問(wèn)題描述

Table ALogID UserId Date00001 0001 05-0100002 0002 05-0200003 0003 05-0200004 0004 05-0200005 0003 05-0300006 0001 05-03 00007 0002 05-03Table BUserId Status0001 Active0002 Active 0003 Active0004 InactiveTable CUserId Province0001 Yunnan0002 Fujian 0003 Fujian0004 Beijing

以上為數(shù)據(jù)庫(kù)中的三張表，通過(guò)UserID關(guān)聯(lián)。表A為用戶(hù)登錄信息表以LogID為主鍵；表B儲(chǔ)存用戶(hù)活躍狀態(tài)，表C儲(chǔ)存用戶(hù)地理位置信息?，F(xiàn)在想根據(jù)表A中的日期分組得到其他狀態(tài)的數(shù)目累加和，預(yù)期返回結(jié)果為：

Date Active Inactive Yunnan Fujian Beijing05-01 1 0 1 0 0 05-02 2 1 0 2 1 05-03 3 0 1 2 0

能否用一條SQL語(yǔ)句實(shí)現(xiàn)？

問(wèn)題解答

這表業(yè)務(wù)邏輯非常不嚴(yán)密，我也就不嚴(yán)密的給你寫(xiě)一個(gè)了，就當(dāng)你ABC表關(guān)系為多對(duì)一對(duì)一：

select a.date, sum(case when b.status=’Active’ then 1 else 0 end) ’Active’, sum(case when b.status=’Inactive’ then 1 else 0 end) ’Inactive’, sum(case when c.province =’Yunnan’ then 1 else 0 end) ’Yunnan’, sum(case when c.province =’Fujian’ then 1 else 0 end) ’Fujian’, sum(case when c.province =’Beijing’ then 1 else 0 end) ’Beijing’ from a left join b on a.userid=b.user_id join c on a.user_id=c.user_id group by a.date order by a.date;