Rst.FindFirst Problem
Mar 2, 2005
Hello All
I just discovered the reason why my table has not been working the way I want it to. In my code below, I have set my rst to find the first record of the previous month which in itself is correct, however I just discovered that it is picking up the records in ascending order.
Here is my code:
Option Compare Database
Option Explicit
Private Sub Button5_Click()
Dim prevMonth As Integer
Dim curMonth As Integer
Dim prevYear As Integer
Dim curYear As Integer
Dim CurRecordMonth As Integer
Dim rst As Recordset
Dim rst2 As Recordset
Dim db As Database
Dim monthText As Variant
Set db = CurrentDb
'fill an array with the text for months names
monthText = Array("", "Jan", "Feb", "Mar", "Apr", "May", "Jun", "Jul", "Aug", "Sep", "Oct", "Nov", "Dec")
' find previous and current Year and month. If current month = jan then go back to Dec of the year before
curMonth = Month(Date)
curYear = Year(Date)
prevYear = Year(Date)
prevMonth = Month(Date) - 1
If prevMonth = 0 Then
prevMonth = 12
prevYear = prevYear - 1
End If
' count number of existing records for current year and month
' If DCount("Month", "TTransactions", "month = '" & monthText(curMonth) & "' and year = " & curYear) = 0 Then
If DCount("Month", "tTransactions", "month = " & curMonth & " and year = " & curYear) < 2 Then
'if current month and year does not exist in table
' open table and find last months record
Set rst = db.OpenRecordset("tTransactions", dbOpenDynaset)
rst.FindFirst "month = " & prevMonth & " and year = " & prevYear
' open table again to write a new record
Set rst2 = db.OpenRecordset("tTransactions", dbOpenDynaset)
Do Until rst.NoMatch ' loop through all records meeting the criteria
rst2.AddNew
rst2![TelNo] = rst![TelNo]
rst2!Year = curYear
rst2!Month = curMonth
rst2!Rental = rst!Rental
rst2![Fees] = rst![Fees]
rst2![Vat] = rst![Vat]
rst2.Update
rst.FindNext "month = " & prevMonth & " and year = " & prevYear
Loop
rst.Close
rst2.Close
Set rst = Nothing
Set rst2 = Nothing
End If
Set db = Nothing
End Sub
Function CallButton5()
Call Button5_Click
End Function
I have put in an autonumber to assign sequential numbers to the records as they are entered. This I had hoped would allow me to sort my query by autonumber, however if the findfirst keeps finding the records in ascending order then I'm lost. Is there any way to get round this?
Thank you Kindly
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May 9, 2005
Hello,
I hope I have posted this to the correct forum?
Anyways, I have started porting my database backend to MySQL after a spate of corruptions and speed degrading daily, I have managed to get most things functional but one thing which I can not get working is an odd thing with the FindFirst statement.
I have the following line of code...
rst2.FindFirst "AwaitingStock=True and StockIn=False and DOA=False and Model='" & Trim(Me.Model) & "'"
which works find with Access backend but with MySQL I get the error...
Run-time error '3761':
The decimal field's precision is too small to accept the numeric you attempted to add.
This is obviously not the correct error as I am not trying to add anything!
If I remove the bit about the Model, the code executes fine, also, if I remove all the =True parts and just leave the Model part everything works fine so I guess it is because I am mixing string and integer fields in the search???
Any help greatly appreciated.
Kind regards,
Tom Findlay
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May 1, 2006
When, id do press the 'RecordWeergeven' button, Microsoft Visual Basic returns with a error. Method or data member not found. I think the error is located bij Me.Keuzelijst0. Do i have to declare a Listbox or create a second recordset. Please help.
Private Sub RecordWeergeven_Click()
'Me.RecordsetClone.Findfirst "[ID] = " & Me.lstList.ItemData(lstList.ListIndex)
'Me.Bookmark = Me.RecordsetClone.Bookmark
Dim rst As ADODB.Recordset
Set rst = Forms![FMR_users].RecordsetClone
rst.FindFirst "usr_id=" & Me.Keuzelijst0 & ""
Forms![FMR_users].Bookmark = rst.Bookmark
DoCmd.Close acForm, "GaNaarRecord"
End Sub
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Mar 25, 2008
Hi, I need some syntax help on the following line of code:
rs.FindFirst (rs![NewRecord] = True And IsNull(rs![Event_No]))
where NewRecord is type boolean and Event_No is string. Ive used this function mainly on strings so am not sure what to do with this one!
thanks!!
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Aug 1, 2005
I am writing an event procedure to check to see if a particular Project number exist in a recordset. I am trying to use the findfirst method and are having some problems. Here is my code.
