表联接列名到列值

Question

我已经看到了许多相似的问题，但是我似乎找不到与我需要做的事情相匹配的问题。

我有以下2表：

表1（ambro_awards_categories）：

id  cat_name    category
1   cat1        Would most likely know how to fondue an ant
2   cat2        Most Likely to survive a nuclear winter
3   cat3        Most Likely to survive a nuclear winter
4   cat4        Category 4
5   cat5        Category 5
6   cat6        Category 6
7   cat7        Category 7
8   cat8        Category 8
9   cat9        Category 9

和表2（ambro_awards）：

id  voter_name          cat1                cat2                cat>>>      date
1   Cavey Charlesworth  Cavey Charlesworth  Cavey Charlesworth  ...         2015-01-18 00:00:00
2   Lynn Wylder         Lynn Wylder         Lynn Wylder         ...         2015-01-18 00:00:00
3   Lynn Wylder         Peter Parker        Batman              ...         2015-01-18 00:00:00
etc

我想做的事情（我完全接受，在那里可能有更好的方法）是将所有内容合并到1个大表中，并加入cat1，cat2等。这样我可以将Category名称修改为更长，更具描述性的名称，而无需延长第二张表。

这意味着要对表2中的cat1，cat2列名称使用表1的cat_name值来执行此操作。

所需的结果将是这样的：

Would most likely know how to fondue an ant (ie "Category 1")
     name 1
     name 2
     name 3
     name 4... etc

Most Likely to survive a nuclear winter (ie "Category 2")
     name 1
     name 2
     name 3
     name 4... etc

Most likeley to succeed (ie "Category 3")
     name 1
     name 2
     name 3
     name 4... etc

之类的东西...

如果仅将多个名称添加到1表中很简单，我会这样做，但认为这可能有点不对劲...希望这样做，但随后遇到了问题...

PJ。

Answer 1

至少可以说，这种设计远非最优，但要回答您的问题，就我所知，您无法连接表，其中第二个表的列名称是一列的值（而且您当然可以不可携带）。 但是，我认为可以用您当前的模式以这种方式编写您想要的内容：

CREATE VIEW ambro_full_data AS
SELECT a.id as aid, c.id as cid, c.cat_name,
    CASE c.cat_name
        WHEN 'cat1' THEN a.cat1
        WHEN 'cat2' THEN a.cat2
        WHEN 'cat3' THEN a.cat3
        WHEN 'cat4' THEN a.cat4
        WHEN 'cat5' THEN a.cat5
        WHEN 'cat6' THEN a.cat6
        WHEN 'cat7' THEN a.cat7
        WHEN 'cat8' THEN a.cat8
        WHEN 'cat9' THEN a.cat9
    END as name,
    c.category,
    a.date
FROM ambro_awards_categories c
  CROSS JOIN ambro_awards a;

例如：

mysql> SELECT name FROM ambro_full_data WHERE cid = 1;
+--------------------+
| name               |
+--------------------+
| Cavey Charlesworth |
| Lynn Wylder        |
| Peter Parker       |
+--------------------+
3 rows in set (0.00 sec)

要么

mysql> SELECT category,name FROM ambro_full_data ORDER BY cid;
+---------------------------------------------+--------------------+
| category                                    | name               |
+---------------------------------------------+--------------------+
| Would most likely know how to fondue an ant | Cavey Charlesworth |
| Would most likely know how to fondue an ant | Lynn Wylder        |
| Would most likely know how to fondue an ant | Peter Parker       |
| Most Likely to survive a nuclear winter     | Lynn Wylder        |
| Most Likely to survive a nuclear winter     | Batman             |
| Most Likely to survive a nuclear winter     | Cavey Charlesworth |
| Most Likely to survive a nuclear winter     | Lynn Wylder        |
| Most Likely to survive a nuclear winter     | Cavey Charlesworth |
| Most Likely to survive a nuclear winter     | Lynn Wylder        |
| Category 4                                  | Cavey Charlesworth |
| Category 4                                  | Lynn Wylder        |
| Category 4                                  | Lynn Wylder        |
| Category 5                                  | Lynn Wylder        |
| Category 5                                  | Lynn Wylder        |
| Category 5                                  | Cavey Charlesworth |
| Category 6                                  | Lynn Wylder        |
| Category 6                                  | Cavey Charlesworth |
| Category 6                                  | Lynn Wylder        |
| Category 7                                  | Cavey Charlesworth |
| Category 7                                  | Lynn Wylder        |
| Category 7                                  | Lynn Wylder        |
| Category 8                                  | Lynn Wylder        |
| Category 8                                  | Lynn Wylder        |
| Category 8                                  | Cavey Charlesworth |
| Category 9                                  | Cavey Charlesworth |
| Category 9                                  | Lynn Wylder        |
| Category 9                                  | Lynn Wylder        |
+---------------------------------------------+--------------------+
27 rows in set (0.00 sec)

但是，当您添加列时，您需要相应地更新视图。

但是，更好的设计是使用如下表：

CREATE TABLE ambro_awards (id ..., voter_name ..., date ...); （请注意，没有catN列）

CREATE TABLE ambro_awards_categories (id ..., category ...); （请注意，无需cat_name列）

CREATE TABLE ambro_awards_and_categories (a_id INTEGER NOT NULL, c_id INTEGER NOT NULL, name VARCHAR(<suitable length here>) NOT NULL, PRIMARY KEY (a_id, c_id), FOREIGN KEY (a_id) REFERENCES ambro_awards (id), FOREIGN_KEY (c_id) REFERENCES ambro_awards_categories (id));

使用您的样本数据，后一个表将如下所示：

a_id    c_id    name
1       1       Cavey Charlesworth
1       2       Cavey Charlesworth
1       3       Cavey Charlesworth
2       1       Lynn Wylder
2       2       Lynn Wylder
2       3       Lynn Wylder
3       1       Lynn Wylder
3       2       Peter Parker
3       3       Batman

这样，您可以根据需要创建任意多个类别，而无需为每个类别创建一列。 然后，您可以执行适当的JOIN来提取与上述相同的数据，例如

SELECT c.category, ac.name FROM ambro_awards_and_categories ac JOIN ambro_awards_categories c ON c.id = ac.c_id;

并且类似的联接可用于从其他表中提取其他数据。 例如，您可以将原始的ambro_awards第1行提取为多行，例如

SELECT a.id, a.voter_name, a.date, ac.name FROM ambro_awards a JOIN ambro_awards_and_categories ac ON ac.a_id = a.id WHERE a.id = 1;

这是SQL中非常标准的做法，如果使用得当，也应该可以解决您的问题。