Question

我有下面的查询。 SOFTWARE_DEVELOPMENT_CYCLE有多行，但我对最新的行感兴趣。

我想重写查询，以便我不使用子查询。我尝试使用DENSE_RANK LAST ORDERY BY，但无济于事。

有人可以提供建议吗？谢谢。

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;

Answer 1

你可以这样做，以避免第二次点击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;

RANK分析函数正在按sl.sdf_lan_id对结果集进行分区。然后，对于每个不同的sl.sdf_lan_id，它会根据sdc.sdf_dc_id的降序为行指定数字排名。这意味着特定sdc.sdf_dc_id具有最大sl.sdf_lan_id的行的RANK为1.外WHERE rnk=1谓词然后仅选择具有该最大值的行。这应该完成与MAX子查询完成相同的事情。

Oracle - 使用max（）重构子查询

1 个答案: