我有以下数据。我想在ID上分区,并且只返回基于Date描述的唯一OrgIDs。 对于ID 14,根据Date在输出中选择Encompass和Premier。对于ID 57,选择Encompass、Example和Premier是因为Example属于独特的OrgID类别。 ID Num Dx Code Describe OrgID Nm1 Name Type Date
我正在尝试一些具有分析函数的示例,并且我已经创建了一个sql来理解count distinct over partition by clause.This是我的。'xyz','abc3'); value2,
count(distinctvalue2) over (partition by value1)
Select distinct CONCAT(User_Name, DATALENGTH(User_Name)) as User_name, Partition, Game, Bet, Num_spins-17' and Bet>0 and credits is not nullSelect datepart(week,
Table2 asCOUNT(DISTINCTCLAIM_ID) OVER (PARTITION BY NPI) AS TOTAL_CLAIMS,
COUNT(DISTINCT CLAIM_ID) OVER (PARTITION BY NPI)where APPROVAL_FLAG = 'APPROVED' as APPROVED_
有没有可能结合OVER(PARTITION BY id)这样的窗口函数来计算不同的值?BY congestion.id_elementCOUNT(DISTINCT congestion.week_nb) OVER(PARTITION BY congestion.id_elementFROM congestionAND congestion.d
我想数数过去30天的distinct day_number。然而,不同的函数不能与over一起使用 count( distinct day_number) OVER (PARTITION BY ACCOUNT ORDER BY DAY_number range 29 PRECEDING) as result
from table tr
FORMAT DELIMITED FIELDS TERMINATED BY '\t' STORED AS TEXTFILE;INSERT OVERWRITE TABLE user PARTITION(logindate=20130120) SELECT DISTINCT(id) FROM a WHERE logindate=20130120;
INSERT OVERWRITE TABLE userPARTITION(logindate=20130121) SELECT DISTINCT<