我有下面的查询。SOFTWARE_DEVELOPMENT_CYCLE有多行,但我对最新的感兴趣。
我想重写查询,这样我就不用子查询了。我曾尝试过DENSE_RANK LAST ORDERY BY,但没有成功。
有人能提出建议吗?非常感谢。
SELECT SOF.VENDOR,
SOF.NAME,
LAN.LANGUAGE,
SOF.VERSION,
SDC.STATUS,
SDC.SOF_DC_ID
FROM SOFTWARE SOF
JOIN SOFTWARE_LANGUAGES SL
ON (SL.SOF_SOF_ID = SOF.SOF_ID)
JOIN LANGUAGES LAN
ON (SL.LAN_LAN_ID = LAN.LAN_ID)
JOIN SOFTWARE_DEVELOPMENT_CYCLE SDC
ON (SDC.SOF_LAN_SOF_LAN_ID = SL.SOF_LAN_ID)
WHERE SDC.SOF_DC_ID IN (SELECT MAX(SDC2.SOF_DC_ID)
FROM SOFTWARE_DEVELOPMENT_CYCLE SDC2
WHERE SDC2.SOF_LAN_SOF_LAN_ID = SL.SOF_LAN_ID)
ORDER BY SOF.VENDOR,
SOF.NAME,
LAN.LANGUAGE,
SOF.VERSION;
您可以这样做,以避免第二次时碰到SOFTWARE_DEVELOPMENT_CYCLE
表
SELECT vendor,
name,
language,
version,
status,
sof_dc_id
FROM (SELECT SOF.VENDOR,
SOF.NAME,
LAN.LANGUAGE,
SOF.VERSION,
SDC.STATUS,
SDC.SOF_DC_ID,
RANK() OVER (PARTITION BY sl.sdf_lan_id
ORDER BY sdc.sdf_dc_id DESC) rnk
FROM SOFTWARE SOF
JOIN SOFTWARE_LANGUAGES SL
ON (SL.SOF_SOF_ID = SOF.SOF_ID)
JOIN LANGUAGES LAN
ON (SL.LAN_LAN_ID = LAN.LAN_ID)
JOIN SOFTWARE_DEVELOPMENT_CYCLE SDC
ON (SDC.SOF_LAN_SOF_LAN_ID = SL.SOF_LAN_ID))
WHERE rnk = 1
ORDER BY VENDOR,
NAME,
LANGUAGE,
VERSION;
CCD_ 2分析函数将结果集划分为CCD_ 3。然后,对于每个不同的sl.sdf_lan_id
,它将根据sdc.sdf_dc_id
的降序为行分配一个数字秩。这意味着对于特定sl.sdf_lan_id
具有最大sdc.sdf_dc_id
的行将具有1的RANK
。外部WHERE rnk=1
谓词然后只选择具有该最大值的行。这应该完成与MAX
子查询相同的事情。