MS SQL Server Online Test

App usage data are kept in the following table:

TABLE sessions
  id INTEGER PRIMARY KEY,
  userId INTEGER NOT NULL,
  duration DECIMAL NOT NULL

Write a query that selects userId and average session duration for each user who has more than one session.

See the example case for more details.

The following table is used in an application that tracks the items its users bought:

TABLE items
  id INTEGER NOT NULL PRIMARY KEY
  name VARCHAR(30) NOT NULL
  dateBought DATE NOT NULL

Finish the ItemsBought procedure so that it returns only items whose dateBought is between @date and one month after @date (both inclusive). For each item it should return:

• The day of the month the item was bought [1-31].
• The number of the month the item was bought [1-12].
• The name of the item.

See the example case here for more details.

The following data definition defines an organization's employee hierarchy.

An employee is a manager if any other employee has their managerId set to this employee's id. That means John is a manager if at least one other employee has their managerId set to John's id.

TABLE employees
  id INTEGER NOT NULL PRIMARY KEY
  managerId INTEGER
  name VARCHAR(30) NOT NULL
  FOREIGN KEY (managerId) REFERENCES employees(id)

Write a query that selects only the names of employees who are not managers.

See the example case for more details.

The following two tables are used to define users and their respective roles:

TABLE users
  id INTEGER NOT NULL PRIMARY KEY,
  userName VARCHAR(50) NOT NULL

TABLE roles
  id INTEGER NOT NULL PRIMARY KEY,
  role VARCHAR(20) NOT NULL

The usersRoles table should contain the mapping between each user and their roles. Each user can have many roles, and each role can have many users.

Modify the provided SQL create table statement so that:

• Only users from the users table can exist within usersRoles.
• Only roles from the roles table can exist within usersRoles.
• A user can only have a specific role once.

See the example case for more details.

An insurance company maintains records of sales made by its employees. Each employee is assigned to a state. States are grouped under regions. The following tables contain the data:

TABLE regions
  id INTEGER PRIMARY KEY
  name VARCHAR(50) NOT NULL

TABLE states
  id INTEGER PRIMARY KEY
  name VARCHAR(50) NOT NULL
  regionId INTEGER NOT NULL
  FOREIGN KEY (regionId) REFERENCES regions(id)

TABLE employees
  id INTEGER PRIMARY KEY
  name VARCHAR(50) NOT NULL
  stateId INTEGER NOT NULL
  FOREIGN KEY (stateId) REFERENCES states(id)

TABLE sales
  id INTEGER PRIMARY KEY
  amount INTEGER NOT NULL
  employeeId INTEGER NOT NULL
  FOREIGN KEY (employeeId) REFERENCES employees(id)  

Management requires a comparative region sales analysis report.

Write a query that returns:

• The region name.
• Average sales per employee for the region (Average sales = Total sales made for the region / Number of employees in the region).
• The difference between the average sales of the region with the highest average sales, and the average sales per employee for the region (average sales to be calculated as explained above).

Employees can have multiple sales. A region with no sales should be also returned. Use 0 for average sales per employee for such a region when calculating the 2nd and the 3rd column.

See the example case for more details.

Each item in a web shop belongs to a seller. To ensure service quality, each seller has a rating.

The data are kept in the following two tables:

TABLE sellers
  id INTEGER PRIMARY KEY,
  name VARCHAR(30) NOT NULL,
  rating INTEGER NOT NULL

TABLE items
  id INTEGER PRIMARY KEY,
  name VARCHAR(30) NOT NULL,
  sellerId INTEGER
  FOREIGN KEY (sellerId) REFERENCES sellers(id)

Write a query that selects the item name and the name of its seller for each item that belongs to a seller with a rating greater than 4. The query should return the name of the item as the first column and name of the seller as the second column.

See the example case for more details.

Information about pets is kept in two separate tables:

TABLE dogs
  id INTEGER NOT NULL PRIMARY KEY,
  name VARCHAR(50) NOT NULL

TABLE cats
  id INTEGER NOT NULL PRIMARY KEY,
  name VARCHAR(50) NOT NULL

Write a query that select all distinct pet names.

See the example case for more details.

Given the following data definition, write a query that returns the number of students whose first name is John.

TABLE students
   id INTEGER PRIMARY KEY,
   firstName VARCHAR(30) NOT NULL,
   lastName VARCHAR(30) NOT NULL

A table containing the students enrolled in a yearly course has incorrect data in records with ids between 20 and 100 (inclusive).

TABLE enrollments
  id INTEGER NOT NULL PRIMARY KEY
  year INTEGER NOT NULL
  studentId INTEGER NOT NULL

Write a query that updates the field 'year' of every faulty record to 2015.

In a company, several different programs insert rows into a messages table which should later be parsed and inserted into appropriate tables. Consider the tables:

TABLE messages
  id INTEGER NOT NULL PRIMARY KEY
  type VARCHAR(30) NOT NULL
  data VARCHAR(100) NOT NULL

TABLE notifications
  id INTEGER NOT NULL PRIMARY KEY
  message VARCHAR(100) NOT NULL

TABLE alerts
  id INTEGER NOT NULL PRIMARY KEY
  shortMessage VARCHAR(15) NOT NULL

Create a procedure called ParseMessages, in the default schema, which should do the following for all rows in the messages table:

• Insert all rows where the type is 'notification' to the notifications table with the message field being set to the data field of messages table.
• Insert all rows where the type is 'alert' to the alerts table with the shortMessage field being set to the first 15 characters of the data field of messages table.

In both cases the id field should be the same as in the messages field.

See the example case here for more details.

A new social network site has the following data tables:

Select data that will be returned by the following SQL query:

SELECT users.name, COUNT(*) as count FROM users
LEFT JOIN friends
ON users.id = friends.user1 OR users.id = friends.user2
WHERE users.sex = 'f'
GROUP BY users.id, users.name;