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create reference number based on old one...I'm trying to create a function that will create a "reference number" which is one higher then the latest one added. The reference number consist of "2009000100" which is the year (2009) followed by a 6 figur digit "000100" The code must create the new reference number as : "2009000101" 'oldKenmerk = latest used reference number Dim i As Integer Dim numberPart As String numberPart = Mid(oldKenmerk, 5, 6) Dim sNumPart As String = "" For i = 1 To 6 If Mid(numberPart, i, 1) = "0" Then sNumPart = sNumPart & Mid(numberPart, i, 1) Else Exit For End If Next numberPart = Replace(numberPart, sNumPart, "") If Mid(oldKenmerk, 1, 4) = Year(Today) Then newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 Else 'the file is the first of the year newKenmerk = Year(Today) & "000100" End If I don't think my code will work when you go from for example "2009000199" to "2009000200" Regards Marco Co wrote:
Show quoteHide quote > Hi All, Why do you use "Today"? Above you write you want to use the year in the > I'm trying to create a function that will create a "reference number" > which is one higher then the latest one added. > The reference number consist of "2009000100" > which is the year (2009) followed by a 6 figur digit "000100" > The code must create the new reference number as : "2009000101" > > 'oldKenmerk = latest used reference number > Dim i As Integer > Dim numberPart As String > numberPart = Mid(oldKenmerk, 5, 6) > Dim sNumPart As String = "" > > For i = 1 To 6 > If Mid(numberPart, i, 1) = "0" Then > sNumPart = sNumPart & Mid(numberPart, i, 1) > Else > Exit For > End If > Next > numberPart = Replace(numberPart, sNumPart, "") > > If Mid(oldKenmerk, 1, 4) = Year(Today) Then > newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 > Else > 'the file is the first of the year > newKenmerk = Year(Today) & "000100" > End If > > I don't think my code will work when you go from for example > "2009000199" to "2009000200" String. What should happen with "2009999999"? If you assume always a 4 digit year and ignore an overflow of the counter, the simplest solution is newnumber = (clng(oldnumber) + 1).tostring Armin
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On 17 mei, 13:41, "Armin Zingler" <az.nos***@freenet.de> wrote: Armin,> Co wrote: > > Hi All, > > I'm trying to create a function that will create a "reference number" > > which is one higher then the latest one added. > > The reference number consist of "2009000100" > > which is the year (2009) followed by a 6 figur digit "000100" > > The code must create the new reference number as : "2009000101" > > > 'oldKenmerk = latest used reference number > > Dim i As Integer > > Dim numberPart As String > > numberPart = Mid(oldKenmerk, 5, 6) > > Dim sNumPart As String = "" > > > For i = 1 To 6 > > If Mid(numberPart, i, 1) = "0" Then > > sNumPart = sNumPart & Mid(numberPart, i, 1) > > Else > > Exit For > > End If > > Next > > numberPart = Replace(numberPart, sNumPart, "") > > > If Mid(oldKenmerk, 1, 4) = Year(Today) Then > > newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 > > Else > > 'the file is the first of the year > > newKenmerk = Year(Today) & "000100" > > End If > > > I don't think my code will work when you go from for example > > "2009000199" to "2009000200" > > Why do you use "Today"? Above you write you want to use the year in the > String. What should happen with "2009999999"? > > If you assume always a 4 digit year and ignore an overflow of the counter, > the simplest solution is > > newnumber = (clng(oldnumber) + 1).tostring > > Armin your solution leaves me with "2009145" instead of "2009000145" MArco Co wrote:
>> newnumber = (clng(oldnumber) + 1).tostring I correctly get "2009000145" here.>> > > your solution leaves me with "2009145" instead of "2009000145" Armin
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"Armin Zingler" <az.nospam@freenet.de> wrote in message Wouldn't you need a formatting string to get that? Your post did not news:OojrYRv1JHA.1712@TK2MSFTNGP03.phx.gbl... > Co wrote: >>> newnumber = (clng(oldnumber) + 1).tostring >>> >> >> your solution leaves me with "2009145" instead of "2009000145" > > I correctly get "2009000145" here. > > > > Armin include any. -- Mike Family Tree Mike wrote:
