我有一张包含各种信息的表格,我需要选择以下值:
1)有cod_anag_prov = 0或= 2
2)具有计数(1)> 1
然后为每个与第1点和第2点相关的单个记录设置一个标志为1,并在所有事件中都有最小计数(1)。
我想到了使用dense_rank函数并编写了这个:
SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
MAX_CNT,
MIN(MAX_CNT) KEEP (DENSE_RANK FIRST ORDER BY MAX_CNT) OVER (PARTITION BY PDRA) OCCORRENZA_MINORE
FROM
(SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
COUNT(1) AS MAX_CNT
FROM STM_VOLUME_AGGR
WHERE (COD_ANAGR_PROV = 0
OR COD_ANAGR_PROV = 2)
GROUP BY PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF
HAVING COUNT(1)>1
ORDER BY PDRA);
到目前为止(我认为)我已经能够做到我之前说的话。
现在,如果我有这样的结果:
34624200 1905 201305 6 6
34624200 83 201305 13 6
34624200 93 201305 14 6
34439201 1 201305 11 2
34439201 6 201305 2 2
我希望将行的标志设置为1:
34624200 1905 201305 6 6
34439201 6 201305 2 2
我怎么能这样做?!
我知道我做的事情要复杂得多,但现在我的大脑正在融化xD(我对SQL很新)...
更新1:
好的,我已经完成了,但我当然需要优化它。费用为3.300.000:S
这是我的解决方案:
SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
MIN(MAX_CNT),
NUMERO_OCCORRENZE FROM
(SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
MAX_CNT,
MIN(MAX_CNT) KEEP (DENSE_RANK FIRST ORDER BY MAX_CNT) OVER (PARTITION BY PDRA) NUMERO_OCCORRENZE
FROM
(SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
COUNT(1) AS MAX_CNT
FROM STM_VOLUME_AGGR
WHERE (COD_ANAGR_PROV = 0
OR COD_ANAGR_PROV = 2)
GROUP BY PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF
HAVING COUNT(1)>1
ORDER BY PDRA))
GROUP BY
PDRA, COD_DISTRIBUTORE_STARGAS, ANNOMESE_RIF, NUMERO_OCCORRENZE
HAVING MIN(MAX_CNT)=NUMERO_OCCORRENZE
;
答案 0 :(得分:0)
您的查询需要简化,但是, 快速获胜;
SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
MAX_CNT,
row_number() over(partititon by PDRA order by MAX_CNT) rank_id
FROM
(SELECT PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF,
COUNT(1) AS MAX_CNT
FROM STM_VOLUME_AGGR
WHERE (COD_ANAGR_PROV = 0
OR COD_ANAGR_PROV = 2)
GROUP BY PDRA,
COD_DISTRIBUTORE_STARGAS,
ANNOMESE_RIF
HAVING COUNT(1)>1
ORDER BY PDRA)
答案 1 :(得分:0)
如果我理解这种情况发生,当OCCORRENZA_MINORE = MAX_CNT:
UPDATE STM_VOLUME_AGGR
SET flag = 1
WHERE (PDRA, COD_DISTRIBUTORE_STARGAS, ANNOMESE_RIF) IN
(SELECT PDRA, COD_DISTRIBUTORE_STARGAS, ANNOMESE_RIF
FROM ( your_query )
WHERE OCCORRENZA_MINORE = MAX_CNT)
AND (COD_ANAGR_PROV = 0 OR COD_ANAGR_PROV = 2)
答案 2 :(得分:0)
前面提到的一个组或另一个你通过" case"填充的列。以及查询的最后两列之间的比较(大小写等于1然后0结束)。