多项选择题
In which two cases would you use an outer join?（）
A. The tables being joined have NOT NULL columns.
B. The tables being joined have only matched data.
C. The columns being joined have NULL values.
D. The tables being joined have only unmatched data.
E. The tables being joined have both matched and unmatched data.
F. Only when the tables have a primary key/foreign key relationship.
相关考题
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单项选择题
The PRODUCTS table has these columns: PRODUCT_ID NUMBER(4) PRODUCT_NAME VARCHAR2(45) PRICE NUMBER(8,2) Evaluate this SQL statement: SELECT * FROM PRODUCTS ORDER BY price, product _ name; What is true about the SQL statement?（）
A. The results are not sorted.
B. The results are sorted numerically.
C. The results are sorted alphabetically.
D. The results are sorted numerically and then alphabetically. -
单项选择题
Examine the structure of the EMPLOYEES, DEPARTMENTS, and TAX tables. For which situation would you use a nonequijoin query?（）
A. To find the tax percentage for each of the employees.
B. To list the name, job id, and manager name for all the employees.
C. To find the name, salary, and department name of employees who are not working with Smith.
D. To find the number of employees working for the Administrative department and earning less then 4000.
E. To display name, salary, manager ID, and department name of all the employees, even if the employees do not have a department ID assigned. -
单项选择题
Examine the structure of the EMPLOYEES and DEPARTMENTS tables: EMPLOYEES EMPLOYEE_ID NUMBER DEPARTMENT_ID NUMBER MANAGER_ID NUMBER LAST_NAME VARCHAR2(25) DEPARTMENTS DEPARTMENT_ID NUMBER MANAGER_ID NUMBER DEPARTMENT_NAME VARCHAR2(35) LOCATION_ID NUMBER You want to create a report displaying employee last names, department names, and locations. Which query should you use to create an equi-join?（）
A. SELECT last_name, department_name, location_id FROM employees , department ;
B. SELECT employees.last_name, departments.department_name, departments.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
C. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE manager_id = manager_id;
D. SELECT e.last_name, d.department_name, d.location_id FROM employees e, departments D WHERE e.department_id = d.department_id;
