| 字段 | 描述 |
|---|---|
| id | 主鍵 |
| date | 日期 |
| product | 產(chǎn)品 |
| channel | 渠道 |
| amount | 銷售額 |
表數(shù)據(jù)
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051726328010100000', '2021-05-04', '產(chǎn)品1', '京東', '8899.0000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051726328010100001', '2021-05-04', '產(chǎn)品2', '京東', '99.0000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100010', '2021-05-04', '產(chǎn)品1', '天貓', '230.0000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100011', '2021-05-04', '產(chǎn)品2', '天貓', '9.9000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100011', '2021-05-04', '產(chǎn)品3', '線下門店', '10.1000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100000', '2021-05-04', '產(chǎn)品1', '其它', '10');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100099', '2021-05-04', '產(chǎn)品2', '其它', '20000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100033', '2021-05-01', '產(chǎn)品1', '其它', '20000');
INSERT INTO "report1"("id", "date", "product", "channel", "amount") VALUES ('2105051727068010100044', '2021-05-01', '產(chǎn)品3', '線下門店', '12345');
如果你看到這裏請稍稍思考下，一開篇我説過我們的數(shù)據(jù)是按 日期|產(chǎn)品|渠道|銷售額 這樣按行存儲的，以上截圖大家一看就懂，然後再看看開篇的報表截圖，我想大家可以同我一樣可以分析出以下幾點：
一部分是前一日產(chǎn)品銷售明細
然後一部分是前一日產(chǎn)品渠道產(chǎn)品合計
最後一部分是按渠道做的月統(tǒng)計
一部分是前一日的數(shù)據(jù)
另一部分則是月份匯總數(shù)據(jù)
最後一部分則是所有渠道的產(chǎn)品合計、日合計、月合計
好了，問題來了，如何做呢，我是這麼想的：首先要很清楚的是你的sql大致分兩大部分(兩個子查詢)
一部分是前一日的數(shù)據(jù)另一部分則是月份匯總數(shù)據(jù)
最後需要將兩部分數(shù)據(jù)做聯(lián)表查詢，這樣太贊了，似乎完成了報表的80%，至於最後一行的求總，這裏先賣個關(guān)子哈~
第一部分數(shù)據(jù)(前一日的數(shù)據(jù))
我想我們立馬能做的第一部分sql恐怕就是行專列吧(似乎這是最容易實現(xiàn)的😄)
select channel, sum(case product when '產(chǎn)品1' then amount end) as c1, sum(case product when '產(chǎn)品2' then amount end) as c2, sum(case product when '產(chǎn)品3' then amount end) as c3 from report1 group by channel ;
sql似乎沒什麼問題，但是我們少了一列，對那就是按渠道日合計，當然如果您對postgresql窗口函數(shù)熟悉的話，這裏實現(xiàn)的方式估計你已經(jīng)猜到了(窗口over函數(shù)),上sql...
select
channel,
day_sum,
sum(case product when '產(chǎn)品1' then amount end) as c1,
sum(case product when '產(chǎn)品2' then amount end) as c2,
sum(case product when '產(chǎn)品3' then amount end) as c3
from
( select *,sum(amount) over (partition by channel) as day_sum from report1 where date=to_date('2021-05-04','yyyy-MM-dd') ) as t1
group by t1.channel ,t1.day_sum;
哈哈，上圖的day_sum估計大家很熟悉了吧，哈哈哈~
看來已經(jīng)成功地完成了日數(shù)據(jù)部分，這裏可能的難點可能就兩點
sum)+分組(group by)做行專列(當然postgresql也有其他很好用的行專列擴展，這裏就不介紹啦~)另一個是使用窗口函數(shù)(over)對明細提前做 按渠道的窗口匯總，這樣渠道日合計(行）的數(shù)據(jù)就有啦~想想是不是很容易😂，接下來我們看看第二部分數(shù)據(jù)怎麼獲取~
第二部分數(shù)據(jù)(月份匯總數(shù)據(jù))
月份匯總的數(shù)據(jù)看似簡單的可怕，如果您熟練掌握postgresql中的日期處理的話估計分分鐘就能搞定，這裏就不耍大刀了，直接放出sql，哈哈哈😄
select