Show quoteHide quote > "Armin Zingler" <az.nospam@freenet.de> wrote in message Did you try it? I don't know how "000" in the _middle_ can be left out> news:OojrYRv1JHA.1712@TK2MSFTNGP03.phx.gbl... >> Co wrote: >>>> newnumber = (clng(oldnumber) + 1).tostring >>>> >>> >>> your solution leaves me with "2009145" instead of "2009000145" >> >> I correctly get "2009000145" here. >> > > > Wouldn't you need a formatting string to get that? Your post did not > include any. without formatting. Only leading zeros would need a format string. Dim newnumber, oldnumber As String oldnumber = "2009000144" newnumber = (CLng(oldnumber) + 1).ToString MsgBox(newnumber) '=> "2009000145" Maybe I missed the point. I read the OP's first post again but I'm still looking for the problem. He clearly said he has a 4-digit year followed by a 6-digit number in a string. The number is to be incremented. Nowhere I read that the current year should be taken into account. Armin Hi Armin,
May be his Dunglish (Dutch English) is more easier for me then for you but the OP wrote. "which is the year (2009) followed by a 6 figure digit "000100" I understood from it a year in a string plus a 6 figur digit in a string. As I write this I know that you are more strict then me and you are right it is not exactly written, but this kind of sentenced do you not understand. I know. (I can see from words as "newKenmerk" that Co is a Dutch or maybe Afrikaans speaker, "newKenmerk" is English combined with Dutch which even is not Dunglish.) Cor Cor Ligthert[MVP] wrote:
> Hi Armin, That's what I understand, too.> > May be his Dunglish (Dutch English) is more easier for me then for > you but the OP wrote. > > "which is the year (2009) followed by a 6 figure digit "000100" > > I understood from it a year in a string plus a 6 figur digit in a > string. > As I write this I know that you are more strict then me and you are ?? I still don't understand what's wrong with my solution.> right it is not exactly written, but this kind of sentenced do you > not understand. I know. <quote> I'm trying to create a function that will create a "reference number" which is one higher then the latest one added. The reference number consist of "2009000100" which is the year (2009) followed by a 6 figur digit "000100" The code must create the new reference number as : "2009000101" </quote> His reference number is "2009000100" which is a 4 digit year followed by a 6 digit number. The new reference number is to be "2009000101". I can read it another dozend times, but that's exactly what my code does. Which part do I misunderstand? I don't get it. Armin
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"Armin Zingler" <az.nospam@freenet.de> wrote in message I interpreted his desire, that, if the code ran on 31 December 2009 beyond news:%23QTT4gy1JHA.1712@TK2MSFTNGP03.phx.gbl... > Cor Ligthert[MVP] wrote: >> Hi Armin, >> >> May be his Dunglish (Dutch English) is more easier for me then for >> you but the OP wrote. >> >> "which is the year (2009) followed by a 6 figure digit "000100" >> >> I understood from it a year in a string plus a 6 figur digit in a >> string. > > That's what I understand, too. > >> As I write this I know that you are more strict then me and you are >> right it is not exactly written, but this kind of sentenced do you >> not understand. I know. > > ?? I still don't understand what's wrong with my solution. > > > <quote> > I'm trying to create a function that will create a "reference number" > which is one higher then the latest one added. > The reference number consist of "2009000100" > which is the year (2009) followed by a 6 figur digit "000100" > The code must create the new reference number as : "2009000101" > </quote> > > His reference number is "2009000100" which is a 4 digit year followed by a > 6 > digit number. The new reference number is to be "2009000101". I can read > it > another dozend times, but that's exactly what my code does. > Which part do I misunderstand? I don't get it. > > > Armin > midnight, that the code needed to handle the change of year. I now understand your interpretation. I had assumed in your post that you were only describing the index part, not including the first four digits. Sorry about my confusion. -- Mike Family Tree Mike wrote:
Show quoteHide quote >> <quote> Yes, I left the year unchanged because he didn't state that it also has to >> I'm trying to create a function that will create a "reference number" >> which is one higher then the latest one added. >> The reference number consist of "2009000100" >> which is the year (2009) followed by a 6 figur digit "000100" >> The code must create the new reference number as : "2009000101" >> </quote> >> >> His reference number is "2009000100" which is a 4 digit year >> followed by a 6 >> digit number. The new reference number is to be "2009000101". I can >> read it >> another dozend times, but that's exactly what my code does. >> Which part do I misunderstand? I don't get it. >> >> >> Armin >> > > > I interpreted his desire, that, if the code ran on 31 December 2009 > beyond midnight, that the code needed to handle the change of year. I now > understand your interpretation. I had assumed in your post > that you were only describing the index part, not including the first > four digits. Sorry about my confusion. be changed. I also asked why he uses "Today" in his code - unanswered til now. Maybe that's the missing part. :) Armin Armin,
