select last record in a one to many relationship

I need to select the last record from the child table in a one to many relationship.

I have found some solutions in Experts Exchange but have not been able to translate them to my database.

My simple test database has two tables

tblIncident has two fields, IncidentID (PK, AutoNumber) and LastName (Text)
tblStatus has three fields, StatusID (PK, AutoNumber), IncidentID (Foreign Key, Number), Comments (Memo)

tblStatus may have many records related to one IncidentID in tblIncident.

I want to SELECT only the last record from tblStatus per IncidentID based on the highest number StatusID

Of course I can add a date field to tblStatus if that is necessary
GetLastStatus1.accdb
Wayne MarkelOwnerAsked:
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PatHartmanCommented:
Create a query that selects only the StatusID and IncidentID
Select Max(StatusID) as MaxStatusID, IncidentID
From tblStatus
Group By IncidentID;

This will get you the highest StatusID for each IncidentID.  Remember that "last" has no meaning in a relational database.  If you are positive that status values get inserted in order then this is sufficient.  However, if they might get entered out of order, you will need to add a date field and use that.

Then in your main query, include this query between tblIncident and tblStatus.  This query will NOT be updateable because of the aggregation in the totals query.

Alternatively, you can use a subselect to find the maximum value for statusID.  This method may or may not result in an updateable query.

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Rey Obrero (Capricorn1)Commented:
here is how to write the query


SELECT tblIncident.IncidentID, tblIncident.LastName,S4.Comment,S4.StatusID
FROM tblIncident
INNER Join
(SELECT S.StatusID, S.IncidentID, S.Comment
FROM tblStatus AS S
Inner Join
(SELECT Max(S2.StatusID) AS MaxOfStatusID, S2.IncidentID
FROM tblStatus AS S2
GROUP BY S2.IncidentID) As S3
ON S.IncidentID=S3.IncidentID And  S.StatusID= S3.MaxOfStatusID) As S4
On tblIncident.IncidentID=S4.IncidentID
Wayne MarkelOwnerAuthor Commented:
Excellent.  The solution has been tested in my production database.
Rey Obrero (Capricorn1)Commented:
:-0
Wayne MarkelOwnerAuthor Commented:
The solution from Rey Obrero also worked.  Thanks to both who contributed.
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