sql count distinct values

ON SC.[C#]=C.[C#] GROUP BY [S#] HAVING COUNT(*)=COUNT(DISTINCT [S#])) 5. query the number of students who have selected the course -- Implementation code: SELECT students = COUNT (DISTINCT [S #]) FROM SC 6. query the student ID and organization of more than 5 Electiv

1. Select year (ordertime) annual, sum total sales from order form group by year (Ordertime) 2, monthly select years (Ordertime), Month ( ordertime) month, sum (total) Sales totals from order form GROUP by year (Ordertime), Month (Ordertime 3, daily Select year (ordertime), Month ( ordertime) month, day (ordertime), sum (total) Sales totals from order form GROUP by year (Ordertime), month (Ordertime), Day (Ordertime) Also daily can be this: select CONVERT (char (8), ordertime,112) dt,sum (total)

Label: 1. Select Year (ordertime) annual, sum total sales from order Form GROUP by year (Ordertime) 2, monthly Select Year (ordertime), Month (ordertime) months, sum (total) sales totals from order Form Group by year (Ordertime), month (Ordertime 3, daily Select year (ordertime), Month (ordertime), Day (ordertime), sum (total The sales total from order Form Group by year (Ordertime), month (ordertime), Day (ordertime) can also be done in addition daily: Select CONVERT (char (8), ordertime,112

As mentioned on the previous page,COUNT is one of the functions. Because of its wide use, we have made a special offer here to discuss it. Basically,Count allows us to count how many of the data in the table is selected. Its syntax is:Select COUNT ("Field name")From "table name";For example, if we want to find out how

T_ SQLInCountFunction CountFunction Definition visibleMsdn. Definition:Count ({[[All | distinct] expression] | *}) SoCountThere are two ways to useCount (expression)AndCount (*)Returns the value of a column for a table. 1.Count (*)The number of rows in the returned table. It does not filterNullAnd duplicate rows. 2.Count (expression)Will filter outNullV

Tags: style blog http color using AR strong data SP1, AnnualSelect Year(Ordertime) years,sum(total) Sales totals fromOrder FormGroup by Year(ordertime)2, monthlySelect Year(Ordertime) years,Month(ordertime) month,sum(total) Sales totals fromOrder FormGroup by Year(ordertime),Month(Ordertime3, DailySelect Year(Ordertime) years,Month(ordertime) month, Day(Ordertime) Day,sum(total) Sales totals fromOrder FormGroup by Year(ordertime),Month(ordertime), Day(Ordertime) In addition also daily c

records in the table have filtered out the records whose category is 3, which are not included in group processing. Supplement: complete execution sequence of SQL select statements The complete execution sequence of the SQL SELECT statement:1. The from clause assembles data from different data sources;2. The WHERE clause filters records based on specified conditions;3. The group by clause divides da

also be extended to handle unknown. The result of And:true and unknown is unknown,false and unknown result is false,unknown and unknown result is unknown. The result of Or:true or unknown is true,false and unknown result is unknown,unknown and unknown result is unknown. The result of not:not unknown is unknown. * If the WHERE clause predicate evaluates to a tuple false or unknown, then the tuple is not written to the result set. SQL uses the special

$ Cid $ _ GET [id]; $cid = $_GET['id']; Reply content: divide by 10 and use group by and count aggregate again. Let's make SQL into a dynamic language. It's really true-I'm a cainiao and can write it at will ··· DECLARE @ grade TABLE ( [Id] [int] IDENTITY (1, 1) not null, [CId] [int] not null, [SId] [int] not null, [Sgrade] [int] NOT NULL ); Insert into @ grade (cId, sId, sgrade)

Reply content:Divided by 10 after the group by count once again aggregate up and quickly make SQL dynamic language it, really-I am a rookie, write and write ... DECLARE @grade TABLE ( [ID] [int] IDENTITY (*) Not NULL, [CID] [int] not NULL, [SId] [INT] Not NULL, [Sgrade] [INT] Not NULL ); INSERT into @grade (cid,sid,sgrade) VALUES (1,0,0); INSERT into @grade (cid