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CHAPTER 5 SE LF JOINS
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Figure 5-2. Data model for members coaching members
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The relationship line in Figure 5-2 can be read in a clockwise direction to say that a particular member might coach several other members or none (0..n). In the other direction, we can read that a particular member might have one coach or none (0..1). Relationships between classes can be represented by foreign keys, as discussed in 1. We can insert a foreign key column (Coach) in the Member table, as shown in Figure 5-3. In this case, the foreign key doesn t refer to a different table but back to the Member table itself. The values in the Coach column must already exist as a value in the primary key field of the table, MemberID. This ensures that only valid member IDs for existing members are inserted in the Coach column. The first row in the table in Figure 5-3 says that Melissa McKenzie is coached by member 153 (Brenda Nolan).
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Figure 5-3. Foreign key column Coach added to the Member table
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C HA PTER 5 S ELF JOINS
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With this table, we now can answer several different types of questions, such as: What are the names of the coaches Who is Jane Gilmore s coach Is anyone being coached by someone with a higher handicap Are any women being coached by men None of these questions can be answered by using simple select or project operations. What we require is a self join on the Member table. The easiest way to think of a self join is to see how we make one.
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Creating a Self Join
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Remember the definition of a join between two tables (from 3): a Cartesian product (all combination of rows from each table) followed by a select comparing a column in each of the tables. For a self join, we think of two copies of the same table. In Figure 5-4, we see part of the Cartesian product between two copies of the Member table. To distinguish the different bits of the product, I ve given the first copy an alias m and the second an alias c (you ll see why in a minute). In the small section of the Cartesian product visible in Figure 5-4, we see the first row (Melissa) from copy m paired with rows from copy c.
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Columns from first copy of Member (m)
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Columns from second copy of Member (c)
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m.Coach
c.MemberID
Figure 5-4. Cartesian product between two copies of the Member table
For queries about coaching, the interesting rows from the Cartesian product are those where the value of Coach from m is the same as MemberID from c. In Figure 5-4, you can see that the third line contains information about Melissa (from the m copy of Member) and information about her coach (from the c copy of Member). Now you can see why I chose the
CHAPTER 5 SE LF JOINS
aliases m for member and c for coach. Choosing helpful aliases can make understanding self joins much easier. The rows we would like to select from the Cartesian product are those satisfying m.Coach = c.MemberID. This is the join condition required to find information about members and their coaches. The SQL is shown in Listing 5-1.
Listing 5-1. Self Join on Member Table to Find Information About Members and Their Coaches SELECT * FROM Member m INNER JOIN Member c ON m.Coach = c.MemberID
The table resulting from the self join is shown in Figure 5-5 (some of the headings of the columns are truncated, as it was getting rather wide).
Information about a member (from m)
Information about a member s coach (from c)
Join condition: m.Coach=c.MemberID
Figure 5-5. Self join on Member table to retrieve information about members and
their coaches
None of these ideas are going to help us if our database has not been designed properly in the first place. However, once we recognize that there is a self relationship involved (members coach other members), and that relationship has been implemented correctly with a foreign key (Coach), then the trickiest part has been done. With this understanding, it is a simple job to create the self join, as shown in Figure 5-5.
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