问题描述
当我将一个值粘贴到组合框中时,在加载时粘贴它时不显示。尽管在不向表单传递参数的情况下插入值时,一切正常。如何解决? 第一张图显示了问题本身。第二个表明一切都将在不向表单传递参数的情况下工作!而在下拉列表本身,一切都显示了
' Populating a list of values
If Not Me!PostField.RowSource = "" Then Me!PostField.RowSource = ""
Me!PostField.ColumnCount = 2
Me!PostField.ColumnWidths = "0; 25"
Set QueryRecordset = CurrentDb.OpenRecordset("SELECT tblPost.PostCode,tblPost.PostName FROM tblPost;")
Do Until QueryRecordset.EOF
Me!PostField.RowSource = Me!PostField.RowSource & QueryRecordset("PostCode").Value & ";" & QueryRecordset("PostName").Value & ";"
QueryRecordset.MoveNext
Loop
QueryRecordset.Close
' Inserting user data
QueryObj = CurrentDb.QueryDefs("qrsCurrentEmployeeInfo")
QueryObj![EmployeeCode] = CInt(Me.OpenArgs)
Set QueryRecordset = QueryObj.OpenRecordset
Do Until QueryRecordset.EOF
Me!Title.Caption = QueryRecordset("LastName").Value & " " & QueryRecordset("FirstName").Value & " " & QueryRecordset("MiddleName").Value & " (Ðåäàêòèðîâàíèå)"
Me!LastNameField.Value = QueryRecordset("LastName").Value
Me!FirstNameField.Value = QueryRecordset("FirstName").Value
Me!MiddleNameField.Value = QueryRecordset("MiddleName").Value
Me!SexField.Value = QueryRecordset("Pol").Value
Me!IndividualTaxnumberField.Value = QueryRecordset("InduvialNumber").Value
Me!SalaryField.Value = QueryRecordset("Salary").Value
Me!CompanyField.Value = QueryRecordset("CompanyCode").Value
Me!LegalAddressField.Value = QueryRecordset("LegalAddress").Value
Me!ActualAddressField.Value = QueryRecordset("ActualAddress").Value
Me!PhoneField.Value = QueryRecordset("Phone").Value
Me.MailField.Value = QueryRecordset("[E-Mail]").Value
Me!PostField.Value = QueryRecordset("EmployeePost").Value
QueryRecordset.MoveNext
Loop
解决方法
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