postgre——case、union、小計總計（GROUP BY ROLLUP）寫法

阿新 • • 發佈：2019-01-14

//將幾張不同表結構的資料全部合併在一起，使用了case、union，同時實現小計總計

//小計總計使用的 GROUP BY ROLLUP函式，GROUP BY ROLLUP使用雙括號將欄位括起來，是將這些欄位分為一組，只會產生一個總計

select case when t_all.temp_comm is null then '總計' else t_all.temp_comm end as temp_comm,
commodity_name,guige,t_bi_commbaseinfo.comm_type_id,comm_type_name,t_all.temp_bill,t_all.temp_date,
sum(temp_inCount) as temp_inCount,temp_inPrice as temp_inPrice,
sum(temp_inCount*temp_inPrice) as temp_inTotal,
sum(temp_inCount*temp_inPrice*(1-temp_inTax)) as temp_inTatolTax,
sum(temp_inCount*temp_inTax*temp_inPrice) as temp_inTaxTatal,
sum(temp_outCount) as temp_outCount,temp_outPrice as temp_outPrice,
sum(temp_outCount*temp_outPrice) as temp_outTotal,
sum(temp_retailPrice*temp_inCount) as temp_inSaleTotal,
sum((temp_retailPrice-temp_inPrice)*temp_inCount) as temp_outDivInTotal,
sum(temp_inCount-temp_outCount) as temp_outDivIn,
sum(temp_inCount*temp_inPrice) as temp_divTotal
from ((SELECT receipt_num AS temp_bill,shd.commodity_id as temp_comm,sh.check_date AS temp_date,
				(CASE WHEN shd.order_real_inprice IS NOT NULL THEN shd.order_real_inprice ELSE 0 END) AS temp_inPrice,0 as temp_outPrice,
				(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
				(CASE WHEN shd.in_tax IS NOT NULL THEN shd.in_tax ELSE 0 END) AS temp_inTax,
				(case when shd.real_rece_count is not null then shd.real_rece_count else 0 end) as temp_inCount,0 as temp_outCount
			FROM t_bi_receivedetail AS shd LEFT JOIN t_bi_receive AS sh ON sh.receipt_id = shd.receipt_num
			LEFT JOIN t_bi_commbaseinfo on shd.commodity_id=t_bi_commbaseinfo.commodity_id
			WHERE sh.check_date BETWEEN '2015/7/1' AND '2016/12/10' AND receipt_status = '2'
		)UNION(SELECT thd.return_com_receipt_num AS temp_bill,thd.commodity_id as temp_comm,th.check_date AS temp_date,
					0 AS temp_inPrice,(CASE WHEN thd.return_com_price IS NOT NULL THEN thd.return_com_price ELSE 0 END) AS temp_outPrice,
					(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
					0 AS temp_inTax,0 as temp_inCount,(case when thd.return_com_count is not null then return_com_count else 0 end) as temp_outCount  
				FROM t_bi_returncommdetail AS thd LEFT JOIN t_bi_returncomm AS th ON thd.return_com_receipt_num = th.return_com_receipt_num
				LEFT JOIN t_bi_commbaseinfo on thd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE th.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND th.status = '2')
		UNION(SELECT lyd.entertain_recei_num AS temp_bill,lyd.commodity_id as temp_comm,ly.check_date AS temp_date,0 as temp_inPrice,
					(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
					0 as temp_outPrice,0 as temp_inTax,0 as temp_inCount,0 as temp_outCount
				FROM t_bi_entaindetail AS lyd LEFT JOIN t_bi_entainment AS ly ON lyd.entertain_recei_num = ly.entertain_recei_num
				LEFT JOIN t_bi_commbaseinfo on lyd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE ly.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND ly.status IN ('2', '3')
			)UNION(SELECT bsd.break_rece_num AS temp_bill,bsd.commodity_id as temp_comm,bs.check_date AS temp_date,0 as temp_inPrice,0 as temp_outPrice,
					(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
					0 as temp_inTax,0 as temp_inCount,0 as temp_outCount
