如何计算每天的总和 [英] How to calculate the sum per day

问题描述

HI



我目前正在为我的客户创建报告，并想知道如何计算用户处理的记录数量每天。





我的桌子是Stage1





id（int），product（Nvarchar），date（datetime），userid（int），stage（int）



我的查询是什么选择用户输入的产品数量？



i需要获得用户每天通过的产品数量



选择Sum（userid），Workcompleted from Table group by workcompleted

HI

I am currently creating a report for my client and would like to know how would i be able to calculate the amount of records processed by a user for each day.


My table is Stage1


id(int),product(Nvarchar),date(datetime),userid(int),stage(int)

what would my query be to select the amount of products entered by a user?

i need to get the count of products passed per day by a user

Select Sum(userid),Workcompleted from Table group by workcompleted

推荐答案

Select Sum(Workcompleted ) from Table where userid = @userid 

会做它，假设workcompleted是r的数量ecords处理



如果你只是想记录那么记录

would do it, assuming workcompleted is the number of records processed

if you just want to counmt the records then

select count(*) from Table where userId = @userId

尝试下面的T-SQL代码：

try below T-SQL code :

CREATE TABLE #TEMP
(
	ITEM	VARCHAR(50),
	AMOUNT	INT,
	INDATE  DATETIME,
	USERID	INT
)

INSERT INTO #TEMP VALUES('WATCH',500,'02/13/2013',1)
INSERT INTO #TEMP VALUES('BELT',300,'02/13/2013',1)
INSERT INTO #TEMP VALUES('WALLET',600,'02/13/2013',2)
INSERT INTO #TEMP VALUES('TROUSERS',1600,'02/13/2013',2)

SELECT USERID,INDATE,SUM(AMOUNT) AS TOTALAMOUNT
FROM #TEMP
GROUP BY USERID,INDATE

DROP TABLE #TEMP

选择sum（amt），来自tablex的id，其中date =''2012-10-10''和id = 1 group by id

select sum(amt),id from tablex where date=''2012-10-10'' and id= 1 group by id

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