channel,sum(amount) as month_sum from report1
where
date>=date(date_trunc('month',to_date('2021-05-04','yyyy-MM-dd'))) and date date(date_trunc('month',to_date('2021-05-04','yyyy-MM-dd')) + '1 month')
group by
channel
報表數(shù)據(jù)最終求解
現(xiàn)在，我們將求解的兩部分數(shù)據(jù)按渠道channel字段做inner join合并以上兩部分數(shù)據(jù)，合并后的數(shù)據(jù)大致是這樣子的
這個是sql
select
ttt.channel,
sum(ttt.day_sum) as day_sum,
sum(ttt.month_sum) as month_sum,
sum(ttt.c1) as c1,
sum(ttt.c2) as c2,
sum(ttt.c3) as c3
from (
select tt1.*,tt2.month_sum from
(
select
channel,
day_sum,
sum(case product when '產(chǎn)品1' then amount end) as c1,
sum(case product when '產(chǎn)品2' then amount end) as c2,
sum(case product when '產(chǎn)品3' then amount end) as c3
from
( select *,sum(amount) over (partition by channel) as day_sum from report1 where date=to_date('2021-05-04','yyyy-MM-dd') ) as t1
group by t1.channel ,t1.day_sum
) as tt1 left join
(
select channel,sum(amount) as month_sum from report1 where date>=date(date_trunc('month',to_date('2021-05-04','yyyy-MM-dd'))) and date date(date_trunc('month',to_date('2021-05-04','yyyy-MM-dd')) + '1 month') group by channel
) as tt2 on tt1.channel = tt2.channel
) ttt
GROUP BY ttt.channel
order by channel asc
看，匯總的數(shù)據(jù)已經(jīng)有了,已經(jīng)可以算作是最終結(jié)果了（如果你需要報表系統(tǒng)來計算匯總行數(shù)據(jù)的話），當然 ，我們的報表系統(tǒng)過於繁瑣(不是不能做，而是太麻煩)，需要你將做好的菜喂給它吃，這時，該怎麼辦呢。。。，哈哈哈 我們似乎忘記了很久不用的rollup函數(shù)(一開始我也沒發(fā)現(xiàn)有這麼個函數(shù)哈哈)，試試看吧
select
ttt.channel,
sum(ttt.day_sum) as day_sum,
sum(ttt.month_sum) as month_sum,
sum(ttt.c1) as c1,
sum(ttt.c2) as c2,
sum(ttt.c3) as c3
from (
select tt1.*,tt2.month_sum from
(
select
channel,
day_sum,
sum(case product when '產(chǎn)品1' then amount end) as c1,
sum(case product when '產(chǎn)品2' then amount end) as c2,
sum(case product when '產(chǎn)品3' then amount end) as c3
from
( select *,sum(amount) over (partition by channel) as day_sum from report1 where date=to_date('2021-05-04','yyyy-MM-dd') ) as t1
group by t1.channel ,t1.day_sum
) as tt1 left join
(
select channel,sum(amount) as month_sum from report1 where date>=date(date_trunc('month',to_date('2021-05-04','yyyy-MM-dd'))) and date date(date_trunc('month',to_date('2021-05-04','yyyy-MM-dd')) + '1 month') group by channel
) as tt2 on tt1.channel = tt2.channel
) ttt
group by rollup(ttt.channel)
order by channel asc
數(shù)是對的，意味著我們成功了~😂
如果您肯下功夫?qū)W，postgresql世界有很多精彩的東西，當然也有一些東西對比mysql顯得繁瑣些，不過本著學(xué)習(xí)的心態(tài)，我們縂能剋服這些，同時我們還是能做出超出我們自身能力範疇的東西的，哈哈，各位加油哦~
下章，我將講一講如何實現(xiàn)通過sql實現(xiàn)前端合并單元格的效果，是不是很神奇(我保證你全網(wǎng)搜不到), 希望不翻車，哈哈哈~
到此這篇關(guān)于postgresql高級應(yīng)用之行轉(zhuǎn)列匯總求和的實現(xiàn)思路的文章就介紹到這了,更多相關(guān)postgresql行轉(zhuǎn)列匯總求和內(nèi)容請搜索腳本之家以前的文章或繼續(xù)瀏覽下面的相關(guān)文章希望大家以后多多支持腳本之家！