That is where I wrote you are more strict then me, I was reading between the lines that the number exist from a year plus a serial number. That means that he has to start every year again. I took for the year the Now.year in my example, however that can be of course every variable from four digits separated from the serial number. But strictly is your example is in my opinion a correct answer on the question from the Op and in nothing wrong. Cor
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"Co" <vonclausow***@gmail.com> wrote in message You will need to worry about what happens when your code is restarted, but news:72cfbffb-a039-4c56-9933-fc1893851acc@g19g2000vbi.googlegroups.com... > Hi All, > I'm trying to create a function that will create a "reference number" > which is one higher then the latest one added. > The reference number consist of "2009000100" > which is the year (2009) followed by a 6 figur digit "000100" > The code must create the new reference number as : "2009000101" > > 'oldKenmerk = latest used reference number > Dim i As Integer > Dim numberPart As String > numberPart = Mid(oldKenmerk, 5, 6) > Dim sNumPart As String = "" > > For i = 1 To 6 > If Mid(numberPart, i, 1) = "0" Then > sNumPart = sNumPart & Mid(numberPart, i, 1) > Else > Exit For > End If > Next > numberPart = Replace(numberPart, sNumPart, "") > > If Mid(oldKenmerk, 1, 4) = Year(Today) Then > newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 > Else > 'the file is the first of the year > newKenmerk = Year(Today) & "000100" > End If > > I don't think my code will work when you go from for example > "2009000199" to "2009000200" > > Regards > Marco the following should work. As your subject says, I based my code on taking in an old id from which I create the new one. Function NewID(ByVal s As String) As String Dim oldid As Integer = Integer.Parse(s.Substring(4)) Dim oldyear As Integer = Integer.Parse(s.Substring(0, 4)) Dim id As Integer Dim y as Integer = DateTime.Now.Year ' checks if the year has changed If (oldyear <> y) Then id = 1 Else id = oldid + 1 End If Return String.Format("{0}{1:d6}", y, id) End Function -- Mike
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On 17 mei, 13:48, "Family Tree Mike" <FamilyTreeM***@ThisOldHouse.com> Thanks Mike,wrote: > "Co" <vonclausow***@gmail.com> wrote in message > > news:72cfbffb-a039-4c56-9933-fc1893851acc@g19g2000vbi.googlegroups.com... > > > > > Hi All, > > I'm trying to create a function that will create a "reference number" > > which is one higher then the latest one added. > > The reference number consist of "2009000100" > > which is the year (2009) followed by a 6 figur digit "000100" > > The code must create the new reference number as : "2009000101" > > > 'oldKenmerk = latest used reference number > > Dim i As Integer > > Dim numberPart As String > > numberPart = Mid(oldKenmerk, 5, 6) > > Dim sNumPart As String = "" > > > For i = 1 To 6 > > If Mid(numberPart, i, 1) = "0" Then > > sNumPart = sNumPart & Mid(numberPart, i, 1) > > Else > > Exit For > > End If > > Next > > numberPart = Replace(numberPart, sNumPart, "") > > > If Mid(oldKenmerk, 1, 4) = Year(Today) Then > > newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 > > Else > > 'the file is the first of the year > > newKenmerk = Year(Today) & "000100" > > End If > > > I don't think my code will work when you go from for example > > "2009000199" to "2009000200" > > > Regards > > Marco > > You will need to worry about what happens when your code is restarted, but > the following should work. As your subject says, I based my code on taking > in an old id from which I create the new one. > > Function NewID(ByVal s As String) As String > Dim oldid As Integer = Integer.Parse(s.Substring(4)) > Dim oldyear As Integer = Integer.Parse(s.Substring(0, 4)) > Dim id As Integer > Dim y as Integer = DateTime.Now.Year > > ' checks if the year has changed > If (oldyear <> y) Then > id = 1 > Else > id = oldid + 1 > End If > > Return String.Format("{0}{1:d6}", y, id) > End Function > > -- > Mike that works great. MArco Try this.
