

然后在你的PHP中你会做类似的事情（免责声明：未经测试的代码）：

$i = 0;
echo "<table><tr><th>Age range?</th><th>Use Amazon?</th><th>Use Facebook?</th><th>Interested in Toys?</th></tr><tr>";
foreach ($results as $column) {
    echo "<td>".$column."</td>"; 
    if ($i != 0 && $i % 4) // Just arbitrarily chose 4, but it would be however many columns you have
        echo "</tr><tr>";

    $i++;
}
echo "</tr></table>";

Answer 2

解决此问题的另一种方法是分别为每个问题创建临时表，最后创建一个mysql查询以一次性水平获取所有数据。

http://www.tutorialspoint.com/mysql/mysql-temporary-tables.htm

MySQL Query以某种方式显示信息

2 个答案: