Data Modelling and Databases Exercise dates: March 22/March 23, 2018 Ce Zhang, Gustavo Alonso Last update: March 26, 2018.

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1 Data Modelling and Databases Exercise dates: March 22/March 23, 2018 Ce Zhang, Gustavo Alonso Last update: March 26, 2018 Spring Semester 2018 Head TA: Ingo Müller Assignment 4: SQL This assignment will be discussed during the exercise slots indicated above. If you want feedback for your copy, hand it in during the lecture on the Wednesday before (preferably stapled and with your address). You can also annotate your copy with questions you think should be discussed during the exercise session. If you have questions that are not answered by the solution we provide, send them to Renato In this exercise you will write SQL queries and execute them on the databases that you have to set up following the instructions found in the course website. For reference, the database schema is provided where primary keys are underlined and foreign keys are in italic. 1 Employee Database The following queries refer to the employee database. For reference, the schema of this database is shown below. departments: (dept_no, dept_name) dept_emp: (emp_no, dept_no, from_date, to_date) dept_manager: (emp_no, dept_no, from_date, to_date) employees: (emp_no, birth_date, first_name, last_name, gender, hire_date) salaries: (emp_no, salary, from_date, to_date) title: (emp_no, title, from_date, to_date) 1.1 Average salaries per department The following three queries try to compute the average salary of all current employees per department and to display department names and salary averages in descending order of the average salary. Mark the correct solution:

2 SELECT d. dept_name, AVG ( s. salary ) AS avgs FROM employees e JOIN salaries s ON e. emp_no = s. emp_no JOIN dept_ emp de ON de. emp_no = s. emp_no JOIN departments d ON d. dept_no = de. dept_no WHERE s. to_date > NOW () AND de. to_date > NOW () GROUP BY d. dept_ name ORDER BY avgs DESC SELECT d. dept_name, AVG ( s. salary ) AS avgs FROM employees e JOIN salaries s ON e. emp_no = s. emp_no JOIN dept_ emp de ON de. emp_no = s. emp_no JOIN departments d ON d. dept_no = de. dept_no GROUP BY d. dept_ name ORDER BY avgs DESC SELECT e. dept_no, AVG ( s. salary ) AS avgs FROM employees e JOIN salaries s ON e. emp_no = s. emp_no JOIN dept_ emp de ON de. emp_no = s. emp_no WHERE s. to_date > NOW () AND de. to_date > NOW () ORDER BY avgs DESC 1.2 Salaries Consider the following query. Select the statement which describes what the query is meant to do. SELECT e. first_name, e. last_name, d. dept_name, s. salary, avgst. avgs FROM employees e JOIN salaries s ON s. emp_no = e. emp_no JOIN dept_ manager dm ON dm. emp_no = e. emp_no JOIN departments d ON d. dept_no = dm. dept_no JOIN ( SELECT de. dept_no, AVG ( s. salary ) AS avgs FROM employees e JOIN salaries s ON e. emp_no = s. emp_no JOIN dept_ emp de ON de. emp_no = s. emp_no WHERE s. to_date > NOW () AND de. to_date > NOW () GROUP BY de. dept_no ) avgst ON avgst. dept_no = dm. dept_no WHERE dm. to_date > NOW () AND s. to_date > NOW () AND s. salary > avgst. avgs It returns names of managers (first_name and last_name) who earn more than the average salary of all employees that they have ever managed. It also returns department s name (dept_name), manager s salary and department s average salary.

3 It returns names of employees (first_name and last_name) who earn the most in their department. It also returns department s name (dept_name), employees salary and department s average salary. It returns names of managers (first_name and last_name) who earn more than the average salary of the current employees that they manage. It also returns department s name (dept_name), manager s salary and department s average salary. It returns names of employees (first_name and last_name) who earn no more than the average salary of the current employees of their same department. It also returns department s name (dept_name), and department s average salary. 1.3 Employee Names (Substrings) Which query or queries return the correct count of employees in the database whose first name starts with the letter B? SELECT COUNT ( first_ name ) FROM employees WHERE first_ name LIKE 'B%' SELECT COUNT ( first_ name ) FROM employees WHERE SUBSTRING ( first_name,0,1) IN ('B') SELECT COUNT ( first_ name ) FROM employees WHERE first_ name LIKE 'B' OR first_ name LIKE 'b' SELECT COUNT ( first_ name ) FROM employees WHERE SUBSTRING ( first_name,1,1) = 'B' 1.4 First employees query Write a query that returns a list of the 10 first employees (first name (first_name) and last name (last_name)) which the company hired (hire_date). The list should be sorted in alphabetical order of the employees last names. Hint: use the LIMIT keyword to constrain the number of output records.

