多項(xiàng)選擇題

Which two statements are true about constraints? （）

A.The UNIQUE constraint does not permit a null value for the column.
B.A UNIQUE index gets created for columns with PRIMARY KEY and UNIQUE constraints.
C.The PRIMARY KEY and FOREIGN KEY constraints create a UNIQUE index.
D.The NOT NULL constraint ensures that null values are not permitted for the column.

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單項(xiàng)選擇題

What is true about joining tables through an equijoin?（）

A.You can join a maximum of two tables through an equijoin.
B.You can join a maximum of two columns through an equijoin.
C.You specify an equijoin condition in the SELECT or FROM clauses of a SELECT statement.
D.To join two tables through an equijoin, the columns in the join condition must be primary key and foreign key columns.
E.You can join n tables (all having single column primary keys) in a SQL statement by specifying a minimum of n-1 join conditions.

單項(xiàng)選擇題

Click the Exhibit button to examine the structure of the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables.
Two new departments are added to your company as shown: DEPARTMENT_ID DEPARTMENT_NAME MGR_ID LOCATION_ID
9998 Engineering 123
9999 Administrative Boston
You need to list the names of employees, the department IDs, the department names, and the cities where the departments are, even if there are no employees in the departments and even if the departments are not yet assigned to a location. You need to join the EMPLOYEES, DEPARTMENTS, and LOCATIONS tables to retrieve this information.
Which statement do you execute to retrieve this information?（）

A.SELECT e.last_name, d.department_id, d.department_name, l.city FROM departments d RIGHT OUTER JOIN employees e ON d.department_id = e.department_id RIGHT OUTER JOIN locations l ON d.location_id = l.location_id;
B.SELECT e.last_name, d.department_id, d.department_name, l.city FROM departments d FULL OUTER JOIN employees e ON d.department_id = e.department_id FULL OUTER JOIN locations l ON d.location_id = l.location_id;
C.SELECT e.last_name, d.department_id, d.department_name, l.city FROM departments d LEFT OUTER JOIN employees e ON d.department_id = e.department_id LEFT OUTER JOIN locations l ON d.location_id = l.location_id;
D.SELECT last_name, department_id, department_name, city FROM departments d NATURAL JOIN employees e NATURAL JOIN locations l;