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Working with Subqueries
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In this chapter, I discussed how you can use subqueries to query and modify data. The subqueries we looked at, for the most part, relied on the use of predicates to define the subquery condition. In this Try This exercise, you will create a number of SELECT statements that include WHERE clauses. Those clauses will each include a predicate that defines a subquery, allowing you to access data from more than one table. You will also modify data by using an UPDATE statement that contains subqueries in the SET clause and the WHERE clause. For this exercise, as with previous Try This exercises, you will be using the INVENTORY database. You can download the Try_This_12.txt file, which contains the SQL statements used in this exercise.
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1. Open the client application for your RDBMS and connect to the INVENTORY database. 2. The first SELECT statement that you ll create allows you to return the name and number of
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CDs that are produced by MCA Records. Enter and execute the following SQL statement:
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SELECT CD_TITLE, IN_STOCK FROM COMPACT_DISCS WHERE LABEL_ID IN ( SELECT LABEL_ID FROM CD_LABELS WHERE COMPANY_NAME = 'MCA Records' );
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This statement uses a subquery to return the LABEL_ID value for MCA Records, which is stored in the CD_LABELS table. The value is then used in the IN predicate to compare it to the LABEL_ID values in the COMPACT_DISCS table. Your query should return four rows.
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SQL: A Beginner s Guide
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3. In the next statement, you will use an EXISTS predicate to define a subquery. The predicate
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determines whether the COMPACT_DISCS table contains any rows with a CD_TITLE value of Out of Africa. Enter and execute the following SQL statement:
SELECT COMPANY_NAME FROM CD_LABELS l WHERE EXISTS ( SELECT * FROM COMPACT_DISCS d WHERE l.LABEL_ID = d.LABEL_ID AND CD_TITLE = 'Out of Africa' );
The statement will return the name of the company that produces the Out of Africa CD, which in this case is MCA Records. The MCA Records row in the CD_LABELS table is the only row that evaluates to true for the subquery in the EXISTS predicate.
4. In the next statement you create, you ll determine the distributor names for those CDs in
which the LABEL_ID value in the CD_LABELS table is equal to any LABEL_ID values returned by the subquery. Enter and execute the following SQL statement:
SELECT COMPANY_NAME FROM CD_LABELS WHERE LABEL_ID = ANY ( SELECT LABEL_ID FROM COMPACT_DISCS WHERE IN_STOCK > 30 );
The subquery returns only those LABEL_ID values for rows that contain an IN_STOCK value greater than 30. When you execute this statement, the names of only three companies should be returned.
5. Now you ll create a SELECT statement that uses a comparison predicate to define a subquery.
The subquery returns the LABEL_ID value (from the CD_LABELS table) for Capitol Records. That value is then compared to the LABEL_ID values in the COMPACT_DISCS table. Enter and execute the following SQL statement:
SELECT CD_TITLE, IN_STOCK FROM COMPACT_DISCS WHERE LABEL_ID = ( SELECT LABEL_ID FROM CD_LABELS WHERE COMPANY_NAME = 'Capitol Records' );
This statement should return only two rows.
6. Now let s redo the statement in step 5 and turn it into a comma-separated join. Remember
that you should assign correlation names to the tables to simplify the code. Also remember
12:
Using Subqueries to Access and Modify Data
that the WHERE clause should include an equi-join condition that matches up LABEL_ID values. Enter and execute the following SQL statement:
SELECT FROM WHERE AND CD_TITLE, IN_STOCK COMPACT_DISCS d, CD_LABELS l d.LABEL_ID = l.LABEL_ID COMPANY_NAME = 'Capitol Records';
As you can see, this statement is a lot simpler than the subquery used in the preceding statement, and it returns the same results.
7. In the next statement that you ll create, you will use a nested subquery to return values to
the outer subquery. Enter and execute the following SQL statement:
SELECT ARTIST_NAME FROM ARTISTS WHERE ARTIST_ID IN ( SELECT ARTIST_ID FROM ARTIST_CDS WHERE COMPACT_DISC_ID IN ( SELECT COMPACT_DISC_ID FROM COMPACT_DISCS WHERE CD_TITLE = 'Past Light' ) );
The inner subquery returns the COMPACT_DISC_ID value for the Past Light CD. The outer subquery then uses this value to determine the ARTIST_ID value for that CD. This value is then used in the main SELECT statement, which returns one value: William Ackerman. He is the artist on the Past Light CD.
8. Now we re going to move on to using subqueries in an UPDATE statement. However,
let s first take a look at the table we re going to update, which is the COMPACT_DISC_ TYPES table. In order to know what to update, we re going to use values from the COMPACT_DISCS table and the MUSIC_TYPES table to help identify the IDs used in the COMPACT_DISC_TYPES table. Enter and execute the following SQL statement:
SELECT FROM WHERE AND AND CD_TITLE, TYPE_NAME COMPACT_DISCS d, COMPACT_DISC_TYPES t, MUSIC_TYPES d.COMPACT_DISC_ID = t.COMPACT_DISC_ID t.MUSIC_TYPE_ID = m.TYPE_ID CD_TITLE = 'Kojiki'; m
In this statement, you join three tables to return the CD_TITLE value and TYPE_NAME value for the Kojiki CD. The CD is classified as New Age.
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