MySQL通过匹配First Table中的行然后休息

Question

伙计们，我很难想出一个能为我提供以下输出的SQL：

ID|CITY_SLUG|BAIRRO|NEIGHBORHOOD_BAIRRO|CIDADE|UF|STATE_SLUG
4444|4444|Porto Alegre do Norte
5671|5671|Porto Alegre do Piauí
7994|7994|Porto Alegre
9919|9919|Porto Alegre do Tocantins
12538|12538|Linha Porto Alegre
12538|12538|Centro|63310|Linha Porto Alegre|RS|rs
7994|7994|Aberta dos Morros|13460|Porto Alegre|RS|rs
4444|4444|Centro|38981|Porto Alegre do Norte|MT|mt
5671|5671|Centro|39387|Porto Alegre do Piauí|PI|pi
9919|9919|Setor Central|32994|Porto Alegre do Tocantins|TO|to
4444|4444|Centro|38981|Porto Alegre do Norte|MT|mt
5671|5671|Centro|39387|Porto Alegre do Piauí|PI|pi
7994|7994|Aberta dos Morros|13460|Porto Alegre|RS|rs

这个想法是首先列出城市，然后列出匹配城市名称的邻域，这是我的sql：

Select 
CITY.ID, 
CITY.SLUG as CITY_SLUG,  
NEIGHBORHOOD.NAME as BAIRRO, 
NEIGHBORHOOD.SLUG AS NEIGHBORHOOD_BAIRRO, 
CITY.NAME as CIDADE,  
STATE.ABB AS UF,  
STATE.SLUG AS STATE_SLUG  
from CITY
LEFT OUTER join STATE on CITY.ID_STATE=STATE.ID
LEFT OUTER join NEIGHBORHOOD on NEIGHBORHOOD.ID_CITY = CITY.ID
WHERE (CITY.NAME LIKE '%PORTO ALE%' )

Answer 1

正如@Barmar所提到的，首先获取具有匹配的城市名称的行，然后创建一个UNION，其行包含具有匹配城市名称的邻域。添加相关逻辑以忽略具有NULL值的列：

Select 
CITY.ID, 
CITY.SLUG as CITY_SLUG,  
NEIGHBORHOOD.NAME as BAIRRO, 
NULL,
NULL,
NULL,
NULL
from CITY
LEFT OUTER join STATE on CITY.ID_STATE=STATE.ID
LEFT OUTER join NEIGHBORHOOD on NEIGHBORHOOD.ID_CITY = CITY.ID
WHERE (CITY.NAME LIKE '%PORTO ALE%' )
UNION
Select 
CITY.ID, 
CITY.SLUG as CITY_SLUG,  
NEIGHBORHOOD.NAME as BAIRRO, 
NEIGHBORHOOD.SLUG AS NEIGHBORHOOD_BAIRRO, 
CITY.NAME as CIDADE,  
STATE.ABB AS UF,  
STATE.SLUG AS STATE_SLUG  
from CITY
LEFT OUTER join STATE on CITY.ID_STATE=STATE.ID
LEFT OUTER join NEIGHBORHOOD on NEIGHBORHOOD.ID_CITY = CITY.ID
WHERE (CITY.NAME LIKE '%PORTO ALE%' )