How to add primary key for two columns in same table. if possible then give simple example.

List the names of all Staff members who hold a faculty position and have a salary of $45,000 or less. Show their names as their first name, followed by a space, and then their last name but sort the list alphabetically by last name and then first name. Hint: Check out the Position column in the Staff table. All staff members who do not hold a faculty position will get a bonus this year equal to 5% of their salary. List the names, current salary, and bonus amount for each Staff member (non-faculty) who is due a bonus. Show your list in order with the person receiving the highest bonus first.

Using the same query: SELECT major FROM student WHERE gender = 'F'; How many clauses are there in this query? How many tables are referenced? How many columns are referenced?

Some rows of the table STUDENT are shown below:   CODE NAME DEPARTMENT GPA YEAR 291 ALEX PSI 3.1 1992 938 MICHELE PHY 2.3 1992 931 JHON MD 3.3 2001 182 JOE MD 3.4 2002 190 REY PHY 2.0 2001 330 RON PSI 3.9 2001   You want the name all departments that have a department GPA average of 2004 greater than the global GPA average. Which query gives you the desired result?     a. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT);   b. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 AND AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT) GROUP BY DEPARTMENT;   c. SELECT DEPARTMENT FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT GPA FROM STUDENT GROUP BY DEPARTMENT);   d. SELECT * FROM STUDENT WHERE YEAR = 2004 GROUP BY DEPARTMENT HAVING AVG(GPA) > (SELECT AVG(GPA) FROM STUDENT);

For this problem create a (temporary) table called instructor_course_nums. Write a procedure that accepts an instructor ID as input. The procedure calculates the total number of course sections taught by that instructor, and adds a tuple to the temporary table consisting of the instructors ID number, name, and total courses taught - call these attributes: ID, name, and tot_courses. If the instructor already has an entry in the table,then the procedure makes sure the total number of courses taught in the temporary table is up-to-date. You must name your procedure: Written in PostgreSQL

JOIN TABLES One row in the PAT_ENC table represents one patient encounter. One row in the ORDER MED table represents one medication order. One patient encounter can have many medication orders but one medication order can only have one patient encounter. In other words, the cardinality of this PAT_ENC to ORDER_MED relationship is one-to- many. You start a query with PAT_ENC. You then add ORDER_MED using an inner join. What is true about the granularity of the result set before and after adding the ORDER_MED table? SELECT ONE OF THE FOLLOWING A. The granularity stays at one row per patient encounter. B. The granularity stays at one row per medication order. C. The granularity changes from one row per patient encounter to one row per medication order on an encounter. D. The granularity changes from one row per medication order to one row per patient encounter.

The PA T_ENC_HSP table has 100 rows each, representing the hospital and counter. The ADMISSION _PROV_ID column identifies encounters admitting provider. There are five unique known values in ADMISSION _PROV_ID and separately, there are encounters with a null admitting provider. Use the following query to answer the question below: Select ADMISSION _PROV_ID FROM PAT_ENC_HSP GROUP BY ADMISSION_PROV_ID   How many rows will be in the results? Select one of the following:A. 0 B. 5 C . 6 D. 100 E. 500

SQL Code for  Extend the code in (Perform a LEFT OUTER JOIN between the tables Wicketkeeping and Batting. You may select any column(s). Look at the result and state how many wicketkeepers have not played as batsmen (no need to write code for this part ) To show the name of the fielders who have played 7 times as many matches as the wicketkeeper(s) in 5).

A Pivot Table is the column name with its table name. O True O False

a. Write a query to get all unique first names that start with the letter W from the student table.  Output the names in all upper case. Sort the results in ascending order. Write a query to get the first 3 characters of all last names that start with an T from the student table and sort them in descending order.