Mysql错误的计算

我修复了最后一个问题，但现在有其他lil位：我edt我的查询以这种格式。但是，我仍然有一个大问题 - PU_1天& PU_7day计算不正确：在每行中增加“1”（true时为3，当true为1时为1）。我如何解决这个问题？ :(Mysql错误的计算

SELECT u.date, 
     u.des_channel, 
     u.des_type, 
     u.country, 
     count(distinct(u.id)) as Reg_n, 
     sum(if((u.date=bp.date)*(u.id=bp.user_id),bp.o_outcome,0)) as P_$, 
     sum(if((u.date=bp.date)*(u.id=bp.user_id),1,0)) as P_n, 
     count(distinct(if((u.date=bp.date)*(u.id=bp.user_id),bp.user_id,0))) as PU_1day, 
     count(distinct(if(truncate(timestampdiff(hour,u.datetime,bp.datetime)/24,0)<7,bp.user_id,0))) as PU_7day 
FROM mayadata.users u 
left join mayadata.billing_pays bp 
     on u.id=bp.user_id 
WHERE u.country in ('TH','ZA','ID','IN','NG','MY') and 
     truncate(timestampdiff(hour,u.datetime,bp.datetime)/24,0)<7 and 
     u.date>='2011-01-01' 
GROUP BY u.date, u.des_channel, u.des_type

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2016-01-20 Ruslan Podgorny

减去一个我的意思.. 。它会计算'0'的值作为一个独特的值，你也可以尝试用'NULL'来切换你的'0'，看看是否有帮助，我不记得MySQL是否将NULL计数为一个不同的值。我知道在其他RDBMS中，NULL不被视为“DISTINCT”计数。 – JNevill

根据JNevill ... COUNT（）将不包含NULL。你可以COUNT（IF（无论，1，NULL））... – wally

哦，这些家伙，听起来很不错，希望它的作品！ –

你可以尝试在PU_1day & PU_7day计算表达式从更换的第二个参数空象下面这样：从计算

SELECT u.date, 
     u.des_channel, 
     u.des_type, 
     u.country, 
     count(distinct(u.id)) as Reg_n, 
     sum(if((u.date=bp.date)*(u.id=bp.user_id),bp.o_outcome,0)) as P_$, 
     sum(if((u.date=bp.date)*(u.id=bp.user_id),1,0)) as P_n, 
     count(distinct(if((u.date=bp.date)*(u.id=bp.user_id),bp.user_id,null))) as PU_1day, 
     count(distinct(if(truncate(timestampdiff(hour,u.datetime,bp.datetime)/24,0)<7,bp.user_id,null))) as PU_7day 
FROM mayadata.users u 
left join mayadata.billing_pays bp 
     on u.id=bp.user_id 
WHERE u.country in ('TH','ZA','ID','IN','NG','MY') and 
     truncate(timestampdiff(hour,u.datetime,bp.datetime)/24,0)<7 and 
     u.date>='2011-01-01' 
GROUP BY u.date, u.des_channel, u.des_type

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2016-01-20 14:01:58

雅，其工作，谢谢！ –

Mysql错误的计算

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