$("#txtdoe").datepicker({ minDate: 0 });
我得到的错误没有给出一个或多个必需参数的值。参数在保存时工作正常,但无法更新。
答案 0 :(得分:0)
您的Field BankName和ReferenceNo
旁边有一个额外的@<强> @ BANKNAME = @名称强> 的 @ ReferenceNo = @ REF 强>
myconnection() 'open connection
sql = "UPDATE tblbankposting SET BankEntryDate=@date,BankName=@name,BankCode=@code,AccountNo=@acc"
sql &= ",TransactionType=@type,Amount=@amount,BankCharges=@charges,ReferenceNo=@ref,Description=@des "
sql &= "WHERE TransactionID=@id"
command = New OleDbCommand(sql, connection)
With command
.Parameters.Add(New OleDbParameter("@id", OleDbType.VarChar, 30, "TransactionID"))
.Parameters.Add(New OleDbParameter("@date", OleDbType.DBDate, 30, "BankEntryDate"))
.Parameters.Add(New OleDbParameter("@name", OleDbType.VarChar, 30, "BankName"))
.Parameters.Add(New OleDbParameter("@code", OleDbType.VarChar, 30, "BankCode"))
.Parameters.Add(New OleDbParameter("@acc", OleDbType.VarChar, 30, "AccountNo"))
.Parameters.Add(New OleDbParameter("@type", OleDbType.VarChar, 30, "TransactionType"))
.Parameters.Add(New OleDbParameter("@amount", OleDbType.Double, 10, "Amount"))
.Parameters.Add(New OleDbParameter("@charges", OleDbType.Double, 10, "BankCharges"))
.Parameters.Add(New OleDbParameter("@ref", OleDbType.VarChar, 30, "ReferenceNo"))
.Parameters.Add(New OleDbParameter("@des", OleDbType.VarChar, 30, "Description"))
'set parameter values
.Parameters("@id").Value = txttxnid.Text
.Parameters("@date").Value = CDate(dtpbankentrydate.Value.Date)
.Parameters("@name").Value = cbobanks.Text
.Parameters("@code").Value = txtbankcode.Text
.Parameters("@acc").Value = cboaccno.Text
.Parameters("@type").Value = cbotxntype.Text
.Parameters("@amount").Value = CDbl(txtamount.Text)
.Parameters("@charges").Value = CDbl(txtbankcharges.Text)
.Parameters("@ref").Value = txtrefno.Text
.Parameters("@des").Value = txtdescription.Text
End With