mysql命令通過匹配第一個錶行再休息

Question

人，我有一個很難拿出一個SQL會爲我提供以下的輸出：

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 

的想法是先列出城市然後列出匹配的城市名稱的鄰里，這是我的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提到的，首先提取排有匹配的城市名稱，然後創建一個UNION具有鄰爲匹配城市名行。添加相關邏輯來忽略具有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%')