record Number and Customer Number

[Guest]

I have a customer file and it has a recordid as primary key autogenerate
and customer number is also generating by the DMax("[zfileno]","zfile")+1.
at one time recordid and zfileno was same but after delete of some records,
now the recordid and zfileno is different. when I am using lable print from a
form selected record it was printing fine when the recordid and zfileno was
same when it become diff then it is not printing the selected record as on
selection I have strWhere=[Recordid]
what is the best way to print the same record from the form.

 

[Joseph Meehan]

I have a customer file and it has a recordid as primary key
autogenerate
and customer number is also generating by the
DMax("[zfileno]","zfile")+1. at one time recordid and zfileno was
same but after delete of some records, now the recordid and zfileno
is different. when I am using lable print from a form selected record
it was printing fine when the recordid and zfileno was same when it
become diff then it is not printing the selected record as on
selection I have strWhere=[Recordid]
what is the best way to print the same record from the form.

More information please. What are the relationships and table
structures?

What is the sql for the selection to print. How are you "selecting"
records to print?

 

[Guest]

Provided you include the RecordID field in the label report's RecordSource
you can still use it to print the current record even though it differs from
the customer number. The RecordID does not need to appear on the label. The
code in the Click event procedure of a button on the form would be something
like this:

Dim strWhere As String

strWhere = "RecordID = " & Me.RecordID

' first make sure record is saved
RunCommand acCmdSaveRecord
' print the label
DoCmd.OpenReport "YourLabelReport", WhereCondition:= strWhere

You could also do it by using the customer number as this also uniquely
identifies the record, in which case you'd use something like:

strWhere = "[CustomerNumber] = " & Me.[CustomerNumber]

Ken Sheridan
Stafford, England