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Queries Involving a Self Join
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With the joined table in Figure 5-5 as our base, we can answer all sorts of questions with quite simple select and project operations. Whenever I need to do queries involving self joins, I usually perform the join first (retaining all the rows and columns as in Figure 5-5), because the answers are usually pretty obvious when I have the joined table (or a quick sketch of the columns) in front of me. Let s see how this works with a few questions.
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C HA PTER 5 S ELF JOINS
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What Are the Names of the Coaches
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Looking at Figure 5-5, it is clear that the names of the coaches are in the columns coming from the c part of the join. We just want a list of the names in the columns c.LastName and c.FirstName. This is a simple project operation (that is, a subset of the columns). All we need to do is alter the first line of the query in Listing 5-1 to retrieve just those two columns. We don t want the names repeated, so we use the keyword DISTINCT in the SQL statement, as shown in Listing 5-2. For the data in Figure 5-5, this will return Brenda Nolan and William Cooper.
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Listing 5-2. Finding the Names of the Coaches SELECT DISTINCT c.FirstName, c.LastName FROM Member m INNER JOIN Member c ON m.Coach = c.MemberID
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Who Is Being Coached by Someone with a Higher Handicap
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To find out who is being coached by someone with a higher handicap, we need to compare the handicap of the member (m.Handicap) with the handicap of that member s coach (c.Handicap). We want to find the rows where the latter is greater than the former. This is a select operation (that is, retrieving a subset of rows). What is required is a WHERE clause to just retrieve those rows from the result of the join (Listing 5-1, Figure 5-5), as shown in Listing 5-3.
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Listing 5-3. Finding Rows for Members Being Coached by Someone with a Higher Handicap SELECT * FROM Member m INNER JOIN Member c ON m.Coach = c.MemberID WHERE m.Handicap < c.Handicap
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For the data in Figure 5-5, this will retrieve the data in the last four rows. (You don t have to be a great golfer to be a good coach!) Having done the join and selected the appropriate rows, we can then choose which columns we want to appear in our final result.
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List the Names of All the Members and the Names of Their Coaches
Listing the names of members and their coaches sounds pretty trivial, but if we are not careful, we can get it wrong. We have our joined table in Figure 5-5, and a first thought might be just to project the four columns containing the names. However, there are only 10 rows in the joined table, whereas there are 20 members in the Member table. The issue here is that not all the members have coaches. We looked at this in the section on outer joins in 3.
CHAPTER 5 SE LF JOINS
To recap, let s go back to the Cartesian product of the two copies of the Member table, but look at some rows involving a member with no coach, as shown in Figure 5-6.
Figure 5-6. Part of the Cartesian product between two copies of the Member table
The join condition (m.Coach = c.MemberID) is never satisfied for a member with a Null in the Coach field, so all those members will be missing from our joined table. We just need to be careful to understand what we really want. Do we want a list of all the members with coaches (10 rows), or a list of all the members along with their coach s name if they have one (20 rows) If it s the latter, we need an outer join. We need to see the name of each member (from the m copy of the Member table), along with the name of his coach, if any, (from the c copy). The SQL for this outer join is shown in Listing 5-4.
Listing 5-4. Names of Members Along with the Names of Their Coach (If Any) SELECT m.FirstName, m.LastName, c.FirstName, c.LastName FROM Member m LEFT OUTER JOIN Member c ON m.Coach = c.MemberID
Recall from 3 that for a left outer join, where there is no matching row from the right-hand table, those columns will be filled with Nulls. Figure 5-7 shows some of the rows from the left outer join in Listing 5-4.
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