ACC2000: How to Query Records Within Range of Each Other
ID: Q197219
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The information in this article applies to:
SUMMARY
Moderate: Requires basic macro, coding, and interoperability skills.
You can use a query to return records that have duplicate values within a
field or whose values, between records within that field, are less than a
specified amount. For example, you may want to return a list of all the
orders that were placed for the same customer within two days of each
other.
MORE INFORMATION
The following example looks for multiple orders for the same customer
within two days of each other. It uses the sample database Northwind.mdb.
In order to compare a field with itself across multiple records, you need
to include the table more than once in the query. This way, you can compare
a field from the first instance of the table to the same field in the
second instance of that same table.
CAUTION: Following the steps in this example will modify the sample
database Northwind.mdb. You may want to back up the Northwind.mdb file
and perform these steps on a copy of the database.
How to Create the Query Described in the "Summary" Section
- Create the following query and include the Orders table twice:
Query: Catalog
---------------------------------------------------------------
Type: Select Query
Join: Orders.[CustomerID] <-> Orders_1.[CustomerID]
Field: Orders.*
Table: Orders
Field: DateDiff("d",[Orders].[OrderDate],[Orders_1].[OrderDate])
Criteria: Between -2 And 2
Field: OrderID
Table: Orders
Criteria: <> [Orders_1].[OrderID]
- To test this query, click Run on the Query menu.
Explanation of the Query
The query includes the Orders table twice so that you can compare fields to
themselves but in different records.
Because you want to return records for the same customer with different
OrderIDs where the orders are within two days of each other, you need to
make three comparisons.
The first comparison is to keep the CustomerIDs the same. To do this,
simply join the CustomerID field from the first instance of the table
(Orders) to CustomerID in the second instance of the table (Orders_1.)
The second comparison is trickier. To compare days between orders for the
same customer, you can use the DateDiff() function. The function in the
query calculates the number of days between the OrderDate from the first
instance of the table and OrderDate in the second instance of the table.
The criterion limits the records to plus or minus two days. The second
table should translate to "all the other records in the same table." There
is one catch with this.
The third comparison is to handle that catch. The query so far will return
all the records that have OrderDate within two days of each other for the
same customer. However, that will include orders that have the same
OrderID. In other words, the second table not only contains all other
records; it contains the same record itself. Therefore, you need to limit
the records to the ones that have different OrderIDs. That is why you
should include the OrderID field from the Orders table and apply the
criteria "<>[Orders_1].[OrderID]" (without the quotation marks).
You may go further and return all the orders made to the same customer by
the same employee within one or two days of each other. To do this, create
an additional join between the EmployeeID field from the first table to the
EmployeeID field in the second table.
NOTE: If you are only searching for duplicate records, you may want to use
the Microsoft Access Find Duplicates Wizard. To do so, on the Queries tab,
click New. When you click Query Wizards, you will see the Find Duplicates
Wizard.
Additional query words:
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Keywords : kbdta QryHowto
Version : WINDOWS:2000
Platform : WINDOWS
Issue type : kbhowto
Last Reviewed: July 15, 1999