在BigQuery中,count distinct + case when / with group by rollup的问题是指如何使用count distinct函数结合case when语句和group by rollup子句进行数据分析。 首先,count distinct函数用于计算某一列中不重复值的数量。它可以用于统计唯一值的个数,例如统计某个表中不同用户的数...
select date, count(distinct case when type='completed' then email.id else null end )as success, count(distinct case when type='no_completed' then email.id else null end )as fail, round(count(distinct case when type='no_completed' then email.id else null end )/ count(distinct case when...
count(case when )的用法 今天接触到了 count(case when ……)的用法,挺棒的。 表数据: 编号 性别 语句: SELECTCOUNT(DISTINCTtel)gender_count, COUNT(DISTINCTCASEWHENgender='male'THENtelEND)male_count, COUNT(DISTINCTCASEWHENgender='female'THENtelEND)female_countFROMexample_dataset http://stackoverflow...
SELECT count(distinct CASE WHEN IS_ARRIVED = 1 THEN BUYER_NICK ELSE NULL END) AS BUYER_NICK FROM marketing_summary WHERE (STORE_ID = ‘xxxx’ AND DISPATCH_TIMESTAMP >= ‘2020-09-22 00:00:00.0’ AND DISPATCH_TIMESTAMP <= ‘2020-09-25 23:59:59.0’); 若时间在同一分区中可顺利执行,若...
但是有时往往需要添加不同的条件已经去重的统计以上语句就不能满足需求。 解决方案为: 1.添加条件的统计方案: COUNT(CASEWHEN条件THEN1ELSENULLEND) xxxGROUPBY分组字段 2.添加条件并去重的统计方案: COUNT(DISTINCTCASEWHEN条件THEN去重字段END) xxxGROUPBY分组字段 ...
count(case ……) count(distinct) image.png selecta.购买时间,count(distinct a.用户id)当日首次购买用户数,count(distinctcasewhentimestampdiff(month,a.购买时间,b.购买时间)<=1then a.用户idelsenullend)as此月复购用户数,count(distinctcasewhentimestampdiff(month,a.购买时间,b.购买时间)=3then a.用户...
总是以end以及group by在所有的连接之后必须保留
1）请将 CASE WHEN 逻辑落到表中（字段 A），再在 kylin 中对字段 A 进行 COUNT DISTINCT 聚合函数...
实例2: select count(DISTINCT case when cd.LikeShareUserId = '101148' and cd.IsDelete = 1 then cd.BusinessId end) as DeleteArticleNum, count(case when cd.ToLikeShareUserId = '101148' and cd.LikeShareType = 1 and cd.IsLike=1 and cd.IsDelete = 1 then cd.LikeShareUserId end) AS Chang...
count(distinct a.用户id) as活跃用户数, count(distinct when 时间间隔=1 then 用户id else null end) as 次日留存数, count(distinct when 时间间隔=1 then 用户id else null end) as 次日留存数 / count(distinct a.用户id) as次日留存率,