我有一个名为HMS_PGHL8_HQID_MAP_DEMO
的表和名为UDMTOOL_STG
的数据库。
以下查询会提取我所需的值。
SELECT udm_main.RECORD_ID AS record_id ,udm_main.HMS_PGH_NODE_ID AS hms_pgh_node_id,
udm_main.HMS_PGH_NODE_NAME AS hms_pgh_node_name ,udm_main.HQ_NODE_ID AS hq_node_id, udm_main.HQ_NODE_NAME AS hq_node_name,
udm_main.NODE_TYPE_IND as node_type_ind,UPDATED_ON,UPDATED_BY,
INSERT_TS FROM UDMTOOL_STG.HMS_PGHL8_HQID_MAP_DEMO udm_main
WHERE op_code <> 'D' OR op_code IS NULL
ORDER BY RECORD_ID,HMS_PGH_NODE_ID,HQ_NODE_ID
GROUP BY 1,2,3,4,5,6,7,8,9;
现在我需要得到总行值:
请注意,下面不起作用,
SELECT MIN(udm_main.RECORD_ID AS) record_id ,udm_main.HMS_PGH_NODE_ID AS hms_pgh_node_id,
udm_main.HMS_PGH_NODE_NAME AS hms_pgh_node_name ,udm_main.HQ_NODE_ID AS hq_node_id, udm_main.HQ_NODE_NAME AS hq_node_name,
udm_main.NODE_TYPE_IND as node_type_ind,UPDATED_ON,UPDATED_BY,
INSERT_TS FROM UDMTOOL_STG.HMS_PGHL8_HQID_MAP_DEMO udm_main
WHERE op_code <> 'D' OR op_code IS NULL
ORDER BY RECORD_ID,HMS_PGH_NODE_ID,HQ_NODE_ID
GROUP BY 1,2,3,4,5,6,7,8,9;
请求您的帮助以解决问题。
答案 0 :(得分:3)
您可以使用QUALIFY
过滤MIN / MAX:
SELECT udm_main.RECORD_ID,udm_main.HMS_PGH_NODE_ID AS hms_pgh_node_id,
udm_main.HMS_PGH_NODE_NAME AS hms_pgh_node_name,udm_main.HQ_NODE_ID AS hq_node_id, udm_main.HQ_NODE_NAME AS hq_node_name,
udm_main.NODE_TYPE_IND as node_type_ind,UPDATED_ON,UPDATED_BY,
INSERT_TS
FROM UDMTOOL_STG.HMS_PGHL8_HQID_MAP_DEMO udm_main
WHERE op_code <> 'D' OR op_code IS NULL
GROUP BY 1,2,3,4,5,6,7,8,9
QUALIFY udm_main.RECORD_ID = MIN(udm_main.RECORD_ID) OVER ()
OR udm_main.RECORD_ID = MAX(udm_main.RECORD_ID) OVER ()
ORDER BY RECORD_ID