[英]How do I inner join 2 SQL tables, but only take the first result from the second table?
[英]How do I join a table to this result?
我有一个表作为T1和表作为T2如下所示:
T1
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id | price | email
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1 | $1000 | jacky@domain.com
2 | $2000 | angle@domain.com
3 | $3000 | kevin@domain.com
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T2
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id | master | country | key | value
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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
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如何获得此结果:
T3
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id | price | price_US | price_ES | price_JP | email | email_US | email_ES | email_JP
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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
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还是可以在PHP中获得此结果?
T4
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id | price | email | more_info
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1 | $1000 | jacky@domain.com | [array (rows...)]
2 | $2000 | angle@domain.com | [array (rows...)]
3 | $3000 | kevin@domain.com | [array (rows...)]
-------------------------------------------------------
任何想法?
还是可以得到如下结果?
T5(国家/地区结果的美国)
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id | price | email
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1 | $399 | jacky@domain.us
2 | $360 | angle@domain.us
3 | $3000 | kevin@domain.com
-------------------------------------------------------
T6(国家/地区排名结果)
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id | price | email
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1 | $820 | jacky@domain.jp
2 | $2000 | angle@domain.com
3 | $3000 | kevin@domain.com
-------------------------------------------------------
这种类型的数据变换的是枢转 。 MySQL没有枢轴函数,但是您可以使用带有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
编辑#1,如果您只想使用联接而不是聚合函数,那么您的查询将与此类似:
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'
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,则将使用以下内容。 对于T6
,请用JP
替换US
:
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);
我认为尝试赋予表之间的关系,然后重试。
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