問(wèn)題描述

問(wèn)題解答

其實(shí)最好寫(xiě)明你的表結(jié)構(gòu)，以下答案基于你提供的有限信息：

select district as 行政區(qū)，count(1) as 小區(qū)數(shù) -- 我默認(rèn)你每個(gè)小區(qū)時(shí)一條記錄，且無(wú)重復(fù), sum(if(idNB = 1 ,1 ,0)) as 高檔小區(qū)數(shù) -- 假設(shè)高檔小區(qū)的idNB標(biāo)記為1from table_name group by district其實(shí) sum(if(idNB = 1 ,1 ,0)) 也可以替換成count(idNB = 1 or null)回答2：

mysql不支持分析函數(shù):

select t1.district, (select count(t2.xiaoqu) from table t2 where t2.district=t1.district) count_xiaoqu, (select count(t2.idNB) from table t2 where t2.district=t1.district) count_idNBfrom table t1

分析函數(shù)的寫(xiě)法:

select district, count(xiaoqu) over (district) count_xiaoqu, count(idNB) over (district) count_idNBfrom table回答3：

我這邊說(shuō)下我的思路吧，使用MySQL將區(qū)內(nèi)的高端小區(qū)和非高端小區(qū)統(tǒng)計(jì)出來(lái)

select district,idNB,count(*) from xx GROUP BY district,idNB

然后區(qū)內(nèi)小區(qū)的總數(shù)再由服務(wù)端這邊自己處理計(jì)算。