				FROM t_bi_breakdetail AS bsd LEFT JOIN t_bi_break AS bs ON bsd.break_rece_num = bs.break_rece_num
				LEFT JOIN t_bi_commbaseinfo on bsd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE bs.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND bs.status = '2'
			)UNION(SELECT fcd.back_reciept_num AS temp_bill,fcd.commodity_id as temp_comm,fc.check_date AS temp_date,0 as temp_inPrice,0 as temp_outPrice,
					(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
					0 as temp_inTax,0 as temp_inCount,0 as temp_outCount
				FROM t_bi_returnwaredetail AS fcd LEFT JOIN t_bi_returnwarehose AS fc ON fcd.back_reciept_num = fc.back_reciept_num
				LEFT JOIN t_bi_commbaseinfo on fcd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE fc.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND fc.status IN ('2', '3', '4')
			)UNION(SELECT zdd.deliever_rece_num AS temp_bill,zdd.commodity_id as temp_comm,zd.check_date AS temp_date,0 as temp_inPrice,0 as temp_outPrice,
					(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
					0 as temp_inTax,0 as temp_inCount,0 as temp_outCount
				FROM t_bi_deliverdetail AS zdd LEFT JOIN t_bi_deliver AS zd ON zdd.deliever_rece_num = zd.deliever_rece_num
				LEFT JOIN t_bi_commbaseinfo on zdd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE zd.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND zd.status IN ('5', '3', '4')
			)UNION(SELECT tgd.group_reciept AS temp_bill,tgd.commodity_id as temp_comm,tg.check_date AS temp_date,0 as temp_inPrice,(case when tgd.groupbuy_price is not null then tgd.groupbuy_price else 0 end) as temp_outPrice,
				(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,	
				0 as temp_inTax,0 as temp_inCount,(case when tgd.groupbuy_count is not null then tgd.groupbuy_count else 0 end) as temp_outCount
				FROM t_bi_hqgroupdetail AS tgd LEFT JOIN t_bi_hqgroup AS tg ON tgd.group_reciept = tg.group_reciept
				LEFT JOIN t_bi_commbaseinfo on tgd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE tg.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND tg.status IN ('2', '3', '4')
			)UNION(SELECT ttgd.group_return_reciept AS temp_bill,ttgd.commodity_id as temp_comm,ttg.check_date AS temp_date,(case when ttgd.groupbuy_price is not null then ttgd.groupbuy_price else 0 end) as temp_inPrice,
				(case when t_bi_commbaseinfo.retail_price is not null then t_bi_commbaseinfo.retail_price else 0 end) AS temp_retailPrice,
				0 as temp_outPrice,0 as temp_inTax,(case when ttgd.groupbuy_count is not null then ttgd.groupbuy_count else 0 end) as temp_inCount,
				0 as temp_outCount
				FROM t_bi_returnhqgroupdetail AS ttgd LEFT JOIN t_bi_returnhqgroup AS ttg ON ttg.group_return_reciept = ttgd.group_return_reciept
				LEFT JOIN t_bi_commbaseinfo on ttgd.commodity_id=t_bi_commbaseinfo.commodity_id
				WHERE ttg.check_date BETWEEN '2015/6/1' AND '2016/12/10' AND ttg.status IN ('2', '3', '4')) 
) as t_all
LEFT JOIN t_bi_commbaseinfo on t_bi_commbaseinfo.commodity_id=t_all.temp_comm
LEFT JOIN t_bi_commtype on t_bi_commtype.comm_type_id=t_bi_commbaseinfo.comm_type_id
GROUP BY ROLLUP((t_all.temp_comm,commodity_name,guige,t_bi_commbaseinfo.comm_type_id,comm_type_name,t_all.temp_bill,t_all.temp_date,
 temp_inPrice,temp_outPrice))
order by temp_comm