Private Sub Command3_Click()
Dim db As DAO.Database
Dim rs As DAO.Recordset, ProjectNo As String, SqlStr As String, StrProjectNo As String
Set db = CurrentDb()
Set rs = db.OpenRecordset("tblTrackingSheetFrm", dbOpenTable)
StrProjectNo = Me![ProjectNumber]
rs.FindFirst StrProjectNo
If rs.NoMatch Then
Forms![frmProjectCriteria].Visible = False
DoCmd.SetWarnings WarningsOff
DoCmd.OpenQuery "(1)qryDeletetblTrackingSheetFrm"
DoCmd.OpenQuery "(1A)qryDeletetblTrackingSheetTMP"
DoCmd.OpenQuery "(2)qryAppendProjectTasks"
DoCmd.OpenQuery "(3)qryMaketblLaborActuals"
DoCmd.OpenQuery "(3A)qryUpdatetblTrackingSheetTMP"
DoCmd.OpenQuery "(4)qryDeletetblMaterialActualsTMP"
DoCmd.OpenQuery "(5)qryAppendEquipment"
DoCmd.OpenQuery "(6)qryAppendInventory"
DoCmd.OpenQuery "(7)qryAppendPayables"
DoCmd.OpenQuery "(8)qryAppendPurchaseOrder"
DoCmd.OpenQuery "(9)qryUpdateMaterialActuals"
DoCmd.OpenQuery "(A)qryAppendtblTrackingSheetFrm"
DoCmd.SetWarnings WarningsOn
DoCmd.OpenForm "frmTrackingSheet", acNormal
Else
MsgBox " Project worksheet already opened by another user."
rs.Close
End If
End Sub
What this does is check to see if another user has a project open and if so doesnt allow that user to access that project. I am getting the following error when I execute the procedure on the findfirst Code line.
Runtime error 3251 Operation is not supported by this object type.
Can someone take a look and see what I am doing wrong.
Any help is greatly appreciated.
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Feb 23, 2005
Hello,
I have a procedure that searches a table (recordset 1) for values in a lookup table (recordset 2) using FindFirst / FindNext. The routine worked fine until recently, I now have two issues;
I have included a new country in the lookup table - Cote D'Ivoire, now I get an error message when it reaches this name. I guess it's the " ' " that is causing the problem but don't know how to get round it!!??
I have modified the program so I can select the field I want to search from a form (thanks John) but it won't accept the field name as it's not part of the recordset e.g. rstTempTable.findfirst "[Field] = etc. How can I pass the chosen field from the form to the recordset?
Thanks in advance .....
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Sep 27, 2005
Need a little help with a record selector.
I ask it to find a record and bookmark it. No problem.
If record doesnt exist I get the value from a control and run a Insert Into command into my table creating a new record with that case #.
Now how do I modify the code below to make the new record just inserted into the bookmarked record. See sample code below. I'm not too practiced when it comes to this recordset business.
Any help is appreciated
Private Sub FindTheRecord()
' Find the record that matches the control.
Dim rs As Object
Dim Answer As String
Dim aSQL As String
Set rs = Me.Recordset.Clone
rs.FindFirst "[CaseNo] = " & Str(Nz(Me![CaseNo]))
If rs.NoMatch Then
Answer = MsgBox("No Matching Case Number Found." & vbCrLf & "Would you like to start a new" & vbCrLf & "record using this case number?", vbYesNo)
If Answer = 6 Then
DoCmd.SetWarnings False
aSQL = "Insert Into Main ([CaseNo])Values ([Forms]![frmMain]![CaseNo]);"
DoCmd.RunSQL aSQL
DoCmd.SetWarnings True
DoCmd.GoToRecord acDataForm, "frmMain", acLast
Code: Original - Code ' does not recognize the last record just added to the table, is there a command I can use here? ' does not recognize the last record just added to the table, is there a command I can use here?
Else
MsgBox "Action Cancelled"
CaseNum = ""
CaseNumYear = ""
DoCmd.GoToControl "CaseNum"
End If
Else
Me.Bookmark = rs.Bookmark
Call EnableControls
End If
End Sub
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Nov 15, 2013
I can't seem to figure out the proper syntax for the FindFirst method. I am using several variations of this effort:
Dim dbs As dao.Database
Dim rst As dao.Recordset
Set dbs = CurrentDb
Set rst = dbs.OpenRecordset("tblInvoice", dbOpenDynaset)
rst.FindFirst "rst!ID = frmInvoice!txtID"
I get an error message that says that Access doesn't recognize rst.ID as a valid field. But, it most certainly is. I tried substituting tblInvoice but got the same error.
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Mar 8, 2014
I am trying to locate a record on a multirecord form by using Findfirst method. Here is the code.
Dim UA1 As String, UAE1 As String, UA2 As String, UAE2 As String, UA3 As String, UAE3 As String, apost as string, repl as string
apost = "'"
repl = "''"
UA1 = Nz(Forms(ParName).Form.NAME, " ")
UAE1 = Replace(UA1, apost, repl)
[Code] ....
This code sometimes works and sometimes it does not work. The field CNO is a text field of 5 characters but contains the card numbers that is numeric data or nothing.
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