select DISTINCT(cctv_id), cctv.[name] from points_cctv left join cctv on points_cctv.[cctv_id] = cctv.[id]
我想要计算points_cctv where points_cctv.[cctv_id] = cctv.当我插入count(points_cctv.[cctv</
noiclt22815||K:CMS||O:REgetPerspectiveList||A0:googleP||A1:yahooP||A2:gmail||A3:test||A4:hello||A16:CCTVBarco||A17:CCTV: Corridor CC||A18:CCTV: DR Andy Warhol||A19:CCTV: DR Gaudi (Analog)||A20:CCTV: DR Miro||A21:CCTV: Entrance CC||A22:CCTV</e
我有一些密码:AS FROM cctv1_details AS ALFET JOIN cctv3_details AS B ON (A.CDE_WR = B.CDE_WR) AND A.CDE_dist = B.CDE_distCREATE ORREPLACE TEMPORARY VIEW cctv_join_2 (SELECT *
FROM c