問(wèn)題描述

原始數(shù)據(jù)表如下

idnamemoneytime1mike62016-09-01 12:58:002mike102016-09-01 13:52:563leo102016-09-02 00:05:054mike62016-09-03 08:06:05

希望轉(zhuǎn)制后的數(shù)據(jù)表如下

name2016-09-012016-09-022016-09-03mike1606leo0100

以demo為例，可以說(shuō)說(shuō)這類(lèi)問(wèn)題的解決思路嗎？比如統(tǒng)計(jì)兩個(gè)月的數(shù)據(jù)，那么就會(huì)有60行轉(zhuǎn)為60列，是否存在效率的瓶頸

問(wèn)題解答

回答1：

select sum(money) s,name,date_format(time,’%Y-%m-%d’) d from table group by name,d;然后根據(jù)這里轉(zhuǎn)換

回答2：

非要用sql做的話，得寫(xiě)存儲(chǔ)過(guò)程，然后創(chuàng)建個(gè)臨時(shí)表了。沒(méi)必要，把數(shù)據(jù)取出來(lái)后用服務(wù)端去做吧。效率高很多。

回答3：

create table tt1(id int ,name VARCHAR(100),money FLOAT,time DATETIME)insert into tt1select 1,’mike’,6,’2016-09-01 12:58:00’ UNIONselect 2,’mike’,10,’2016-09-01 13:52:56’ UNIONselect 3,’leo’,10,’2016-09-02 00:05:05’ UNIONselect 4,’mike’,6,’2016-09-03 08:06:05’select name,sum(case when datediff(time,’2016-09-01’)=0 then money else 0 end) as ’2016-09-01’ ,sum(case when datediff(time,’2016-09-02’)=0 then money else 0 end) as ’2016-09-02’ ,sum(case when datediff(time,’2016-09-03’)=0 then money else 0 end) as ’2016-09-03’from tt1 GROUP BY name回答4：

@ch21st 給的就是典型的行轉(zhuǎn)列SQL寫(xiě)法因?yàn)橹恍枰頀呙枰淮危虼艘话闱闆r下不存在性能問(wèn)題，除非是特別大的表。

這種寫(xiě)法本身只是一個(gè)解決思路，如果列不固定的話，可以在應(yīng)用端通過(guò)php或java動(dòng)態(tài)生成sql語(yǔ)句。

回答5：

寫(xiě)個(gè)生成腳本的sql就完了嘛

select 'select name,' union select concat('sum(case when datediff(time,’',date(time),'’)=0 then money else 0 end) as ’',date(time),'’,') a from tt1 group by a union select ' from tt1 group by name;';PS：union中間段最后一個(gè)逗號(hào)記得去掉回答6：

SELECT name, sum( if(date_format(time,’%Y-%m-%d’) = ’2016-09-01’, money, 0 ) ) AS ’2016-09-01’, sum( if(date_format(time,’%Y-%m-%d’) = ’2016-09-02’, money, 0 ) ) AS ’2016-09-02’, sum( if(date_format(time,’%Y-%m-%d’) = ’2016-09-03’, money, 0 ) ) AS ’2016-09-03’FROM money_table GROUP BY name;