postgre——case、union、小計總計（GROUP BY ROLLUP）寫法

//將幾張不同表結構的資料全部合併在一起，使用了case、union，同時實現小計總計//小計總計使用的 GROUP BY ROLLUP函式，GROUP BY ROLLUP使用雙括號將欄位括起來，是將這些欄位分為一組，只會產生一個總計select case when t_a

GridView自動求和、求平均值小計

轉自清清月兒gridview72絕技，略有改動前臺：唯一的花頭就是設定ShowFooter="True" private double sum = 0;//取指定列的資料和，你要根據具體情況對待可能你要處理的是int protected void GridView1_R

46、求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）。

closed else while spl 判斷語句 stat 條件執行 ret 思路：循環或者遞歸都有個結束條件和執行條件。用&&短路與代替。 //短路與&&；就是只有前一個條件滿足才可以去判斷第二個條件。 //遞歸的出口

namespace、struct、enum、union、string（day01）

mic str 歷史數組 hello 科學練習 line 空字符串一 C++概述 1 C++歷史背景 1）C++的江湖地位 jave C C++ C# python 2）C++之父：Bjarne Stroustrup(1950--) 1979,Cpre,為C語言

結構體、聯合體、類：struct、union、class

區別：類的成員預設是私有的private &n

【類和物件】求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）

題目描述：求1+2+3+...+n，要求不能使用乘除法（數學公式)、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）. 分析：這道題雖然常用的方式都給禁掉了，但是可以使用我們c++學過的類和物件，通過使用類和靜態來完成。常規情況下，從1+到n，

劍指offer-求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）

題目如下：求1+2+3+…+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）本題考的知識點就兩點 && 運算子 false && true =>

Leetcode演算法——64、最小路徑之和（minimum path sum）

給定一個 m*n 的非負整數矩陣，找到一條路徑，從左上角到右下角，要求這條路徑上的所有陣列之和最小。備註：每次只能向下或向右移動。 Input: [ [1,3,1], [1,5,1], [4,2,1] ] Output: 7 Explanation: Because

題目描述求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）。

//求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字 //及條件判斷語句（A?B:C）。 public class Solution { public int Sum_Solution(int n) { //if

劍指offer---47求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）

求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）。 public class Solution { public int Sum_Solution(int n) { // if(

C++實現：求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）

程式碼： class Solution { public: class Sum { public: Sum() { s_count++; s_sum += s_count;

求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）。Java實現

前幾天跟同學交流聽說一個比較有意思的oj題目，具體描述如下：題目描述：求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）。輸入：輸入可能包含多個測試樣例。對於每個測試案例，輸

最小的k個數、第k小的數（利用快排，堆排序）

快排：（若求top M個元素，則只需把下面的程式裡的k換乘n-M即可） //利用快排求最小的k個數，第k小的數 void GetLeastKNum(int *input,int n,int *output,int k) { if(input==NU

求1+2+3+...+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）（劍指offer）

今天在牛客網上遇到這樣一道題目，求1+2+3+…+n，要求不能使用乘除法、for、while、if、else、switch、case等關鍵字及條件判斷語句（A?B:C）一般大家在看到這種題目的時候是完全沒有頭緒的，但是我們可以從多方面進行思考，首先我們應該想

購物車實現清空小計總計

<!DOCTYPE html><html><head><meta charset="UTF-8"><title></title><script src="../js/angular.min.js"

關於hive 子查詢、union 、left join

建表語句： create table tb_in_base ( id bigint, devid bigint, devname string ) partitioned by (j

java題目：求 1+2+...+n,要求不能用除法、 for、 while、 if、 else、 switch、 case 等關鍵字及條件判斷語句（ A?B:C）。

這是劍指offer的46道面試題，由於面試官諸多條件的限制，本文采用遞迴的手法完成程式碼 public class Sum { public static void main(String[] args) { System.out

【離散數學】最大元素、最小元素、極大元素、極小元素、上界、下界、最小上界（上確界）、最大下界（下确界）

設(A, ≤)是一偏序集合，B是A的子集。最大元素、最小元素：(1)元素b∈B是B的最大元素，如果對每一元素x∈B，x≤b(2)元素b∈B是B的最小元素，如果對每一元素x∈B，b≤x即：對於每一

最長連續公共子串、最長公共子串（可以非連續）、最長回文串（連續）、最長回文串（可以不連續）、最長遞增數組的求解

鑲嵌 wid 方法數量子串進行遞增動態動態規劃問題：最長連續公共子串、最長公共子串（可以非連續）、最長回文串（連續）、最長回文串（可以不連續）、最長遞增數組、長方形鑲嵌最多的求解方法：上述問題有相似性，都可以采用動態規劃進行求解。（1）最長連續公共子串：

what's the python之可叠代對象、叠代器與生成器（附面試題）

urn for循環生成器出現擁有移動 iterable 文本協議可叠代對象字符串、列表、元祖、集合、字典都是可叠代的，數字是不可叠代的。（可以用for循環遍歷取出內部元素的就是可叠代的）如何查看一個變量是否為可叠代： from collections im

postgre——case、union、小計總計（GROUP BY ROLLUP）寫法

相關推薦