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CHAPTER 6 MORE THAN ON E RELA TION SHIP BE TWEEN TABLES
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Calculus Approach to Two Relationships Between Tables
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You can use different ways to construct the query to retrieve all the information about a team (members names, team name, and manager s name) for a report like the one in Figure 6-6. I find the idea of two joins quite intuitive, but other people prefer to take a calculus approach. I have reproduced the two tables in Figure 6-8. Now let s see how we can pick a member and find out what team he is in and who the manager is for that team.
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team s manager
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Team
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Figure 6-8. Finding a team member (William Cooper), his team s name, and the name of the
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Without needing to think about joins, we can find the information we require. We need information from three rows. Let s look at one specific case. One row (m) from the Member table will give us the name of a member (William Cooper in Figure 6-8). We need to find the row (t) in the Team table for that member s team (m.Team = t.TeamName). Then we need another row in the Member table (m2) for the manager of the team (t.Manager = m2.MemberID). The calculus expression to find all such rows is shown in Listing 6-5.
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Listing 6-5. Calculus Expression for Information About Members, Their Team, and Their Team s Manager {m,t,m2 | Member(m), Team(t), Member(m2) AND m.Team = t.TeamName AND t.Manager = m2.MemberID}
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The SQL equivalent is in Listing 6-6.
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CHAPTER 6 MORE THAN ONE RELATION SHIP BE TWEEN TA BLES
Listing 6-6. SQL Statement for Information About Members, Their Team, and Their
Team s Manager
SELECT * FROM Member m, Team t, Member m2 WHERE m.Team = t.TeamName AND t.Manager = m2.MemberID
Listing 6-6 is equivalent to the SQL in Listing 6-3. The WHERE clause is the equivalent of a join between Member (m) and Team (t) on m.Team = t.TeamName, and another join between Team and another copy of Member (m2) on t.Manager = m2.MemberID.
Business Rules
The data model in Figure 6-1 shows the two relationships between members and teams: members can belong to teams, and members can manage teams. When we implement these relationships with foreign keys, the constraints that are placed on the data are quite simple. A member can be in only a team that exists in the Team table, and a team can be managed only by someone in the Member table. Other constraints are likely to apply in various situations. For example, we might have the additional constraints that a team can have no more than four members and/or the manager must be a member of the team (or not). Referential integrity alone cannot address these rules. Relational database products will usually provide some way to enforce such constraints. Large systems such as SQL Server and Oracle provide triggers. Triggers are actions that take place at a specified time (for example, when inserting or updating a record). The trigger will check and reject any changes that do not obey the rules. In Access and other products, it is not possible to apply such constraints to the tables themselves. However, you can attach macros, which can do some sort of checking, to input forms. We won t look in detail at how such constraints are implemented in various products, but we will look at how queries can help find any instances where the constraints are not satisfied. Although this is finding the problem after it has occurred, variations of these queries would form a basis for any trigger or macro that you would need to write to enforce the constraints. Let s look at finding teams whose managers are not members of the team. My mind often goes blank when faced with a query like this, and in that case, I always take a calculus approach. This means picturing the tables involved and imagining the type of instance I am seeking. Take a look at Figure 6-9. In Figure 6-9, TeamA s manager is 239, and we can see from the Member table that member 239 is not a member of any team. If we had a constraint that managers must belong to the team, TeamA would not obey it.
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