4 Write a query that determines for each department the average salary of anyone starting or receiving a salary change in 1988 from that department. It should display the department number, its name and its average. 1.6 Current managers Return the first (first_name) and last name (last_name) of all current managers, their title, their salary, and the department name (dept_name) for which they are responsible. Hint: Use the NOW() function to obtain the current date and the from_date and to_date fields in the salary and titles relations to find the corresponding records that are valid at the current date. 2 ZVV Database The following queries refer to the ZVV database. For reference, the schema of this database is shown below. stops: (stop_id, stop_name, stop_lat, stop_lon) stop_times: (trip_id, departure_time, arrival_time, stop_id, stop_sequence) trips: (trip_id, tram_number, trip_headsign)

5 2.1 Trips and tram lines Given the following query which is the statement that describes best the operation it performs. SELECT COUNT (*) AS count_trips, tram_ number FROM trips GROUP BY tram_ number ORDER BY count_ trips DESC Finds how many tram lines (tram_number) are in a trip and orders its result in descending order. Finds the number of trips each tram line (tram_number) makes and orders its result by number of trips in descending order. Finds and counts the number of tram lines (tram_number) and orders them in descending order. Finds the trips each tram line (tram_number) makes and orders its result by number of trips in ascending order. 2.2 Busiest tram stations The following queries try to compute the top 10 stations according to the number of times that any tram stops at that station. SELECT s. stop_id, s. stop_ name FROM stop_ times st, stops s WHERE s. stop_id = st. stop_id GROUP BY st. stop_id, s. stop_ name ORDER BY count_ stations DESC SELECT count_stations, s. stop_ name FROM stop_ times st, stops s WHERE s. stop_id = st. stop_id GROUP BY st. stop_id ORDER BY count_ stations DESC LIMIT 10 SELECT COUNT (*) AS count_stations, s. stop_ name FROM stop_ times st, stops s WHERE s. stop_id = st. stop_id GROUP BY st. stop_id, s. stop_ name ORDER BY count_ stations DESC LIMIT 10 SELECT COUNT (*) AS count_stations, s. stop_ name FROM stop_ times st, stops s WHERE s. stop_id = st. stop_id GROUP BY st. stop_id ORDER BY count_ stations DESC

6 3 TPC-H Database The following queries refer to the TPC-H database. For reference, the schema of this database is shown below. Customer: (customerid, customername, nationid) Nation: (nationid, nationname, regionid) Orders: (orderid, customerid, orderdate, orderpriority) Orderline: (orderid, olid, partid, supplierid, olquantity, oldiscount, olreturnflag, olshipdate, olcommitdate, olreceiptdate) Part: (partid, partname, partbrand, parttype, partretailprice) Region: (regionid, regionname) Supplier: (supplierid, suppliername, nationid) Supplypart: (partid, supplierid, supplypartcost) 3.1 Customers from Japan Write a query which returns how many customers come from 'JAPAN' (nationname). 3.2 Nations from multiple regions Given the following query that tries to list all nations in the regions of 'AMERICA', 'AFRICA' or 'ASIA'. SELECT nationname FROM nation, region WHERE Mark which of the following predicates makes the query perform the task correctly. regionname = ' AMERICA ' OR regionname = ' AFRICA ' OR regionname = ' ASIA '

7 regionname = ' AMERICA ' AND regionname = ' AFRICA ' AND regionname = ' ASIA ' nation. regionid = region. regionid AND ( regionname = ' AMERICA ' OR regionname = ' AFRICA ' OR regionname = ' ASIA ' ) nation. regionid = region. regionid AND ( regionname = ' AMERICA ' AND regionname = ' AFRICA ' AND regionname = ' ASIA ' ) 3.3 Exclusive supplier How many parts are supplied by the supplier with suppliername = 'Supplier# ' but not supplied by the supplier with suppliername = 'Supplier# '?

8 3.4 First order Consider the following query that retrieves the date (orderdate) of the first order made by customer with name (customername) 'Customer# ': SELECT orderdate FROM orders, customer WHERE orders. customerid = customer. customerid AND customername = ' Customer # ' ORDER BY orderdate ASC LIMIT 1 Choose the correct statements: No join is actually needed. We could use only the data from the orders table. The join between the two relations is done on the WHERE clause. The extra predicate checking for 'Customer# ' on the WHERE clause is redundant as we are going to retrieve it anyway. The ORDER BY statement is actually not needed to get the data needed. The ORDER BY and LIMIT statements help us getting the actual first order.

9 3.5 Customers with local shipments Find the number of customers that have made at least one order where at least one part is supplied by the same nation as the customer.

10 3.6 Revenue and profit margin Find the total revenue and profit margin (in percent) per year for all orders and orderlines. Revenue per orderline is defined as olquantity * partretailprice * (1 - oldiscount). Profit revenue - costs margin is calculated as revenue where cost per orderline is olquantity * supplypartcost. Sort the final result in ascending order of the year. Hint: use EXTRACT(YEAR FROM orderdate) to extract the year.

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