use variable instead of table name - ms-access

Question

First Thanks for your comments. The array for Names if the tables names like, Name1, Name2, Name3,.. but we have different names.
However, May I need to explain by other way to be understand.
What I want to do is compare old DataBase with new DataBase (Both in MS Access). Each has many tables (>20), and each table has many fields like “Number”, “ Name”, etc, and put the differences in “Output” Table.
Then to make code short as possible, I want to use Loop and use Variable refer to Tables names. I put all Tables Names in below table:
enter image description here
TablesNames
Old_DB_Tables_Names | New_DB_Tables_Names
Items_List | Items_List
Items_Price | Items_Price
Items_Specs | Items_Specs
Items_Size | Items_Size
Items_Warehouse | Items_Warehouse
Items_Spare | Items_Spare
Items_Quantity | Items_Quantity
:
: :
:
Dim TablesNames As ADODB.Recordset
Dim Items_List As ADODB.Recordset
Dim Items_Price As ADODB.Recordset
Dim Items_Specs As ADODB.Recordset
Dim Items_Size As ADODB.Recordset
Dim Items_Wearhouse As ADODB.Recordset
Dim Items_Spare As ADODB.Recordset
Dim Items_Quantity As ADODB.Recordset
:
:
Dim I as Integer
I=0
Table1=TablesNames.item("Old_DB_Tables_Names").value
Table2=TablesNames.item("New_DB_Tables_Names").value
While TablesNames .EOF=False
if Table1.fields(I) <> Table2.fields(I) then
Output.Fields.Item("Number").Value Table2.Fields.Item("Number ").Value
Output.Fields.Item("Name").Value Table2.Fields.Item("Name ").Value
:
:
End If
I=I+1
Wend
The issue VBA not accept statement [Table1.fields(I) <> Table2.fields(I)].
Is it: we can’t refer to a table by its name, and to any of its fields and get a value from a record in that field?
If you have anther way, Please let me know.

Answer 1

Ok, let's start with the first part:
Dim TablesNames As ADODB.Recordset
Dim Items_List As ADODB.Recordset
Dim Items_Price As ADODB.Recordset
Dim Items_Specs As ADODB.Recordset
Dim Items_Size As ADODB.Recordset
Dim Items_Wearhouse As ADODB.Recordset
Dim Items_Spare As ADODB.Recordset
Dim Items_Quantity As ADODB.Recordset
What it looks like you are trying to do from the rest of your code:
Bring all the information from each table into your code.
What you are actually doing:
Creating empty recordset variables, to store table data and then never storing any table data. i.e. TableNames, Items_List, etc. are all variables, and they currently have nothing to do with anything, at all, they are completely empty and have no meaning to the rest of your code in their current states.
Next we have:
Table1=TablesNames.item("Old_DB_Tables_Names").value
Table2=TablesNames.item("New_DB_Tables_Names").value
What it looks like you are trying to do from the rest of your code:
Store instances of your tables using the name of the table.
What you are actually doing:
Accessing the value of a field in the recordset TableNames and storing that value in a variable named Table1. From the name, I assume you are expecting this to be a TableDef, but you are using it as either a string or a value type, but I can't tell because I'm not seeing where you declared it. If you didn't declare it, you need to type "Option Explicit", without the quotes, at the top of every single one of your modules, right now, seriously, stop reading and go do that.
While TablesNames .EOF=False
If you had set the recordset of TableNames in your code, this will work, as long as you don't have a space before .EOF
if Table1.fields(I) <> Table2.fields(I) then
What it looks like you are trying to do from the rest of your code:
Access a field from the TableDef stored in the variable Table1.
What you are actually doing:
Nothing, since, from your code, Table1 is either a string or value type and there is no method nor property named "fields" for any string or value types, you aren't doing anything.
Finally we have
Output.Fields.Item("Number").Value Table2.Field`enter code here`s.Item("Number ").Value
Output.Fields.Item("Name").Value Table2.Fields.Item("Name ").Value
What?
All of this leads me to believe that you haven't grasped the syntactical structure of VBA. While many of us here would like to help you, in most cases, we aren't going to be able to just give you an answer that will work completely, with out some sort of modification by you.