function GetNextNumber(byval RefNum as String) as string dim Y as integer = cint(val(left(RefNum,4))) dim N as Integer = cint(val(Mid(RefNum,5)))+1 if Year(Today) > Y then ' check for new year Y = Year(Today) : N = 101 end if if (N < 101) then N = 101 ' check minimum value RefNum = Y.ToString.trim().PadLeft(4,"0"c) + _ N.ToString.trim().PadLeft(6,"0"c) Return Refnum end function Test it with this: sub test(byval s as string) dim nf as string = GetNextNumber(s) console.writeline("refnum: {0, -15} next => {1,-15}",s, nf) end sub Sub Main() test("200900212") test("200900000") test("200801121") test("200900199") test("") ' expecting first reference of year console.readkey() End Sub refnum: 200900212 next => 2009000213 refnum: 200900000 next => 2009000101 refnum: 200801121 next => 2009000101 refnum: 200900199 next => 2009000200 refnum: next => 2009000101 -- Show quoteHide quoteCo wrote: > Hi All, > I'm trying to create a function that will create a "reference number" > which is one higher then the latest one added. > The reference number consist of "2009000100" > which is the year (2009) followed by a 6 figur digit "000100" > The code must create the new reference number as : "2009000101" > > 'oldKenmerk = latest used reference number > Dim i As Integer > Dim numberPart As String > numberPart = Mid(oldKenmerk, 5, 6) > Dim sNumPart As String = "" > > For i = 1 To 6 > If Mid(numberPart, i, 1) = "0" Then > sNumPart = sNumPart & Mid(numberPart, i, 1) > Else > Exit For > End If > Next > numberPart = Replace(numberPart, sNumPart, "") > > If Mid(oldKenmerk, 1, 4) = Year(Today) Then > newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 > Else > 'the file is the first of the year > newKenmerk = Year(Today) & "000100" > End If > > I don't think my code will work when you go from for example > "2009000199" to "2009000200" > > Regards > Marco I don't know if I've sent this twice
Co A variation on the string format in this case \\\\ Dim a = "2009000199" a = (CInt(a) + 1).ToString a = Now.Year.ToString & a.Substring(4) /// Cor Show quoteHide quote "Co" <vonclausow***@gmail.com> wrote in message news:72cfbffb-a039-4c56-9933-fc1893851acc@g19g2000vbi.googlegroups.com... > Hi All, > I'm trying to create a function that will create a "reference number" > which is one higher then the latest one added. > The reference number consist of "2009000100" > which is the year (2009) followed by a 6 figur digit "000100" > The code must create the new reference number as : "2009000101" > > 'oldKenmerk = latest used reference number > Dim i As Integer > Dim numberPart As String > numberPart = Mid(oldKenmerk, 5, 6) > Dim sNumPart As String = "" > > For i = 1 To 6 > If Mid(numberPart, i, 1) = "0" Then > sNumPart = sNumPart & Mid(numberPart, i, 1) > Else > Exit For > End If > Next > numberPart = Replace(numberPart, sNumPart, "") > > If Mid(oldKenmerk, 1, 4) = Year(Today) Then > newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 > Else > 'the file is the first of the year > newKenmerk = Year(Today) & "000100" > End If > > I don't think my code will work when you go from for example > "2009000199" to "2009000200" > > Regards > Marco The reason what you want to break it apart first before conversion to
integer is because of potential out of range conditions. In this case, the OP's grandchildren will have a Year 2147 problem :-) Probably using a = (Ctype(a,uint32) + 1).ToString Cor Ligthert[MVP] wrote: Show quoteHide quote > I don't know if I've sent this twice > > Co > > A variation on the string format in this case > > \\\\ > Dim a = "2009000199" > a = (CInt(a) + 1).ToString > a = Now.Year.ToString & a.Substring(4) > /// > > Cor > > "Co" <vonclausow***@gmail.com> wrote in message > news:72cfbffb-a039-4c56-9933-fc1893851acc@g19g2000vbi.googlegroups.com... >> Hi All, >> I'm trying to create a function that will create a "reference number" >> which is one higher then the latest one added. >> The reference number consist of "2009000100" >> which is the year (2009) followed by a 6 figur digit "000100" >> The code must create the new reference number as : "2009000101" >> >> 'oldKenmerk = latest used reference number >> Dim i As Integer >> Dim numberPart As String >> numberPart = Mid(oldKenmerk, 5, 6) >> Dim sNumPart As String = "" >> >> For i = 1 To 6 >> If Mid(numberPart, i, 1) = "0" Then >> sNumPart = sNumPart & Mid(numberPart, i, 1) >> Else >> Exit For >> End If >> Next >> numberPart = Replace(numberPart, sNumPart, "") >> >> If Mid(oldKenmerk, 1, 4) = Year(Today) Then >> newKenmerk = Year(Today) & sNumPart & CInt(numberPart) + 1 >> Else >> 'the file is the first of the year >> newKenmerk = Year(Today) & "000100" >> End If >> >> I don't think my code will work when you go from for example >> "2009000199" to "2009000200" >> >> Regards >> Marco > 
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