sql server 2008 - Query for only one client entry a day -

-

i have table definition this:

create table [dbo].[contact] (     clientnumber [numeric] (20, 0) null,     contactdate [date] null,     name [text] null,     additionalinformation [text] null )

this table filled information single month. each client may have multiple entries single day. meaning there might entries this:

42  2015-12-09  felix       called today 42  2015-12-09  fexclicx    ...

my goal client results table condition there should 1 entry day per client number. not care of (here) 2 entries displayed, top 1.

so far tried achive distinct:

select distinct clientnumber, contactdate, * contact

unfortunately not limit results show client entry specific date once. still appear.

how achive getting entries 1 entry 1 client per day?

i suggest edit table, has id column ( set primary key). if so, can use query i've posted.

i don't know how desire information group by, can't aggregate text columns. 

create table [dbo].[#contact] (     id int identity(1,1),     clientnumber [numeric] (20, 0) null,     contactdate [date] null,     name [text] null,     additionalinformation [text] null )   declare @date datetime set @date = getdate() insert #contact values (1,@date,'first client','info 1') go declare @date datetime set @date = getdate() set @date = dateadd(day,rand()*100,@date) insert #contact values (1,@date,'first client','info 2') go declare @date datetime set @date = getdate() set @date = dateadd(day,rand()*100,@date) print @date insert #contact values (2,@date,'2nd client','info 1') go declare @date datetime set @date = getdate() set @date = dateadd(day,rand()*100,@date) insert #contact values (2,@date,'2nd client','info 2') go declare @date datetime set @date = getdate() set @date = dateadd(day,rand()*100,@date) insert #contact values (2,@date,'2nd client','info 3') go declare @date datetime set @date = getdate() set @date = dateadd(day,rand()*100,@date) insert #contact values (3,@date,'3rd client','info 1')        select * #contact  c1 join (       select clientnumber, max(id) maxid       #contact       group clientnumber ) c2 on  c1.clientnumber = c2.clientnumber , c1.id = c2.maxid

Comments

Post a Comment

Popular posts from this blog

Hatching array of circles in AutoCAD using c# -

-
i beginner in autocad plugins .i have created array of circle using following code. now, have hatch circles in array .how that??how hatch circles @ once? moreover, how can return hatch method , use hatch objects.sorry language.thanks in advance. [assembly: commandclass(typeof(beamsection.class1))] namespace beamsection { public class class1 { //class1 obj = new class1(); [commandmethod("beamatsupport")] public void addlightweightpolyline() { // current document , database document acdoc = application.documentmanager.mdiactivedocument; database accurdb = acdoc.database; promptpointresult ppr = acdoc.editor.getpoint("\nselect starting point "); var ucs = acdoc.editor.currentusercoordinatesystem; point3d startingpt = ppr.value.transformby(ucs); list<double> radius = new list<double>(); radius.add(0.5);
Read more

ios - UITEXTFIELD InputView Uipicker not working in swift -

-
hello trying pop uipickerview view programmatically when user clicks on textfield. have tried doesn't doing anything. isn't working. nothing happening when click textfield class userprofiletableviewcontroller: uitableviewcontroller,uitextfielddelegate,uipickerviewdelegate,uipickerviewdatasource { var itempicker: uipickerview! = uipickerview() @iboutlet weak var gendertxtfield: uitextfield! var gender = ["male","female"] override func viewdidload() { super.viewdidload() gendertxtfield.delegate = self itempicker!.delegate = self itempicker!.datasource = self itempicker!.backgroundcolor = uicolor.blackcolor() self.gendertxtfield.inputview = itempicker } func numberofcomponentsinpickerview(pickerview: uipickerview) -> int{ return 1 } // returns # of rows in each component.. func pickerview(pickerview: uipickerview, numberofrowsincomponent component:
Read more

Python Pig Latin Translator -

-
so have been working on @ codecademy.com while trying learn python. don't know if blind , can't see why won't work or what. here's code won't let me use. function print variable if had typed in , print "empty" if didn't type anything. print "welcome english pig latin translator!" original = raw_input("what's name?") if len(original) > 0 print original else print "empty" if len(original) > 0 else the above lines contains syntax errors. try: if len(original) > 0: else: note added colons, ( : ).
Read more