1

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，以便不使用子查询。我用 DENSE_RANK LAST ORDERY BY 尝试过，但无济于事。

有人可以建议吗？谢谢你。

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;
4

1 回答 1

2

你可以做这样的事情来避免SOFTWARE_DEVELOPMENT_CYCLE第二次打桌子

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;

分析函数将RANK结果集划分为sl.sdf_lan_id。然后对于每个不同的sl.sdf_lan_id，它根据 的降序为该行分配一个数字排名sdc.sdf_dc_id。这意味着sdc.sdf_dc_id对于特定的具有最大的sl.sdf_lan_id行将具有RANK1。然后，外部WHERE rnk=1谓词仅选择具有该最大值的行。MAX这应该完成与您的子查询相同的事情。

于 2012-02-23T16:39:41.673 回答