php - 如何将表连接到此结果？

Question

我有一个 T1 表和 T2 表，如下所示:

T1

-------------------------------------------------------
id | price  | email
-------------------------------------------------------
1  | $1000  | jacky@domain.com
2  | $2000  | angle@domain.com
3  | $3000  | kevin@domain.com
-------------------------------------------------------

T2

-------------------------------------------------------
id | master | country | key   | value
-------------------------------------------------------
1  | 1      | US      | price | $399 
2  | 1      | US      | email | jacky/domain.us 
3  | 1      | ES      | price | $550 
4  | 1      | ES      | email | jacky@domain.es 
5  | 1      | JP      | price | $820 
6  | 1      | JP      | email | jacky@domain.jp 
7  | 2      | US      | price | $360 
8  | 2      | US      | email | angle@domain.us 
-------------------------------------------------------

如何得到这个结果:

T3

----------------------------------------------------------------------------------------------------------------------------
id | price  | price_US  | price_ES  | price_JP  | email            | email_US        | email_ES        | email_JP
----------------------------------------------------------------------------------------------------------------------------
1  | $1000  | $399      | $550      | $820      | jacky@domain.com | jacky@domain.us | jacky@domain.es | jacky@domain.jp
1  | $2000  | $360      | NULL      | NULL      | angle@domain.com | angle@domain.us | NULL            | NULL
1  | $3000  | NULL      | NULL      | NULL      | NULL             | NULL            | NULL            | NULL
----------------------------------------------------------------------------------------------------------------------------

或者我可以用 PHP 得到这个结果吗？

T4

-------------------------------------------------------
id | price  | email             | more_info
-------------------------------------------------------
1  | $1000  | jacky@domain.com  | [array (rows...)]
2  | $2000  | angle@domain.com  | [array (rows...)]
3  | $3000  | kevin@domain.com  | [array (rows...)]
-------------------------------------------------------

有什么想法吗？

编辑1

或者我可以得到如下结果吗？

T5(美国国家/地区的结果)

-------------------------------------------------------
id | price  | email
-------------------------------------------------------
1  | $399   | jacky@domain.us
2  | $360   | angle@domain.us
3  | $3000  | kevin@domain.com
-------------------------------------------------------

T6(日本国家成绩)

-------------------------------------------------------
id | price   | email
-------------------------------------------------------
1  | $820    | jacky@domain.jp
2  | $2000   | angle@domain.com
3  | $3000   | kevin@domain.com
-------------------------------------------------------

Answer 1

这种类型的数据转换是一个枢轴。 MySQL 没有枢轴函数，但您可以使用带有 CASE 表达式的聚合函数来复制它:

select t1.id,
  t1.price,
  max(case when t2.country = 'US' and `key` = 'price' then t2.value end) Price_US,
  max(case when t2.country = 'ES' and `key` = 'price' then t2.value end) Price_ES,
  max(case when t2.country = 'JP' and `key` = 'price' then t2.value end) Price_JP,
  t1.email,
  max(case when t2.country = 'US' and `key` = 'email' then t2.value end) Email_US,
  max(case when t2.country = 'ES' and `key` = 'email' then t2.value end) Email_ES,
  max(case when t2.country = 'JP' and `key` = 'email' then t2.value end) Email_JP
from table1 t1
left join table2 t2
  on t1.id = t2.master
group by t1.id, t1.price, t1.email

参见SQL Fiddle with Demo

编辑#1，如果您只想使用联接而不是聚合函数，那么您的查询将类似于以下内容:

select t1.id,
  t1.price,
  P_US.value Price_US,
  P_ES.value Price_ES,
  P_JP.value Price_JP,
  t1.email,
  E_US.value Email_US,
  E_ES.value Email_ES,
  E_JP.value Email_JP
from table1 t1
left join table2 P_US
  on t1.id = P_US.master
  and P_US.country = 'US'
  and P_US.`key` = 'price'
left join table2 P_ES
  on t1.id = P_ES.master
  and P_ES.country = 'ES'
  and P_ES.`key` = 'price'
left join table2 P_JP
  on t1.id = P_JP.master
  and P_JP.country = 'JP'
  and P_JP.`key` = 'price'
left join table2 E_US
  on t1.id = E_US.master
  and E_US.country = 'US'
  and E_US.`key` = 'email'
left join table2 E_ES
  on t1.id = E_ES.master
  and E_ES.country = 'ES'
  and E_ES.`key` = 'email'
left join table2 E_JP
  on t1.id = E_JP.master
  and E_JP.country = 'JP'
  and E_JP.`key` = 'email'

参见SQL Fiddle with Demo

结果:

| ID | PRICE | PRICE_US | PRICE_ES | PRICE_JP |            EMAIL |        EMAIL_US |        EMAIL_ES |        EMAIL_JP |
------------------------------------------------------------------------------------------------------------------------
|  1 |  1000 |      399 |      550 |      820 | jacky@domain.com | jacky/domain.us | jacky@domain.es | jacky@domain.jp |
|  2 |  2000 |      360 |   (null) |   (null) | angle@domain.com | angle@domain.us |          (null) |          (null) |
|  3 |  3000 |   (null) |   (null) |   (null) |     kevin@domain |          (null) |          (null) |          (null) |

编辑 #2:要获得类似于 T5 和 T6 的结果，您将使用以下内容。对于 T6，将 US 替换为 JP:

select t1.id,
  max(case when `key` = 'price' then value end) price,
  max(case when `key` = 'email' then value end) email
from table1 t1
left join table2 t2
  on t1.id = t2.master
where t2.country = 'US'
group by t1.id
union all
select t1.id,
  t1.price,
  t1.email
from table1 t1
where not exists (select t.id
                  from table1 t
                  left join table2 t2
                    on t.id = t2.master
                  where t2.country = 'US'
                     and t1.id = t.id);

参见SQL Fiddle with Demo